Calculus Exam 1 Review Sheet

NumericalLimitByDelta[function_,{x_,a_},hlist_: {-1,-1/10,-1/100,-1/1000,1/1000,1/100,1/10,1}]:=Block[{table,i,y},table=Sort[Join[N[Table[{x,function}/.x->a+hlist[[i]],{i,1,Length[hlist]/2}],12],{{a,Text["?????"]}},N[Table[{x,function}/.x->a+hlist[[i]],{i,Length[hlist]/2+1,Length[hlist]}],12]]];
Return[TableForm[table,TableHeadings->{None,{Text[TraditionalForm[x]],Text[TraditionalForm[y==function]]}}]]]

NumericalLimitByDelta2[function_,{x_,a_},{h1_,h2_},hlist_: {-1,-1/10,-1/100,-1/1000,1/1000,1/100,1/10,1}]:=Block[{table,i,y},table=Sort[Join[N[Table[{x,function}/.x->a+hlist[[i]],{i,1,Length[hlist]/2}],12],{{a,Text["?????"]}},N[Table[{x,function}/.x->a+hlist[[i]],{i,Length[hlist]/2+1,Length[hlist]}],12]]];
Return[TableForm[table,TableHeadings->{None,{Text[TraditionalForm[h1]],Text[TraditionalForm[h2]]}}]]]

NumericalLimitAbsolute[function_,{x_,a_},belowlist_,abovelist_]:=Block[{table,i,y},table=Sort[Join[N[Table[{x,function}/.x->belowlist[[i]],{i,1,Length[belowlist]}],12],{{a,Text["?????"]}},N[Table[{x,function}/.x->abovelist[[i]],{i,1,Length[abovelist]}],12]]];
Return[TableForm[table,TableHeadings->{None,{Text[TraditionalForm[x]],Text[TraditionalForm[y==function]]}}]]]

TableOfValues[function_,x_,list_]:=Block[{table,k},table=TableForm[Table[N[{x,function}/.x->list[[k]],12],{k,Length[list]}],TableHeadings->{None,{Text[TraditionalForm[x]],Text[TraditionalForm[y==function]]}}];
Return[table];];

Problem 1

Estimate the slope of the tangent line to y=(x^3) at x=2.

Solution

As we have no way (for the moment!) to find the exact value of the slope of the tangent line we can only guess what it should be by constructing a set of increasingly accurate estimates.  The slope between two points ((Subscript[x, 1]),(Subscript[y, 1])) and ((Subscript[x, 2]),(Subscript[y, 2])) is m=(((Subscript[y, 2])-(Subscript[y, 1]))/((Subscript[x, 2])-(Subscript[x, 1]))).  So the slope between the point in question (which is (2,8) as when x=2 we have y=(2^3)=8) and the general point on the curve (x,(x^3)) is (((x^3)-8)/(x-2)).  Making  table of values for x that get closer to 2 (both from above and below) looks like this:

NumericalLimitByDelta2[ (x^3-8)/(x-2),{x,2},{x,Row[ {Text["slope = "], (x^3-8)/(x-2)}]}]

Based on the estimates in the table it seems like the tangent slope (which would correspond to the value x=2 that we can’t actually use) should be 12.   The problem didn’t ask for the actual tangent line, but the line with slope m=12 through (2,8) is y-8=12(x-2).

Problem 2

Estimate the equation of the tangent line to the curve y=x+(Sqrt[x]) at x=4.

Solution

When x=4 we have that y=4+(Sqrt[4])=4+2=6, so the point in question is (4,6).  The only way we have currently to find the tangent slope is by estimation.  The slope between (4,6) and the general point on the curve (x,x+(Sqrt[x])) is m=((Δ y)/(Δ x))=((x+(Sqrt[x])-6)/(x-4)).  If we make a table for values of this “secant slope formula” as x gets closer and closer to 4 it might look like this:

NumericalLimitByDelta2[ (x + Sqrt[x] - 6)/(x - 4), {x, 4}, {x, Row[ {Text[ "slope = "], (x + Sqrt[x] - 6)/(x - 4)}]}]

Based on this it looks like the slope should be 1.25, or (5/4).  So the equation of the tangent line is that of the line through (4,6) with slope (5/4), or y-6=(5/4)(x-4), or y=(5/4)x+1.

It’s worth noting although that we tend to use “nice” numbers in these tables the result is supposed to be the same even if you use less pleasant numbers, provided the numbers still get very close to the “central number” in question.  Another version of the same table with “worse” numbers might look like this:

NumericalLimitByDelta2[ (x + Sqrt[x] - 6)/(x - 4), {x, 4}, {x, Row[ {Text[ "slope = "], (x + Sqrt[x] - 6)/(x - 4)}]},{-1/7,-1/80, -1/97, -1/501, 1/403, 1/217, 1/77, 1/8}]

Even though the numbers in the x column are different the estimates still converge on the true slope of (5/4).

Problem 3

Come up with a sequence of estimates for the length of the curve y=(x^2) on the interval [0,1].

Solution

Recall that the distance between two points ((Subscript[x, 1]),(Subscript[y, 1])) and ((Subscript[x, 2]),(Subscript[y, 2])) is given by d=(Sqrt[((((Subscript[x, 2])-(Subscript[x, 1])))^2)+((((Subscript[y, 2])-(Subscript[y, 1])))^2)]).  To estimate the length of the curve we can break up the interval [0,1] into a number of equal-length pieces, find the points on y=(x^2) at the ends of each piece, compute the distances between the those points, and add them up:

Estimate 1: 1 piece

	“Breaking up” [0,1] into 1 piece gives us the x-values x=0,1.  The curve formula y=(x^2) turns these values into the points (0,0) and (1,0).  The distance between these points is (Sqrt[(((1-0))^2)+(((1-0))^2)])=(Sqrt[1+1])=(Sqrt[2])≈1.414213562, so that is our overall length estimate.
	
Estimate 2: 2 pieces

	Breaking up [0,1] into 2 pieces gives us the x-values x=0,(1/2),1.  The curve formula y=(x^2) turns these values into the points (0,0), ((1/2),(1/4)), and (1,1).  The two corresponding lengths are (Sqrt[((((1/2)-0))^2)+((((1/4)-0))^2)])=(Sqrt[(1/4)+(1/16)])=(Sqrt[(5/16)])=((Sqrt[5])/4) and (Sqrt[(((1-(1/2)))^2)+(((1-(1/4)))^2)])=(Sqrt[(1/4)+(9/16)])=(Sqrt[(13/16)])=((Sqrt[13])/4).  So the overall estimate is ((Sqrt[5])/4)+((Sqrt[13])/4)≈1.460404813.

Estimate 3: 4 pieces

	Breaking up [0,1] into 4 pieces gives us the x-values x=0, (1/4), (1/2), (3/4),1.  The curve formula y=(x^2) translates these into the points (0,0), ((1/4),(1/16)), ((1/2),(1/4)), ((3/4), (9/16)), and (1,1).  The lengths of the 4 segments you get from each point to the next are ((Sqrt[17])/16),(5/16),((Sqrt[41])/16),((Sqrt[65])/16) (work ommitted to save space).  So the overall length estimate is ((Sqrt[17])/16)+(5/16)+((Sqrt[41])/16)+((Sqrt[65])/16)≈1.474280476.
	
Estimate 4: 8 pieces

	Breaking up [0,1] into 8 pieces gives us the x-values x=0,(1/8), (1/4), (3/8), (1/2), (5/8),(3/4), (7/8),1.  The curve formula y=(x^2) traslates these values into the points (0,0), ((1/8),(1/64)), ((1/4),(1/16)), ((3/8),(9/64)), ((1/2), (1/4)), ((5/8), (25/64)), ((3/4),(9/16)), ((7/8),(49/64)), and (1,1).  The lengths of the 8 segements you get by moving from each point to the next are ((Sqrt[65])/64),((Sqrt[73])/64),((Sqrt[89])/64),((Sqrt[113])/64),((Sqrt[145])/64),((Sqrt[185])/64),((Sqrt[233])/64),(17/64).  Adding up these numbers gives you the overall estimate 1.477777985.
	
Obviously this gets more and more computationally intensiveas you break [0,1] into smaller and smaller pieces.  With 50 pieces the estimate is 1.478913043.  The ideal value (which we have no way of computing yet) is ((Sqrt[5])/4)+(1/4)(sinh^(-1))(2)≈1.478942858 (arcsinh is the inverse hyperbolic sine).  You can get this by executing the following pair of Mathematica commands:

problem3curve=ImplicitRegion[ y==x^2 && 0≤x≤1, {x,y}];
ArcLength[problem3curve]

Problem 4

Come up with a sequence of estimates for the area underneath y=(x^2) on [0,1].

Solution

These estimates can be done in a number of ways, the most common of which are based on breaking up the interval [0,1] into equal pieces and using the pieces of the interval as the bases of rectangles which go up to the curve in one fashion or another (maybe their height is given by the curve on the left side of the piece, maybe the right side, possibly the midpoint, etc.).  For my sequence of estimates I will use rectangle heights given by the curve height at the left end of each piece.  As the graph of y=(x^2) is always rising on [0,1] these means my retangles will always undershoot the curve and my estimates will be underestimates.  If you choose right ends to give you the heights your rectangles will always overshoot the curve and you will have an overestimate.  If you use midpoints the rectangles will be partially under the curve and over the curve, so it will be unclear if your estimate is an overshoot or undershoot, but the estimate will be slightly more accurate.

Estimate 1: 1 piece

	“Breaking up” [0,1] into 1 pieces means the base is from 0 to 1 (length 1).  The height would be given the left end, which is the point (0,0) (as y=(x^2)).  So the 1 rectangle has base 1 and height 0, for an area of 1×0=0.
	
Estimate 2: 2 pieces. 

	Breaking up [0,1] into two pieces gives rectangle bases from 0 to (1/2) and (1/2) to 1.  The first rectangle therefore has a base of length (1/2) and a height given by the left end (0,0) (as y=(x^2)); so its area is (1/2)×0=0.  The second rectange has a base of (1/2) and a height given at the left end ((1/2),(1/4)) (as y=(x^2)); so the second rectangle’s area is (1/2)×(1/4)=(1/8).  The total area of the rectangles is therefore 0+(1/8)=(1/8)=.125.
	
Estimate 3: 4 pieces

	Breaking up [0,1] into 4 pieces gives rectangle bases from 0 to (1/4), (1/4) to (1/2), (1/2) to (3/4) and (3/4) to 1.  The 4 rectangles have areas as follows, where the points all come from the formula y=(x^2):
		Rectangle 1: Base is (1/4), height given at (0,0), for area (1/4)×0=0
		Rectangle 2: Base is (1/4), height given at ((1/4),(1/16)), for area (1/4)×(1/16)=(1/64)
		Rectangle 3: Base is (1/4), height given at ((1/2),(1/4)) for area (1/4)×(1/4)=(1/16)
		Rectangle 4: Base is (1/4), height given at ((3/4),(9/16)) for area (1/4)×(9/16)=(9/64)
	So the area estimate is 0+(1/64)+(1/16)+(9/64)=(7/32)=0.21875
	
Estimate 4: 8 pieces: 

	For brevity I will shorten this a bit.  Breaking [0,1] into 8 equal pieces would force each base has a length of (1/8).  The 8 left-hand endpoints would be 0,(1/8),(1/4), (3/8), (1/2), (5/8), (3/4), (7/8).  The curve y=(x^2) would have corresponding heights 0,(1/64), (1/16), (9/64), (1/4),(25/64),(9/16),(49/64) at those eight points.  As each base is (1/8) the total area in one calculation would be

(1/8)×0+(1/8)×(1/64)+(1/8)×(1/16)+(1/8)×(9/64)+(1/8)×(1/4)+(1/8)×(25/64)+(1/8)×(9/16)+(1/8)×(49/64)=(35/128)=.2734375

The true value of the area is (1/3); the area sums do converge to this but do so fairly slowly (with 100 rectangles the value is 0.32835).  We will learn how to get the exact value later in the semester, but you can see how to get it in Mathematica directly using the following two commands:

problem4area=ImplicitRegion[ 0≤ y≤ x^2 && 0≤ x≤ 1, {x,y}];
Area[problem4area]

Problem 4

What do we mean by (lim_(x->a))f(x)? By (lim_(x->(SuperPlus[a])))f(x)?  By (lim_(x->(SuperMinus[a])))f(x)?

Solution

By (lim_(x->a))f(x) we mean the ideal value of f(x) at x=a, based on all of the information for f(x) on both sides of x=a.  It represents the value f(x) “should” have at x=a, and may or may not match the actual value f(a) (if this even exists).

(lim_(x->(SuperPlus[a])))f(x) is the same idea, except rather than based on information on both sides of x=a it is based only on values to the right of a (i.e. x>a).  (lim_(x->(SuperMinus[a])))f(x) uses information only on the left side of x=a (i.e. x<a)

Problem 5

What is the relationship between one-sided limits and the full two-sided limits?

Solution

The full two-sided limit  (lim_(x->a))f(x) will exists if and only if both of the one-sided limits  (lim_(x->(SuperPlus[a])))f(x) and  (lim_(x->(SuperMinus[a])))f(x) exist and are equal in value.  This approach to two-sided limits is often useful in cases there f(x) is a piecewise-defined function with completely different definitions on left and right of a value x=a (sometimes this piecewise definition is implicit, especially in problems  involving absolute values).

Problem 6

What are the three different approaches we have to limits?

Solution

The three approaches we have are numerical estimation (via a table or graph),  the rules-based “algebraic” approach to limits, and continuity-based approaches (i.e. “taking the limit is the same as plugging in”). 

The first pair (table and graphical) can at best give you an estimate for the limit; it is possible to be mislead in either approach.  

The algebraic approach gives you an exact answer (when it applies), and involves breaking down complex expressions into easier ones using the limit rules.

To use continuity you need to know why the function is continuous as a given spot (often because it is a polynomial or well-defined rational function).

Problem 7

Use the table-based approach to estimate the following limits:  (lim_(x->0))(((2^x)-1)/x), (lim_(x->5))(x^2), (lim_(x->0))((1-(cos^2)(x))/(x^2)) (where x is in radian measure), and (lim_(x->2))((Sqrt[(((x-2))^2)])/(x-2)).  If a limit seems to be an exact but unknown number estiamate the limit to 3 places of accuracy

Solution

For  (lim_(x->0))(((2^x)-1)/x) we have:

NumericalLimitByDelta[ (2^x-1)/x,{x,0},{-1/10000,-1/1000,-1/100,-1/10,1/10,1/100,1/1000,1/10000}]

So the limit seems to exist and is abour 0.693. It turns out this limit is exactly ln(2).

For  (lim_(x->5))(x^2) we have

NumericalLimitByDelta[x^2, {x,5}]

It looks like (lim_(x->5))(x^2)=25.

	For  (lim_(x->0))((1-(cos^2)(x))/(x^2)) we have:

NumericalLimitByDelta[ (1-Cos[x]^2)/x^2,{x,0}]

So it appears that in radian measure  (lim_(x->0))((1-(cos^2)(x))/(x^2))=1.  In degree measure you would get (π/180)≈0.01745.
	For (lim_(x->2))((Sqrt[(((x-2))^2)])/(x-2))we have:

NumericalLimitByDelta[ Sqrt[(x-2)^2]/(x-2),{x,2}]

As both sides do not go to the same value (lim_(x->2))((Sqrt[(((x-2))^2)])/(x-2)) does not exist.  The table does tell us that (lim_(x->(SuperPlus[2])))((Sqrt[(((x-2))^2)])/(x-2))=1 and (lim_(x->(SuperMinus[2])))((Sqrt[(((x-2))^2)])/(x-2))=-1.  Absolute value is hidden in this problem as (Sqrt[(((x-2))^2)])=|x-2|.

Problem 8

If f(x) is defined by the following graph, find the following limits: (lim_(x->2))f(x), (lim_(x->1))f(x), (lim_(x->4))f(x), (lim_(x->∞))f(x).

Plot[ 1/(x-2)^2 +x/(x^2+1) + If[x<4,1,2], {x,0,8}, Exclusions->{2,4}, PlotRange-> {-1, 10}, Epilog-> {Black, Opacity[.7], Dashed,InfiniteLine[ {{2,-1},{2,0}}], InfiniteLine[ {{0,2},{1,2}}]}, Ticks-> {Range[0,8], Range[0,10]}]

Solution

(lim_(x->2))f(x)=∞ (saying the limit does not exist would be correct but not the best answer, as saying the limit is infinite gives you more information).

(lim_(x->1))f(x)≈2.5

 (lim_(x->∞))f(x)=2 (as the graph levels off to 2 as you go way out to the right).

(lim_(x->4))f(x) does not exist as the values on the left and right of 4 do no not match.  The limit from above would be about 2.5 and the limit from below about 1.5.

Problem 9

Use the algebraic approach to find (lim_(x->5))(x^2)-3x+2.

Solution

Recall that the algebraic approach involves breaking up complicated limits into simpler ones, so remember to show the breakdown steps.

(lim_(x->5))(x^2)-3x+2=(lim_(x->5))(x^2)-2(lim_(x->5))3x+(lim_(x->5))2=((((lim_(x->5))x))^2)-3(lim_(x->5)) x+2=(5^2)-3×5+2=25-15+2=12

Problem 10

Use the algebraic approach to find (lim_(x->3))(((x^2)-5x+6)/((x^2)+x-12)).

Solution

If you try to evaluate (((x^2)-5x+6)/((x^2)+x-12))at x=3 you will get a “division by 0” error, which means you cannot split up the limit across the fraction as it is currently written.  Factoring the top and bottom will let you proceed, however:

(lim_(x->3))(((x^2)-5x+6)/((x^2)+x-12))=(lim_(x->3))(((x-2)(x-3))/((x-3)(x+4)))=(lim_(x->3))((x-2)/(x+4))=(((lim_(x->3))(x-2))/((lim_(x->3))(x+4)))=((3-2)/(3+4))=(1/7)

The fact that in the limit x cannot be 3 is key here; that is what allows you to cancel the common factor of x-3.

Problem 11

Use the algebraic approach to find (lim_(x->9))((x-9)/((Sqrt[x])-3)).

Solution

Again we would have a “division by 0” error if we tried to plug in x=9, which tells us we need to do some algebra first.  We can use the conjugate trick to get rid of the square root in the denominator:

((x-9)/((Sqrt[x])-3))=(((x-9)((Sqrt[x])+3))/(((Sqrt[x])-3)((Sqrt[x])+3)))=(((x-9)((Sqrt[x])+3))/(((((Sqrt[x])))^2)-3(Sqrt[x])+3(Sqrt[x])-9))=(((x-9)((Sqrt[x])+3))/(x-9))=(Sqrt[x])+3.  So (lim_(x->9))((x-9)/((Sqrt[x])-3))=(lim_(x->9))(Sqrt[x])+3=(Sqrt[(lim_(x->9)) x])+(lim_(x->9)) 3=(Sqrt[9])+3=6

Problem 12

Find (lim_(x->-2))((x+2)/((x^3)+8)).

Solution

If you try to plug in x=-2 both the top and bottom become 0; this suggests there is a common factor of x+2 that can be canceled:

(RowDefault[({{, , }, {x, "+", 2}}), ({{(("x")^2), (("x")^2), (x^2), "-", 2 x, "+", 4}, {(x^3), "+", 0 (x^2), "+", 0 x, "+", 8}, {(x^3), "+", 2 (x^2), (("x")^2), (("x")^2), (("x")^2), (("x")^2)}, {(("x")^2), (("x")^2), -2 (x^2), "+", 0 x, (("x")^2), (("x")^2)}, {(("x")^2), (("x")^2), -2 (x^2), "-", 4 x, (("x")^2), (("x")^2)}, {(("x")^2), (("x")^2), (("x")^2), (("x")^2), 4 x, "+", 8}, {(("x")^2), (("x")^2), (("x")^2), (("x")^2), 4 x, "+", 8}, {(("x")^2), (("x")^2), (("x")^2), (("x")^2), (("x")^2), (("x")^2), 0}})])

So (x^3)+8=(x+2)((x^2)-2x+4), and (lim_(x->-2))((x+2)/((x^3)+8))=(lim_(x->-2))((x+2)/((x+2)((x^2)-2x+4)))=(lim_(x->-2))(1/((x^2)-2x+4)).  This rational function is continuous as it is defined at x=-2; so (lim_(x->-2))(1/((x^2)-2x+4))=(1/((((-2))^2)-2(-2)+4))=(1/12).

Long division steps worked out courtesy of Wolfram|Alpha.

Problem 13

Find (lim_(h->0))(((((2+h))^4)-16)/h)

Solution

If you try to plug in h=0 boththe bottom and top become 0; this suggests we should look for a common factor of h-0=h that can be cancelled.  Using Pascal’s triangle we have that

(((2+h))^4)=1×(2^4)×(h^0)+4×(2^3)×h+6×(2^2)×(h^2)+4×(2^1)×(h^3)+1×(2^0)×(h^4)=16+32h+24(h^2)+8(h^3)+(h^4)

Therefore (lim_(h->0))(((((2+h))^4)-16)/h)=(lim_(h->0))((16+32h+24(h^2)+8(h^3)+(h^4)-16)/h)=(lim_(h->0))((32h+24(h^2)+8(h^3)+(h^4))/h)=(lim_(h->0))32+24h+8(h^2)+(h^3)=32+24×0+8×(0^2)+(0^3)=32, where the limit calculation at the end is justified as the expression (lim_(h->0))32+24h+8(h^2)+(h^3) is a polynomial in h, so taking the limit as h->0 is the same as plugging in h=0.   Remember if you want to use a “plug in” style rule you need to give some kind of justification - that is the “work” in that case.

Problem 14

What are the two fundamental trigonometric limits?

Solution

The two fundamental trigonometric limits are the results (which we derived from the Squeeze Theorem) that in radian measure (lim_(x->0))((sin(x))/x)=1 and (lim_(x->0))((1-cos(x))/(x^2))=(1/2).

Problem 15

Find (lim_(x->0))5sin(x)+((1-cos(x))/(3x))

Solution

Using the algebraic rules we have (lim_(x->0))5sin(x)+((1-cos(x))/(3x))=5 (lim_(x->0))sin(x)+(1/3)(lim_(x->0))((1-cos(x))/x).  The first limit is 0 (as sine is continuous everywhere), so this reduces to (1/3)(lim_(x->0))((1-cos(x))/x).  It looks like we can’t do the second limit but after a second glance it closely resembles one of the fundamental trigonometric limits, and we can get it after a little algebraic manipulation: (1/3)(lim_(x->0))((1-cos(x))/x)=(1/3)(lim_(x->0))x ((1-cos(x))/(x^2))=(1/3)(lim_(x->0))x (lim_(x->0))((1-cos(x))/(x^2))=(1/3)×0×(1/2)=0.  So (lim_(x->0))5sin(x)+((1-cos(x))/(3x))=0/

Problem 16

Find (lim_(x->0)) x cot(x)

Solution

We don’t recognize this limit right away so it is probably a good idea to break down the less-familiar cotangent in terms of sine and cosine. 

(lim_(x->0)) x cot(x)=(lim_(x->0))x ((cos(x))/(sin(x)))=(lim_(x->0))cos(x)(x/(sin(x)))=(lim_(x->0))cos(x)×(lim_(x->0))(((((sin(x))/x)))^(-1))=cos(0)×(1/((lim_(x->0))((sin(x))/x)))=1×(1/1)=1

Problem 17

Define f(x)=({{{, ({{(x^2)-3, x>2}, {x+1, x<2}})}}).   Find the one- and two- sided limits at x=2 using the algebraic approach.

Solution

For the one-sided limit (lim_(x->(SuperMinus[2])))f(x) the values of x are always less than 2; so we can proces f(x) using the bottom branch of the definition.  So (lim_(x->(SuperMinus[2])))f(x)=(lim_(x->(SuperMinus[2])))x+1=(lim_(x->(SuperMinus[2])))x+(lim_(x->(SuperMinus[2])))1=2+1=3.

For the one-sided limit (lim_(x->(SuperPlus[2])))f(x) the values of x are always larger than 2; so we can proces f(x) using the top branch of the definition.  So (lim_(x->(SuperPlus[2])))f(x)=(lim_(x->(SuperPlus[2])))(x^2)-3=(lim_(x->(SuperPlus[2])))(x^2)+(lim_(x->(SuperPlus[2])))-3=((((lim_(x->(SuperPlus[2])))x))^2)-3=(2^2)-3=1.

As the two one-sided limits are not equal the two-sided limit (lim_(x->2))f(x) does not exist.

Problem 18

What does the Squeeze Theorem say?

Solution

The Squeeze Law states that if you have three functions f(x), g(x), and h(x) with f(x)≤g(x)≤h(x) on an interval around x=a (except possibly at x=a) and (lim_(x->a))f(x)=(lim_(x->a))h(x)=L, then (lim_(x->a))g(x)=L.

For us the Squeeze Theorem is the limit rule of last resort as it can be difficult to use.  You will only given the function g(x); it will be up to you to choose the bottom and top functions f(x) and h(x) with identical limits.  That choice process can be complicated, particularly the part about showing f(x)≤ g(x)≤ h(x) (which often uses facts about absolute values or the bounds of trigonometric functions).

Problem 19

What does it mean for a function f(x) to be continuous at x=a?  For f(x) to be continuous on a closed interval [a,b]?  For f(x) to have a removable discontinuity at x=a?  For f(x) to have a jump discontinuity at x=a?  What common classes of functions are continuous?

Solution

f(x) is continuous at x=a if and only if (lim_(x->a))f(x)=f(a).  This can be broken down into pieces; the limit exists, the value f(a) exists, and they are equal. Graphically this means the curve can be drawn through x=a in a single piece - no gaps, holes, or jumps.

f(x) is continuous on the closed interval [a,b] if and only if f(x) is continuous at every point of the open interval (a,b), (lim_(x->(SuperPlus[a])))f(x)=f(a), and (lim_(x->(SuperMinus[b])))f(x)=b.  Graphically this means the graph of y=f(x) can be drawn from x=a to x=b in one piece without lifting your pen from the paper.

f(x) has a removable discontinuity at x=a if f(x) is not continuous at x=a but the single value f(a) can be redefined to make the redefined function continuous at x=a.  Functionally this means that the limit of f(x) exists at x=a but f(a) does not match this limit (or does not exist).  Graphically this means the graph of f(x) would be continuous at x=a except for a hole at x=a; plugging the hole corresponds to redefining f(a).

f(x) has a jump discontinuity at x=a if the two one-sided limits at x=a exist but are not equal (so infinite limits are not allowed).  There is no way to “fix” a jump discontinuity.

Common classes of continuous functions are polynomials (continuous everywhere), sin(x) and cos(x) (continuous everywhere), rational functions (ratios of polynomials, continuous where the denominator is not 0), and (x^(1/n)) (continuous everywhere when n is odd, and for x>0 when n is even).  The composition of continuous functions is continuous as well, so the sine and cosine of continuous functions are also continuous (so sin(3x), cos((x^2)+1), and sin((x^(1/3))) are continuous).

Problem 20

Where is (x^2)tan(x) continuous?  Where is (((x^2)+sin(x))/((x^2)+3x-8)) continuous?  Where is (Sqrt[((x^2)/(1+sin(x)))]) continuous?  Explain.

Solution

(x^2)tan(x)=(x^2)((sin(x))/(cos(x))).  (x^2), sin(x), and cos(x) are continuous everywhere.  So the product/quotient will be continuous everywhere except when the denominator cos(x)=0.  This happens at odd multiples of (π/2), so (x^2)tan(x) is continuous everywhere except odd multiples of (π/2).

For the function (((x^2)+sin(x))/((x^2)+3x-8)), the terms (x^2), sin(x), and (x^2)+3x-8 are all continuous everywhere.  So the numerator (x^2)+sin(x) is continuous everywhere.  Therefore the entire function is continuous as long as the bottom is not 0.  Using the quadratic formula gives as that (x^2)+3x-8=0 only at x=((-3±(Sqrt[(3^2)-4×1×(-8)]))/2)=((-3±(Sqrt[41]))/2).  So the function is continuous everywhere except x=((-3±(Sqrt[41]))/2).

For the function (Sqrt[((x^2)/(1+sin(x)))]), we know that the components (x^2) and 1+sin(x) are continuous everywhere. (Sqrt[x]) is continuous only when x>0.  So if ((x^2)/(1+sin(x))) is both continuous and positive the composite function  (Sqrt[((x^2)/(1+sin(x)))]) will be continuous.  (x^2) is positive when x≠0, and 1+sin(x)≥0 (as sine is trapped between 1 and -1).  So the composite function is definitely going to be continuous as long as x≠0 and 1+sin(x)≠0.  1+sin(x)=0 when sin(x)=-1, which only happens at values of x=((3π)/2)+2π n for integer values of n.  So we know that if x≠0, ((3π)/2)+2π n then the whole function is continuous.  It turns out that the function is also continuous at x=0.  The continuity issue with (Sqrt[x]) at 0 arises because (Sqrt[x]) only exists on one side of 0. But the quantity ((x^2)/(1+sin(x)))≥0 on both the left and right sides of x=0, so the square root is actually OK in this particular case.  you can see this in the graph below:

Plot[{ Sqrt[ x^2/(1+Sin[x])]}, {x,-8,8}, Exclusions->1+Sin[x]==0, PlotLabel->Style[TraditionalForm[Text[y==Sqrt[ x^2/(1+Sin[x])]]],Black], Ticks-> {Range[-3Pi, 3Pi, Pi/2], Range[-5,30,5]}]

Problem 21

Find a value of c such that function f(x)=({{{, ({{2(x^2)-3, x≥1}, {c x+1, x<1}})}}) is continuous.

Solution

For any value of c both branches are polynomials and therefore continuous away from the join at x=1.  To be continuous at x=1 the limit at x=1 must be the same as f(1).  

As 1 is on the branch where x≥ 1 f(1)=2×(1^2)-3=-1.  (lim_(x->(SuperPlus[1])))f(x)=(lim_(x->(SuperPlus[1])))2(x^2)-3=2×(1^2)-3=-1, where the limit process can be justified by the continuity of the polynomial 2(x^2)-3.  (lim_(x->(SuperMinus[1])))f(x)=(lim_(x->(SuperMinus[1])))c x+1=c (lim_(x->(SuperMinus[1])))x+(lim_(x->(SuperMinus[1])))1=c×1+1=c+1.  For all of these to be equal we must have c+1=-1, or c=-2.  So the function f(x)=({{{, ({{2(x^2)-3, x≥1}, {-2 x+1, x<1}})}}) is continuous everywhere.

Problem 22

What does the Intermediate Value Theorem say?  What does it imply for the function f(x)=(x^3) on the interval [2,3]?

Solution

The Intermediate Value Theorem states that if f(x) is continuous on the closed interval [a,b] then for each value W between the values f(a) and f(b) there will be at least 1 value c between a and b for which f(c)=W (the “at least” is important - there can be more than 1 value c in the interval which will give the value W).

As f(x)=(x^3) in continuous on the interval [2,3], the theorem implies that for each value W between f(2)=(2^3)=8 and f(3)=(3^3)=27 there will be at least one number c between 2 and 3 with f(c)=(c^3)=W.  So there must be a number which when cubed is 10 - 10 itself is between 8 and 27, so there is a number c bewtween 2 and 3 with f(c)=(c^3)=10 (which is the definition of the number (10^(1/3))).

A special (and important) case of the Intermediate Value Theorem is where f(a) and f(b) have opposite signs.  Then 0 is automatically between f(a) and f(b), so the IVT predicts the existence of at least one value c between a  and b with f(c).  This case is often use to predict the existence of solutions to hard equations that can’t be solved directly and is one reason why we tend to write such equations in “ = 0 ” format.

Problem 23

Use the Intermediate Value Theorem to show that the equation 3(x^5)-15x+5=0 has at least 3 real solutions.

Solution

The easiest way to approach this is to make a table of values for the continuous function f(x)=3(x^5)-15x+5 as if you were to graph it:

TableForm[ Table[ {x, 3x^5-15x+5}, {x,-4,4}], TableHeadings->{None, TraditionalForm[Style[Text[#], Italic]]& /@ {x, 3x^5-15x+5}}]

As there is a sign change between x=-2 and x=-1 the function f(x)=3(x^5)-15x+5 must hit 0 at least once between -2 and -1, so 3(x^5)-15x+5=0 at least once between -2 and -1.  The same happens as x goes from 0 to 1 and 1 to 2.  So there are at least 3 real solutions to 3(x^5)-15x+5=0.  It turns out there are exactly 3 real solutions, but that requires some extra theorems we don’t have yet - the Intermediate Value Theorem can only tell you an equation has at least so many solutions (1 for each sign change you can find).

Problem 24

Use the Method of Bisections to estimate the unique positive solution to (x^3)+2x-8=0.  Follow the method far enough so you are confident in at least 2 digits after the decimal point.

Solution

The function f(x)=(x^3)+2x-8 is continuous everywhere.  f(1)=-6 and f(2)=2, so the solution must be somewhere between 1 and 2.  The Method of Bisection simply takes an interval where you have a sign change and computes the sign of the function at the midpoint of the interval.  The value of the function at the midpoint will either be 0 (in which case you’ve found the root exactly) or will have the opposite sign from one of the two ends (in which case the root is between the midpoint and that end).  This sounds computationally intensive (and it can be if you need a lot of accuracy) but the only new information you compute at each step is the midpoint of the interval and the value of the function at that midpoint - the rest of the information can be copied down from one step to the next.  For f(x)=(x^3)+2x-8 from x=1 to x=2 the Method of Bisection looks like this:

Bisection[f_,{x_,a_,b_},steps_,digits_: 10]:=Block[{table,c,i,previous,newc},If[(f/.x->a) (f/.x->b)≥0,Return[Text["The method of bisection does not apply as there is no sign change"]]];
table={N[{a,c=(a+b)/2,b,f/.x->a,f/.x->c,f/.x->b},digits]};
Do[previous=table[[-1]];
If[previous[[5]]==0,AppendTo[table,previous]];
If[previous[[4]] previous[[5]]<0,newc=(previous[[1]]+previous[[2]])/2;
AppendTo[table,{previous[[1]],newc,previous[[2]],previous[[4]],f/.x->newc,previous[[5]]}],newc=(previous[[3]]+previous[[2]])/2;
AppendTo[table,{previous[[2]],newc,previous[[3]],previous[[5]],f/.x->newc,previous[[6]]}]];,{i,1,steps}];
Return[TableForm[table,TableHeadings->{None,Text/@{"a","c","b","f(a)","f(c)","f(b)"}}]]]

Bisection[ x^3+2x-8,{x,1,2},11]

So the unique root must be between 1.670166016 and 1.670410156 - we actually get at least 3 places to match at the same time we get 2 (1.670).  The value to 30 places is 1.67024469696273302858324007719.

Generally the Method of Bisection can take up to 4 iterations to get you one more decimal place of confidence (as you halve the interval each time, and ((((1/2)))^4)=(1/16)<(1/10), and (1/10) represents 1 decimal place).  We will introduce a more advanced method later in the course (Newton’s Method) which converges mich more quickly (4 iterations will give you the correct places 1.670244696962)

Problem 25

Find (lim_(x->(SuperPlus[3]))) (2/(x-3))

Solution

The limit of the top is 2 and the limit of the bottom is 0, so the limit does not exist.  There is potential to be able to be more precise about how the limit fails to exist.  As x->(SuperPlus[3]) the values of x must always be greater than 3.  Therefore the values of x-3 are going to 0 but are always positive; that is x-3->(SuperPlus[0]).  2 divided by a tiny positive number becomes a huge positive number, so  (lim_(x->(SuperPlus[3]))) (2/(x-3))=∞.

Two things to note about this sort of argument.  The first is that the form (nonzero/zero) is what guarantees that the limit does not exist - the common limit form (0/0) may or may not exist depending on the problem (we will develop this form in more detail later).  The second is the form (nonzero/zero) does not guarantee that the limit is going to be ∞ or -∞ - it depends very much on how the bottom goes to 0.  If you can make an argument as I did above that the sign of the bottom is always positive or negative then you can justify such a deduction.  But if you can’t then the limit can simply fail to exist.  An example of this is (1/(x sin(1/x))) as x->(SuperPlus[0]).  The bottom does go to 0 but continually flips sign as x->0, so the ratio oscillates between deep negative and large positive numbers and the limit cannot be either -∞ or ∞.

Problem 26

Find (lim_(x->(SuperPlus[2]))) (((x^2)+3x+1)/((x^2)-4)) and (lim_(x->(SuperMinus[2]))) (((x^2)+3x+1)/((x^2)-4)).  What do these limits tell you about the graph of y=(((x^2)+3x+1)/((x^2)-4))?

Solution

As x->2 from either the left or the right the numerator (x^2)+3x+1 goes to (2^2)+3×2+1=11 (as (x^2)+3x+1 is continuous).  Likewise as (x^2)-4 is continuous it will go to 0 as x->2 from either the left or right.  So both one-sided limits fail to exist as they have the form (nonzero/zero).  We can be more precise however.

As x->(SuperPlus[2]) x is always larger than 2.  The expression (((x^2)+3x+1)/((x^2)-4))=(((x^2)+3x+1)/((x+2)(x-2))).  As x->(SuperPlus[2]) we know that x+2->4 and x-2->(SuperPlus[0]).  So the basic form is (11/(4×(SuperPlus[0]))), which must be ∞.  So  (lim_(x->(SuperPlus[2]))) (((x^2)+3x+1)/((x^2)-4))=∞.

As x->(SuperMinus[2]), x is always less than 2.  So while we still know that x+2->4, in this case x-2->(SuperMinus[0]).  So the basic form is (11/(4×(SuperMinus[0]))), which must be -∞.  So  (lim_(x->(SuperMinus[2]))) (((x^2)+3x+1)/((x^2)-4))=-∞

What this means in terms of the graph is that the graph “explodes downwards” on the left side of  x=2 and “blows up” on the right side of x=2 - useful information when the time comes to do detailed graphing.

A good question to ask here is why I didn’t track the “from the + or - side” for the factor x+2.  It is true that as x->(SuperPlus[2]), x+2->(SuperPlus[4]) and that as x->(SuperMinus[2]), x+2->(SuperMinus[4]).  But being a little bit larger or smaller than 4 does not affect the sign of the factor, it has to be positive  - so it doesn’t play any role in the computation (this is also why we didn’t worry about that sort of thing as the top went to 11 - a little bit larger or smaller than 11 doesn’t change things).  It is only when a factor goes to 0 that being a bit larger or smaller can flip the sign, so that’s what we focus on.  There’s nothing wrong with tracking the other non-zero parts but that’s more work for something which won’t make a difference.  In a problem like ((sin(x))/(x^2)) where both the top and bottom go to 0 you might have to track things for both the top and bottom (in this case it goes to ∞ on the right and -∞ on the left).

Problem 27

What does it mean for (lim_(x->a))f(x)=∞?  For (lim_(x->∞))f(x)=L?  For  (lim_(x->∞))f(x)=∞?  For  (lim_(x->-∞))f(x)=L?

Solution

(lim_(x->a))f(x)=∞ means that you can make all of the values for f(x) near x=a as big as you want by restricting the values of x to be close enough (but not equal to) x=a.  Graphically this means the graph of y=f(x) has a vertical asymptote at x=a and explodes uwards on both sides of that asymptote.

(lim_(x->∞))f(x)=L means that you can make the all the values of f(x) as close to L as you want be requiring that the values of x are sufficiently large.  Graphically this means you can force the graph of y=f(x) to be trapped in as thin a band as you like around the horizontal line y=L if you require that you only look at x values that are far enough out to the right (like x≥ 1000 for example - how far out to the right you have to require typically depends on how thin you want the band to be).

(lim_(x->-∞))f(x)=L means that you can make the all the values of f(x) as close to L as you want be requiring that the values of x are sufficiently low negative numbers.  Graphically this means you can force the graph of y=f(x) to be trapped in as thin a band as you like around the horizontal line y=L if you require that you only look at x values that are far enough out to the left (like x≥ -1000 for example - how far out to the left you have to require typically depends on how thin you want the band to be).

(lim_(x->∞))f(x)=∞ means that you can make the values for f(x) as large as you like by requiring that the values of x be sufficiently large.  In terms of the graph of y=f(x) this means the graph blows up as you go way out to the right.

Problem 28

What does it mean for a graph to have a vertical asymptote?  A horizontal asymptote?  An oblique asymptote?

Solution

A graph has x=a as a vertical asymptote means that the graph of y=f(x) explodes upwards or downwards on at least one side of x=a.  This means at least one of the one-sided limits as x->a is either ∞ or -∞. 

A graph has y=b as a horizontal asymptote if the graph gets arbitrarily close to y=b as either x->∞ or x->-∞.

A graph has the oblique asymptote y=m x+b (where m≠0) if the graph of y=f(x) gets arbitrarily close to that of y=m x+b as either x->∞ or x->-∞.

For a rational function ((p(x))/(q(x))) a value x=a for which q(a)=0 and p(a)≠0 will definitely lead to a vertical asymptote (if both p(a) and q(a) are 0 it is possible to have a vertical asymptote there, it depends on the particulars of the two formulas).  Such a function will have a horizontal asymptote (the same asymptote as x->∞ and x->-∞) if the degree of p(x) is less than or equal to the degree of q(x).  It will have an oblique asymptote (the same asymptote as x->∞ and x->-∞)  if the degree of p(x) is exactly 1 more than the degree of q(x).  In that case the oblique asymptote is the quotient in long division.

Problem 29

Find (lim_(x->∞))(((x^2)-3x+3)/(2(x^2)-4x+1))

Solution

As the degree of the top and bottom match the graph of y=(((x^2)-3x+3)/(2(x^2)-4x+1)) has a horizontal asymptote, so the answer must be a number.  The trick in computing such a limit is often to divide the top and bottom of the fraction by the term that is “blowing up the fastest” in the bottom as x->∞.   In this problem the fastest-exploding term in the bottom is (x^2), so we have:

(lim_(x->∞))(((x^2)-3x+3)/(2(x^2)-4x+1))= (lim_(x->∞))(((x^2)-3x+3)/(2(x^2)-4x+1))×((1/(x^2))/(1/(x^2)))= (lim_(x->∞))((1-(3/x)+(3/(x^2)))/(2-(4/x)+(1/(x^2))))=((1-0+0)/(2-0+0))=(1/2)

We are using the fact that (lim_(x->∞))(1/(x^n))=0 whenever n>0.

Problem 30

Find  (lim_(x->∞))(((x^3)+2x-3)/(8(x^2)-7x+6)).

Solution

Applying the underlying trick from Problem 29,  the “fastest exploding term” from the denominator is (x^2).  So we have:

(lim_(x->∞))(((x^3)+2x-3)/(8(x^2)-7x+6))=(lim_(x->∞))(((x^3)+2x-3)/(8(x^2)-7x+6))×((1/(x^2))/(1/(x^2)))=(lim_(x->∞))((x+(2/x)-(3/(x^2)))/(8-(7/x)+(6/(x^2))))=((∞+0-0)/(8-0+0))=∞

Alternatively we can do the division to find the oblique asymptote (the degree of the top is one higher than the bottom):

(RowDefault[({{, , , , }, {8 (x^2), "-", 7 x, "+", 6}}), ({{(("x")^2), (("x")^2), (("x")^2), (("x")^2), (x/8), "+", (7/64)}, {(x^3), "+", 0 (x^2), "+", 2 x, "-", 3}, {(x^3), "-", ((7 (x^2))/8), "+", ((3 x)/4), (("x")^2), (("x")^2)}, {(("x")^2), (("x")^2), ((7 (x^2))/8), "+", ((5 x)/4), "-", 3}, {(("x")^2), (("x")^2), ((7 (x^2))/8), "-", ((49 x)/64), "+", (21/32)}, {(("x")^2), (("x")^2), (("x")^2), (("x")^2), ((129 x)/64), "-", (117/32)}})])

Therefore the graph has a oblique asymptote y=(1/8)x+(7/64).  As this line has positive slope, as x->∞ it goes to ∞.  As the graph of y=(((x^3)+2x-3)/(8(x^2)-7x+6)) gets arbitrarily close to that of y=(1/8)x+(7/64), it goes to ∞ as well.

Long division provided by Wolfram|Alpha.

Problem 31

Find  (lim_(x->∞))(Sqrt[4(x^2)+1])-2x and  (lim_(x->∞))(Sqrt[4(x^2)+x+1])-2x.

Solution

As both of these limits consist of a difference involving a square root the conjugate trick is probably going to be useful.

(lim_(x->∞))(Sqrt[4(x^2)+1])-2x=(lim_(x->∞))(((Sqrt[4(x^2)+1])-2x)/1)×(((Sqrt[4(x^2)+1])+2x)/((Sqrt[4(x^2)+1])+2x))=(lim_(x->∞))(((4(x^2)+1)-4(x^2))/((Sqrt[4(x^2)+1])+2x))=(lim_(x->∞))(1/((Sqrt[4(x^2)+1])+2x))=(lim_(x->∞))(1/((Sqrt[4(x^2)+1])+2x))×((1/x)/(1/x))=(lim_(x->∞))((1/x)/((Sqrt[4+(1/(x^2))])+2))=(0/((Sqrt[4+0])+2))=0

(lim_(x->∞))(Sqrt[4(x^2)+x+1])-2x=(lim_(x->∞))(((Sqrt[4(x^2)+x+1])-2x)/1)×(((Sqrt[4(x^2)+x+1])+2x)/((Sqrt[4(x^2)+x+1])+2x))=(lim_(x->∞))(((4(x^2)+x+1)-4(x^2))/((Sqrt[4(x^2)+x+1])+2x))=(lim_(x->∞))((x+1)/((Sqrt[4(x^2)+x+1])+2x))=(lim_(x->∞))((x+1)/((Sqrt[4(x^2)+x+1])+2x))×((1/x)/(1/x))=(lim_(x->∞))((1+1/x)/((Sqrt[4+(1/x)+(1/(x^2))])+2))=((1+0)/((Sqrt[4+0+0])+2))=(1/((Sqrt[4])+2))=(1/4)

Note that both computations use the “divide by the fastest exploding term from the bottom” idea - the term on bottom which blows up the fastest is essentially (Sqrt[4(x^2)])=2x  (as x->∞ there is no need for an absolute value), and 2x  is the same basic kind of term as x.

Problem 32

What is (lim_(x->(SuperMinus[((π/2))])))tan(x)?  What is  (lim_(x->(((-(π/2)))^(_+))))tan(x)?  What do these limits in combination with the Intermediate Value Theorem tell you about the tangent function?

Solution

As tangent is the ratio ((sin(x))/(cos(x))) it makes sense to look at the values of sine and cosine around (π/2) and -(π/2):

Plot[ {Sin[x], Cos[x]}, {x,-2,2}, Ticks-> {Range[-Pi, Pi, Pi/6], {-1,-1/2,1/2,1}}, PlotLegends->"Expressions", PlotStyle-> {Dashed, Thick}]

As x->(SuperMinus[((π/2))]) from the graphs we have than sin(x)->1 and cos(x)->(SuperPlus[0]) (we could be more specific and say sin(x)->(SuperMinus[1]) but we won’t need to). Therefore as  x->(SuperMinus[((π/2))]) we have tan(x)=((sin(x))/(cos(x)))->(1/(SuperPlus[0]))=∞.  So (lim_(x->(SuperMinus[((π/2))])))tan(x)=∞.

As x->(SuperPlus[(-(π/2))]) from the graphs we have than sin(x)->-1 and cos(x)->(SuperPlus[0]) (we could be more specific and say sin(x)->-(SuperPlus[1]) but we won’t need to). Therefore as  x->(SuperPlus[(-(π/2))]) we have tan(x)=((sin(x))/(cos(x)))->((-1)/(SuperPlus[0]))=-∞.  So (lim_(x->(SuperMinus[((π/2))])))tan(x)=-∞.

The two limits say that between -(π/2) and (π/2) the tangent function hits values as large and positive and as deep negative as you want. Now pick any number y.  Then there will be values values a<b between  -(π/2) and (π/2) such that tan(a)<y<tan(b). tan(x) is continuous on [a,b] (as both a and b are between -(π/2) and (π/2)), so the Intermediate Value Theorem implies that there will be a value c between a and b for which tan(c)=y.  Therefore the tangent function must hit every possible y-value between  -(π/2) and (π/2).  This is why the arctangent function (tan^(-1))(x) is well-defined with domain (-∞,∞) and range (-(π/2),(π/2)) - no matter what number x  you pick, there will be an angle from -(π/2) to (π/2) whose tangent is that number x.

Problem 33

Find all asymptotes of (((x^2)-3x+1)/((x^2)-x-6)) and describe the behavior on either side of any vertical asymptotes.

Solution

As (((x^2)-3x+1)/((x^2)-x-6))=(((x^2)-3x+1)/((x-3)(x+2))), the demoniator will be 0 at x=3,-2 and neither of these numbers make the top 0.  Therefore the graph will have vertical asymptotes at x=3 and x=-2.   Using quick limits to find the behavior on either side of those asymptotes we have:

(lim_(x->(SuperPlus[3])))(((x^2)-3x+1)/((x-3)(x+2)))=(1/((SuperPlus[0])× 5))=∞
(lim_(x->(SuperMinus[3])))(((x^2)-3x+1)/((x-3)(x+2)))=(1/((SuperMinus[0])× 5))=-∞
(lim_(x->-(SuperPlus[2])))(((x^2)-3x+1)/((x-3)(x+2)))=(11/((-5)×(SuperPlus[0])))=-∞
(lim_(x->-(SuperMinus[2])))(((x^2)-3x+1)/((x-3)(x+2)))=(11/((-5)×(SuperMinus[0])))=∞

In addition as the top has the same degree as the bottom there will be a hrizontal asymptote to which the graph will get close both as x->∞ and x->-∞: (lim_(x->∞))(((x^2)-3x+1)/((x^2)-x-6))=(lim_(x->∞))(((x^2)-3x+1)/((x^2)-x-6))×((1/(x^2))/(1/(x^2)))=(lim_(x->∞))((1-(3/x)+(1/(x^2)))/(1-(1/x)-(6/(x^2))))=((1-0+0)/(1-0-0))=1.  So y=1 is the only horizontal asymptote, and the graph gets close to it on both sides.  You can check the behavior in the following graph:

Plot[ (x^2-3x+1)/(x^2-x-6), {x,-10,10}, Exclusions-> x^2-x-6==0, Epilog-> {Dashed, InfiniteLine[ {{3,0},{3,1}}], InfiniteLine[ {{-2,0}, {-2,-2}}], InfiniteLine[ {{0,1},{1,1}}]}]

Problem 34

Find all asymptotes of ((x^3)/((x^2)-x-6)) and describe the behavior on either side of any vertical asymptotes.

Solution

As in Problem 33 there will be vertical asymptotes at x=3 and x=-2 (the top is nonzero at both these values).  Using quick limits to check the behavior on either side of those asymptotes yields:

(lim_(x->(SuperPlus[3])))((x^3)/((x-3)(x+2)))=(27/((SuperPlus[0])× 5))=∞
(lim_(x->(SuperMinus[3])))((x^3)/((x-3)(x+2)))=(27/((SuperMinus[0])× 5))=-∞
(lim_(x->-(SuperPlus[2])))((x^3)/((x-3)(x+2)))=((-8)/((-5)×(SuperPlus[0])))=∞
(lim_(x->-(SuperMinus[2])))((x^3)/((x-3)(x+2)))=((-8)/((-5)×(SuperMinus[0])))=-∞

As the degree of the top is exactly one more than the bottom there is an oblique asymptote which we can find via long division:

(RowDefault[({{, , , , }, {(x^2), "-",  x, "-", 6}}), ({{(("x")^2), (("x")^2), (("x")^2), (("x")^2), x, "+", 1}, {(x^3), "+", 0 (x^2), "+", 0 x, "-", 0 }, {(x^3), "-", (x^2), "-", 6x, (("x")^2), (("x")^2)}, {(("x")^2), (("x")^2), (x^2), "+", 6x, "",  }, {(("x")^2), (("x")^2), (x^2), "-", x, -, 6}, {(("x")^2), (("x")^2), (("x")^2), (("x")^2), 7x, +, 6}})])

So the oblique asymptote as y=x+1.  As this line has positive slope it goes upward, which implies (lim_(x->∞))((x^3)/((x^2)-x-6))=∞ and  (lim_(x->-∞))((x^3)/((x^2)-x-6))=-∞.  As a check we can look at the graph:

Plot[ x^3/(x^2-x-6), {x,-15,15}, Exclusions-> x^2-x-6==0, Epilog-> {Dashed, InfiniteLine[ {{3,0},{3,1}}], InfiniteLine[ {{-2,0}, {-2,1}}], InfiniteLine[ {{0,1},{1,2}}]}, PlotRange-> {-15,15}, AspectRatio-> Automatic]

Problem 35

State the formal definitions for standard two-sided limits, a limit at x=a to infinite, and a limit as x->∞

Solution

(lim_(x->a))f(x)=L if and only if for every ϵ>0 you can find a δ>0 such that  0<|x-a|<δ  implies that |f(x)-L|<ϵ.

(lim_(x->a))f(x)=∞ if and only if for every M>0 you can find a δ>0 such that  0<|x-a|<δ  implies that f(x)>M.

(lim_(x->∞))f(x)=L if and only if for every ϵ>0 you can find a M>0 such that  x>M  implies that |f(x)-L|<ϵ.

Problem 36

Use the formal definition of limit to show that (lim_(x->4))3x+1=13

Solution

Scratch work: |3x+1-13|<ϵ =>|3x-12|<ϵ => 3|x-4|<ϵ =>|x-4|<(ϵ/3)

Formal proof:  Define f(x)=3x+1 and let ϵ>0 be given.  Choose δ=(ϵ/4).  Now assume that 0<|x-4|<δ.  Then the difference |f(x)-13|=|3x+1-13|=|3x-12|=3|x-4|<3×(ϵ/3)=ϵ.  So whenever 0<|x-4|<δ we know that |(3x+1)-13|<ϵ, and (lim_(x->4))3x+1=13.

Problem 37

Use the formal definition of limit to show that (lim_(x->1))(x^2)+4x=5

Solution

Scratch work:  |(x^2)+4x-5|<ϵ  => |(x-1)(x+5)|<ϵ => |x+5|× |x-1|<ϵ => |x-1|<(ϵ/(|x+5|)) .  If we assume that δ≤1 then x is between 0 and 2, so |x+5| is between 5 and 7.  Therefore |x-1|<(ϵ/7) provided δ≤1.

Formal proof: Define f(x)=(x^2)+4x and suppose that ϵ>0 is given.  Choose δ=min(1,(ϵ/7)).  Now assume that 0<|x-1|<δ.  As δ≤1 we know x must be between 0 and 2, and so |x+5| is between 2 and 7.  Therefore the difference |f(x)-13|=|(x^2)+4x-5|=|(x+5)(x-1)|=|x+5|×|x-1|<7|x-1|<7×(ϵ/7)=ϵ.  So whenever  0<|x-1|<δ we have |((x^2)+4x)-5|<ϵ, and therefore (lim_(x->4))3x+1=13.

Note:  This shows that as the functions in the limit get even slightly complicated the choice of δ becomes more difficult (here we had to pick it to be the smaller of two things).  This is why we will only consider basic limits via the definition in Calculus 1; when you take the high-level course Real Analysis you will learn how to work with more complicated examples.

Problem 38

Use the formal definition of limit to show that (lim_(x->3))((|x-3|)/(x-3)) does not exist.

Solution

Note:  This is advanced problem, beyond what is feasible for the exam.  It does illustrate some of the features of limits though.

Showing that a limit does not exist by the definition is tough - you have to show that no matter what limit value L you try, there will be values ϵ>0 such that no matter what δ>0 you try some values of |f(x)-L|≥ϵ.  As the function values for ((|x-3|)/(x-3)) are only 1 and -1 there are three cases: L=1, L=-1, and L≠±1 (i.e. everything else).

Suppose the limit were 1.  Choose the value ϵ=(1/2).  Then no matter what value δ>0 you choose, when you assume 0<|x-3|<δ some of those x-values will be less than 3 (and the corresponding value of ((|x-3|)/(x-3)) is -1).  For those x-values you would have |f(x)-L|=|-1-1|=|-2|=2>ϵ.  So the limit cannot be 1.

Suppose the limit were -1.  Again choose the value ϵ=(1/2). Then no matter what value δ>0 you choose, when you assume 0<|x-3|<δ some of those x-values will be bigger than 3 (and the corresponding value of ((|x-3|)/(x-3)) would be 1). For those x-values you would have |f(x)-L|=|1-(-1)|=|2|=2>ϵ.  So the limit cannot be -1.

Now suppose that L is a number other than 1 or -1.  Out of the two distances from L to 1 and -1 one of them is smaller; choose ϵ to be half that smaller distance (that is ϵ=(1/2)min( |L-1|, |L+1|)).  Now choose any δ>0.  For every x-value with 0<|x-3|<δ consider what |f(x)-L| could be.  f(x) is either 1 or -1.  So |f(x)-L| is either |1-L|=|L-1| or |-1-L|=|L+1|.  But both |L+1| and |L-1| are larger than ϵ=(1/2)min( |L-1|, |L+1|), so |f(x)-L|≥ϵ.  So the limit cannot be a number L≠±1.

Problem 39

Use the formal definition of limit to show that (lim_(x->(SuperPlus[0])))(3/x)=∞

Solution

We need to show that no matter what value M>0 we pick, we can find a value δ such that whenever 0<x-0=x<δ we know that f(x)=(3/x)>M (note the absence of absolute values in the double inequality - this is because the limit is one-sided with x>0).

Scratch work: (3/x)>M => 3>x M -> (3/M)>x (using the fact that both M and x are positive).  So choose δ=(3/M).

Formal proof:  Define f(x)=(3/x) and suppose M>0 is given.  Choose δ=(3/M).  Now assume that 0<x<δ.  Then f(x)=(3/x)>(3/δ)=(3/(3/M))=M.  Therefore we can force (3/x) to be larger than any positive number M by forcing x close enough to 0 on the right (“close enough” being 0<x<δ=(3/M)), so (lim_(x->(SuperPlus[0])))(3/x)=∞.

Problem 40

Use the formal definition of limit to show that (lim_(x->∞))((x+1)/x)=1.

Solution

Scratch work: |((x+1)/x)-1|<ϵ => |1+(1/x)-1|<ϵ => |(1/x)|<ϵ => (1/(|x|))<ϵ => 1<ϵ |x| => |x | > (1/ϵ)

Formal proof:  Define f(x)=((x+1)/x)=1+(1/x) and suppose ϵ>0 is given.  Choose δ=(1/ϵ).  Now assume x>δ.  Then |f(x)-1|=|1+(1/x)-1|=|(1/x)|=(1/(|x|))<(1/δ)=(1/(1/ϵ))=ϵ.  So no matter what number ϵ>0 we are given, whenever x>(1/ϵ) the difference between ((x+1)/x) and 1 is less than ϵ;  therefore (lim_(x->∞))((x+1)/x)=1.