MH1100/MTH112: Calculus I.

Solutions to Problem Set #12.

Problem 1: (Various from Section 6.8 from [Stewart]).

In this problem you are asked to determine a number of limits. If you canfind a more elementary method than L’Hospital’s rule, please use it instead.If L’Hospital’s rule doesn’t apply, explain why.

Solutions.

Uses of L’Hospital’s rule below will be marked LHR.

Solution to 1.

This is an elementary limit only requiring division:

limx→1

x2 − 1

x2 − x= lim

x→1

(x− 1)(x+ 1)

(x− 1)x

= limx→1

(x+ 1)

x= 2.

Solution to 2.

This limit is of indeterminate form of typ 0/0, so we can try L’Hospital’srule:

limx→(π/2)+

cosx

1− sinx= lim

x→(π/2)+

− sinx

− cosx(LHR.)

= limx→(π/2)+

(sinx) ∗ limx→(π/2)+

(1

cosx

)

= −∞.

There are two subtleties here. The first is that L’Hospital’s rule is still valid

when the limit you get from the rule limx→af ′(x)g′(x) equals ±∞. The second

is the fact (intuitive, and straightforward to prove) that limx→a f(x)g(x) =−∞ when limx→a f(x) = L (with L > 0) and limx→a g(x) = −∞.

1

Solution to 3.

Again, this is of type 0/0, and a use of L’Hospital’s rule will solve theproblem. Recall that d

dx [tanx] =1

cos2 x .

limx→0

sin 4x

tan 5x= lim

x→0

4 cos 4x

5 ∗ (1/ cos2(5x)) (LHR.)

=4 cos (4 ∗ 0)

5 ∗ (1/ cos2(5 ∗ 0))

=4

5.

Solution to 4.

This is an indeterminate form of type∞/∞, so can be studied with L’Hospital’srule.

limx→∞

lnx√x

= limx→∞

1/x

1/2 ∗ x−1/2(LHR.)

= limx→∞

2

x1/2= 0.

Solution to 5.

This is an inderterminate form of type −∞/0, so L’Hospital’s rule cannot beapplied. But we can easily understnad the limit by rewriting it as a product:

limx→0+

lnx

x= lim

x→0+

(lnx ∗ 1

x

).

Note that limx→0+ lnx = −∞ while limx→0+1x = +∞. Thus:

limx→0+

lnx

x= −∞.

2

Solution to 6.

This limit is an indeterminate form of type 0/0. (This fact uses the calcu-lation: limt→0(8t − 5t) = 80 − 50 = 1− 1 = 0.) So:

limt→0

8t − 5t

t= lim

t→0

ddt

[8t − 5t

]

ddt [t]

(LHR.)

= limt→0

ln 8 ∗ 8t − ln 5 ∗ 5t

1= ln 8 ∗ 80 − ln 5 ∗ 50

= ln 8− ln 5.

Solution to 7.

This limit is of type 0/0 and you could certainly use L’Hospital’s rule tostudy this limit. There is a more elementary solution, however, which goesas follows.

limx→0

√1 + 2x−

√1− 4x

x

= limx→0

√1 + 2x−

√1− 4x

x∗√1 + 2x+

√1− 4x√

1 + 2x+√1− 4x

= limx→0

(1 + 2x)− (1− 4x)

x(√

1 + 2x+√1− 4x

)

= limx→0

6x

x(√

1 + 2x+√1− 4x

)

= limx→0

6√1 + 2x+

√1− 4x

=6√

1 + 2 ∗ 0 +√1− 4 ∗ 0

= 3.

3

Solution to 8.

This limit can be determined with two uses of L’Hospital’s rule.

limx→0

ex − 1− x

x2= lim

x→0

ex − 1

2x(LHR.)

= limx→0

ex

2(LHR, because of type 0/0.)

=1

2.

Solution to 9.

This is of type 0/0, so L’Hospital’s rule should have something to say:

limx→0

sin−1 x

x= lim

x→0

1/√1− x2

1(LHR.)

=1√

1− (limx→0 x)2

= 1.

Solution to 10.

This limit is of type ∞/∞, so we can proceed with L’Hospital’s rule.

limx→∞

(lnx)2

x= lim

x→∞

2 lnx ∗ (1/x)1

(LHR.)

= 2 limx→∞

lnx

x

= 2 limx→∞

1/x

1(LHR. because of type ∞/∞.)

= 2 ∗ 0

= 0.

4

Solution to 11.

Two uses of L’Hospital’s rule will solve this limit:

limx→1

1− x+ lnx

1 + cosπx= lim

x→1

−1 + 1x

−π sinπx(LHR.)

= limx→1

−1/x2

−π2 cosπx(LHR. because of type 0/0.)

=−1/1

−π2 cosπ

= − 1

π2.

Solution to 12.

You could use L’Hospital’s rule to study the following limit. Another ap-proach is to turn it into the standard limit limt→0

sin tt = 1. To do this we’ll

set t = πx . Then:

limx→∞

x sin(πx

)= π ∗ lim

x→∞

sin(πx

)

π/x

= π ∗ limt→0+

sin t

t= π.

We might as well compare this to the L’Hospital’s rule calculation of thesame limit. When you have two methods available, using both will con-firmyour answer.

limx→∞

x sin(πx

)= lim

x→∞

sin(πx

)

1/x

= limx→∞

− πx2 cos

(πx

)

−1/x2(LHR.)

= limx→∞

π cos(πx

)

= π ∗ cos 0= π.

5

Solution to 13.

We could start by making this limit more familiar in the following way:

limx→0+

sinx lnx = limx→0+

sinx

x∗ limx→0+

lnx

1/x.

This limit will be valid if the second limit exists. The second limit is of type∞/∞, and a standard calculation using L’Hospital’s rule:

limx→0+

lnx

1/x= lim

x→0+

1/x

−1/x2(LHR.)

= limx→0+

(−x)

= 0.

Thus:

limx→0+

sinx lnx = limx→0+

sinx

x∗ limx→0+

lnx

1/x= 1 ∗ 0 = 0.

Solution to 14.

Note that the given limit is of type ∞−∞, which cannot be studied withoutmore analysis. One approach is based on the limit

limx→∞

ex

xLHR.= lim

x→∞

ex

1= ∞.

(Note that this limit is studied in more generality below, in Problem 2.)Once you have this limit you can argue:

limx→∞

(x− lnx) = limx→∞

ln(ex−lnx

)(ln ex = x)

= limx→∞

ln

(ex

x

)

= ∞

because it is a standard fact that if limx→∞ f(x) = ∞, and limx→∞ g(x) =∞, then limx→∞ f(g(x)) = ∞.

6

Solution to 15.

We begin by shifting the dependence on x up into the index, in the standardway.

limx→0+

x√x = lim

x→0+e√x∗lnx

= elimx→0+√x∗lnx (ex is continuous.)

= elimx→0+

ln x

x−1/2

= elimx→0+

1/x

(−1/2)∗x−3/2 (LHR.)

= e−2 limx→0+√x

= e−2∗0

= 1.

Solution to 16.

Again, we begin by shifting the dependence on x up into the index.

limx→∞

x1/x = limx→∞

eln xx

= elimx→∞ln xx (ex is continuous.)

= elimx→∞1/x1 (LHR.)

= e0

= 1.

Solution to 17.

Note that because

limx→∞

2x− 3

2x+ 5= lim

x→∞

1− 3/2x

1 + 5/2x= 1,

and because limx→∞(2x+1) = ∞, this is an indeterminate form of type 1∞.This has no logical meaning, and we need to analyse it further to understandthe limit.

Begin with the standard move: elnx = x.

limx→∞

(2x− 3

2x+ 5

)2x+1

= limx→∞

e(2x+1)∗ln ( 2x−32x+5 )

7

Using the fact that ex is a continuous function, we can pass the limit intothe exponential. This means the limit we need to understand is:

limx→∞

ln(2x−32x+5

)

1/(2x+ 1).

This limit is of type 0/0, so we can study it with L’Hospital’s rule. Thederivative of the numerator is:

d

dx

[ln

(2x− 3

2x+ 5

)]=

2x+ 5

2x− 3∗ d

dx

[2x− 3

2x+ 5

]

=2x+ 5

2x− 3∗ 2 ∗ (2x+ 5)− (2x− 3) ∗ 2

(2x+ 5)2

= 16 ∗ 1

(2x− 3)(2x+ 5).

The derivative of the denominator is:

d

dx

[1

2x+ 1

]= − 2

(2x+ 1)2.

So, using L’Hospital’s rule, the limit we need can be computed:

limx→∞

ln(2x−32x+5

)

1/(2x+ 1)= lim

x→∞

16 ∗ 1(2x−3)(2x+5)

− 2(2x+1)2

= limx→∞

−8 ∗(1 + 1

2x

)2(1− 3

2x

) (1 + 1

2x

)

= −8 ∗ (1 + 0)2

(1− 0)(1 + 0)= −8.

Now that we know this limit we can finish the problem:

limx→∞

(2x− 3

2x+ 5

)2x+1

= elimx→∞ln( 2x−3

2x+5 )1/(2x+1) =

1

e8.

8

Solution to 18.

This limit is of type ∞−∞, so we need to do more work to understand it.

limx→0+

(1

x− 1

ex − 1

)

= limx→0+

ex − (1 + x)

x(ex − 1)

= limx→0+

ex − 1

ex − 1 + xex(LHR, because of type 0/0)

= limx→0+

ex

ex + ex + xex(LHR, because of type 0/0)

=1

1 + 1 + 0

=1

3.

!Problem 2: (#6.8.71 from [Stewart]).

Prove that for any positive integer n ∈ N:

limx→∞

ex

xn= ∞.

Solution

We’ll prove this statement by induction.

The base case of the induction, n = 1, is clear: limx→∞ex

x0 = ∞.

Now assume that for some n ∈ N, limx→∞ex

xn = ∞. We will show thatit follows that limx→∞

ex

xn+1 = ∞ as well.

Note that the limit limx→∞ex

xn+1 is of type ∞/∞. It follows fromL’Hospital’s rule that

limx→∞

ex

xn+1= lim

x→∞

ddx [e

x]ddx [x

n+1]=

1

(n+ 1)limx→∞

ex

xn= ∞

by the induction assumption.

(Note that we are allowed to do this because the precise fourth assump-

tion of L’Hospital’s rule is that the limit limx→∞f ′(x)g′(x) either exists or equals

±∞.)

!

9

Problem 3: (#6.8.72 from [Stewart]).

Prove that for any positive real p ∈ R, p > 0,

limx→∞

lnx

xp= 0.

Solution

Note that:

• limx→∞ lnx = ∞.

• limx→∞ xp = ∞, if p > 0.

Thus the limit is of type∞/∞, and we can apply L’Hospital’s rule to deduce:

limx→∞

lnx

xp= lim

x→∞

ddx [lnx]ddx [x

p]= lim

x→∞

1/x

pxp−1=

1

plimx→∞

1

xp= 0.

!

10

Problem 4: (#6.8.98 from [Stewart]).

For what values of a and b is the following equation true?

limx→0

(sin 2x

x3+ a+

b

x2

)= 0.

Solution:

Start be rewriting the limit:

limx→0

sin 2x+ ax3 + bx

x3= 0.

This limit is of type 0/0. L’Hospital’s rule now tels us to investigate thelimit:

limx→0

ddx

[sin 2x+ ax3 + bx

]

ddx [x

3]= lim

x→0

2 cos 2x+ 3ax2 + b

3x2.

Note that as x → 0, the top line goes to 2+ b while the bottom line goesto 0. Thus we conclude that unless b = −2, the given limit goes to infinity.

If we set b to −2, then this limit is of type 0/0, and we can applyL’Hospital’s rule a second time. We get:

limx→0

2 cos 2x+ 3ax2 − 2

3x2= lim

x→0

−4 sin 2x+ 6ax

6x(LHR.)

= limx→0

−8 cos 2x+ 6a

6(LHR.)

= −4

3+ a.

This is zero if and only if a = 43 .

In summary: The given limit is zero of and only if a = 43 and b = −2.

!

11

Problem 5: (#3.4.8 from [Stewart].)

Evaluate the following limit:

limx→∞

√12x3 − 5x+ 2

1 + 4x2 + 3x3.

Solution:

The square root function is continuous. So we can take the limit towardsinfinity inside the square root:

limx→∞

√12x3 − 5x+ 2

1 + 4x2 + 3x3=

√limx→∞

12x3 − 5x+ 2

1 + 4x2 + 3x3.

This, of course, is only justified if the inside limit exists. The inside limit iscomputed by dividing through by x3:

√limx→∞

12x3 − 5x+ 2

1 + 4x2 + 3x3=

√

limx→∞

12− 5 1x2 + 2 1

x3

3 + 4 1x + 1

x3

=

√12

3= 2 .

!

12

Problem 6: (#3.4.17 and #3.4.18 from [Stewart].)

Evaluate the limits:

(i) limx→∞√9x6−xx3+1

(ii) limx→−∞√9x6−xx3+1 .

Solution to Part (i).

The deduction is:

limx→∞√9x6−xx3+1 = limx→∞

x3!

9− 1x5

x3+1 (Because the limit onlydepends on positive x.)

= limx→∞

!9− 1

x5

1+ 1x3

=

"limx→∞

#9− 1

x5

$

limx→∞#1+ 1

x3

$

= 3 .

Solution to Part (ii).

The deduction is:

limx→−∞√9x6−xx3+1 = − limx→−∞

x3!

9− 1x5

x3+1 (Because the limit onlydepends on negative x.)

= − limx→−∞

!9− 1

x5

1+ 1x3

= −

"limx→−∞

#9− 1

x5

$

limx→−∞#1+ 1

x3

$

= −3 .

!

13

Problem 7: (#3.4.19 from [Stewart].)

Evaluate the limit:limx→∞

(√9x2 + x− 3x

).

Solution:

limx→∞

(√9x2 + x− 3x

)

= limx→∞

((√9x2 + x− 3x

)∗√9x2 + x+ 3x√9x2 + x+ 3x

)

= limx→∞

9x2 + x− (3x)2√9x2 + x+ 3x

= limx→∞

x√9x2 + x+ 3x

= limx→∞

1√9 + 1

x + 3

=1

6.

!

14

Problem 8: (#3.4.20 from [Stewart].)

Evaluate the limit:lim

x→−∞

(x+

√x2 + 2x

).

Solution:

limx→−∞

(x+

√x2 + 2x

)

= limx→−∞

((x+

√x2 + 2x

)∗ x−

√x2 + 2x

x−√x2 + 2x

)

= limx→−∞

x2 −(x2 + 2x

)

x−√x2 + 2x

= limx→−∞

−2x

x−√x2 + 2x

= limx→−∞

−2x

x− (−x) ∗√

1 + 2x

= limx→−∞

−2

1 +√

1 + 2x

= −1.

!

15

Problem 9: (#3.4.21 from [Stewart].)

Evaluate the limit:

limx→∞

(√x2 + ax−

√x2 + bx

).

Solution:

limx→∞

(√x2 + ax−

√x2 + bx

)

= limx→∞

((√x2 + ax−

√x2 + bx

)∗√x2 + ax+

√x2 + bx√

x2 + ax+√x2 + bx

)

= limx→∞

((x2 + ax)− (x2 + bx)√x2 + ax+

√x2 + bx

)

= limx→∞

((a− b)x√

x2 + ax+√x2 + bx

)

= limx→∞

⎛

⎝ (a− b)√1 + a 1

x +√

1 + b 1x

⎞

⎠

=a− b

2.

!

16

Problem 10: (#3.4.22 from [Stewart].)

Does the limit limx→∞ cosx exist? Discuss.

Solution:

This function does not have a limit as x tends to infinity. One explanationis: after every positive M there are points where the function has value 1(whenever x is an even multiple of π), and points where the function hasvalue −1 (whenever x is an odd multiple of π), so the function cannotapproach any particular value.

!

Problem 11: (#3.4.27 from [Stewart].)

Evaluate the limitlimx→∞

(x−√x).

Solution:

If we arrange this as follows:

limx→∞

(√x(√

x− 1))

then we observe that this expression is a product of two functions, each ofwhich goes to infinity. So the limit is ∞.

(Here is an incorrect solution: Observe that x goes to infinity, and√x

goes to infinity as well, so x −√x goes to ∞ − ∞ = ∞. Incorrect. For

example, f(x) = x goes to infinity, and g(x) = x+ 1x goes to infinity as well,

but f(x)− g(x) does not go to infinity.)

!

17

Problem 12: (#3.4.29 and #3.4.30 from [Stewart].)

Evaluate the limits

(i) limx→∞ x sin 1x ,

(ii) limx→∞√x sin 1

x .

Solution to (i).

This problem cannot be determined directly by a product rule

(limx→∞

x)∗(

limx→∞

sin1

x

)

because x → ∞ while sin 1x → 0 as x → ∞. Thus we need to do further

work to understand this limit. We can rearrange it as:

limx→∞

x sin1

x= lim

x→∞

sin 1x

1x

.

A standard trick (compare with problem 20 below) lets us rewrite this:

limx→∞

sin 1x

1x

= limt→0+

sin t

t,

provided that second limit exists. (The intuition is clear: as x → ∞, 1x →

0+.) Of course, that second limit is the standard trigonometric limit, so welearn that:

limx→∞

x sin1

x= 1.

Solution to (ii).

This limit can be determined using Part (i):

limx→∞

√x sin

1

x= lim

x→∞x−1/2 ∗ x sin 1

x

= limx→∞

1√x∗ limx→∞

x sin1

x= 0 ∗ 1= 0.

!

18

Problem 13: (#3.4.41 from [Stewart].)

Find a formula for a function f(x) which has all of the following properties:

limx→±∞ f(x) = 0, limx→0 f(x) = −∞, f(2) = 0,limx→3− f(x) = ∞, limx→3+ f(x) = −∞.

Solution:

This, like a very large part of mathematics, is basically an unpredictableexercise in inspired guessing. Here is an approach to thinking about how toguess the right answer. We begin with two observations:

• If the function is a rational function (that is, a quotient of polyno-mials) then as long as the denominator has degree at least one morethan the numerator, then the first condition will be satisfied.

• The second observation is that it is quite easy to satisfy the require-ments for vertical asymptotes by putting appropriate factors in thedenominator. For example, because both the left and right limits atx = 0 are −∞, we’ll need a factor of x2 in the denominator (dealingwith the minus sign later).

These two observations suggest a guess of the form:

f(x) =??

x2(x− 3).

So what do we need on the top line for a successful outcome? We need, forthe top line:

• A polynomial p(x) of degree less than or equal to 2.

• We need p(0) > 0 (so we get limx→0 f(x) = −∞).

• We need p(3) < 0 (in order to get the required vertical asymptotebehaviour at x = 3).

• We also need that p(2) = 0.

We can easily find a linear function with these properties, the simplest being:

p(x) = 2− x .

So our final guess is:

f(x) =2− x

x2(x− 3).

19

Lets whack that into MATLAB, and see what we get:

−5 0 5−5

−4

−3

−2

−1

0

1

2

3

4

5

As required.

!

20

Problem 14⋆: (#3.4.71 from [Stewart].)

Using the definitions, prove that

limx→∞

f(x) = limt→0+

f (1/t)

and thatlim

x→−∞f(x) = lim

t→0−f (1/t) .

(The meaning of these equations is that when either limit exists, then sodoes the other one, and then the given limits are equal.)

Solution:

We’ll concentrate on the first equation. The second is proved in the sameway with some minor changes. For clarity we’ll split the statement into twodirections.

Case 1: Assume that the limit limx→∞ f(x) exists and equals L. Provelimt→0+ f (1/t) exists as well and equals L.

We’ll start by writing out what we are assuming is true: For every ϵ > 0there exists a positive Mϵ such that x > Mϵ implies |f(x)−L| < ϵ. Call thisassumption (⋆).

Our task is to find a rule δϵ such that whenever 0 < t − 0 < δϵ, then∣∣f(1t

)− L

∣∣ < ϵ.

The rule we need is: δϵ =1Mϵ

. Because if 0 < t < 1Mϵ

, then 1t > Mϵ, so,

by Assumption (⋆),∣∣f(1t

)− L

∣∣ < ϵ.

Case 2: Assume that the limit limt→0+ f (1/t) exists and equals L. Provelimx→∞ f(x) exists as well and equals L.

So, what we are assuming is: For every ϵ > 0, there exists a δϵ > 0, suchthat if 0 < t < δϵ, then

∣∣f(1t

)− L

∣∣ < ϵ. Call this assumption (⋆⋆).

Our task in this case is to find a rule Mϵ > 0 such that whenever x > Mϵ

then |f(x)− L| < ϵ. The rule we need is Mϵ = 1δϵ. For if x > 1

δϵ, then

0 < 1x < δϵ, so by Assumption (⋆⋆),

|f (x)− L| =

∣∣∣∣f(

1

1/x

)− L

∣∣∣∣< ϵ.

!

21