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Calculus 2 : Limits

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Example Questions

Example Question #4 : Limits And Asymptotes

Find the following limit:

$\lim_{x\rightarrow 2}\frac{(x-2)}{x^2-4}$

Possible Answers:

$\frac{1}{4}$

$\frac{3}{7}$

$\frac{2}{9}$

$\frac{1}{2}$

Correct answer:

$\frac{1}{4}$

Explanation:

For this problem we first need to expand the denominator.

$\lim_{x\rightarrow 2}\frac{x-2}{x^2-4}=\frac{x-2}{(x-2)(x+2)}$

We can expand the denominator since x^2-4 is a difference of squares.

From here we can cancel the (x-2) quantity from the numerator and denominator.

The resulting function is as follows:

$\lim_{x\rightarrow2}\frac{1}{x+2}$

Plugging in 2 we get our limit.

$=\frac{1}{2+2}=\frac{1}{4}$

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Example Question #5 : Limits And Asymptotes

Find the following limit:

$\lim_{x\rightarrow 0}\frac{\sqrt{1+x}-\sqrt{1-x}}{x}$

Possible Answers:

$\infty$

$-\infty$

Correct answer:

Explanation:

We will use the following to prove this result.

Assuming that $a,b\ge 0$

$(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})=a-b$ . We will use this result:

we have $({\sqrt{1+x}+\sqrt{1-x}})(\sqrt{1+x}-\sqrt{1-x})$

=1+x-(1-x)=2x

Therefore

$\lim_{x\rightarrow 0}{\frac{\sqrt{1+x}-\sqrt{1-x}}{x} =$

$\lim_{x\rightarrow 0}\frac{2x}{x({\sqrt{1+x}+\sqrt{1-x}})}$

$=\lim_{x\rightarrow 0}\frac{2}{({\sqrt{1+x}+\sqrt{1-x}})}$

$=\frac{2}{2}=1$

this shows the limit is 1.

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Example Question #3 : Limits And Asymptotes

Find the following limit:

$\lim_{x\rightarrow 3}\frac{x^3-27}{\sqrt{3x}-3}$

Possible Answers:

The limit does not exist.

Correct answer:

Explanation:

We will use the following identity to establish this result.

We have

a^3-b^3=(a-b)(a^2+ab+b^2)

and we note that :

$\frac{\sqrt{3x}+3}{\sqrt{3x}+3}=1$

Therefore by multiplying the above equivalency for 1 we get the following:

$\frac{x^3-27}{\sqrt{3x}-3}=\frac{x^3-27}{\sqrt{3x}-3}\frac{\sqrt{3x+3}}{\sqrt{3x+3}}$

and we know that

$(\sqrt{3x}-3)(\sqrt{3x}+3)=3x-9$

$\frac{x^3-27}{\sqrt{3x}-3}\frac{(\sqrt{3x}+3)}{(\sqrt{3x}+3)}= \frac{x^3-27}{\sqrt{3x}-3}=\frac{x^3-27}{3x-9}(\sqrt{3x}+3)= \frac{x^3-27}{3(x-3)}(\sqrt{3x}+3)$

We can rewrite our equation using identities.

x^3-27=x^3-3^3=(x-3)(x^2+3x+9)

This gives :

$\frac{(x^2+3x+9)(\sqrt{3x}+3)}{3}$ and

now taking the limit as x goes to 3, we obtian

x=54

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Example Question #7 : Limits And Asymptotes

Find the following limit:

$\lim_{x\rightarrow 1}\frac{x^7-1}{1-x}$

Possible Answers:

Correct answer:

Explanation:

We note first that we can write:

$x^7-1=(x-1)(x^6+x^5+x^4\cdots +1)$

Therefore our expression becomes in this case:

$\lim_{x\rightarrow 1}\frac{x^7-1}{1-x}=\lim_{x\rightarrow 1}\frac{(x^6+x^5\cdots+1)(x-1)}{1-x}$

Noting now that:

$\frac{x-1}{1-x}=-1$ for all $x\neq 1$ .

Therefore , we have:

$\lim_{x\rightarrow 1}\frac{x^7-1}{1-x}=\lim_{x\rightarrow 1}{-(x^6+x^5\cdots+1)}$

and evaluating now for x=1, we obtain

$\lim_{x\rightarrow 1}\frac{x^7-1}{1-x}=-7$

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Example Question #3 : Limits And Asymptotes

Let the following polynomial:

$p(x)=1+x^2+x^4+\cdots x^{2n}$

What are the vertical asymptotes of

$q(x)= \frac{1}{p(x)}$

Possible Answers:

does not have a vertical asymptote.

$x=1,2,\cdots,n$

There are vertical asymptotes

$x=\pm1,\cdots,\pm n$

Correct answer:

does not have a vertical asymptote.

Explanation:

We first note that the polynomial is defined for all real numbers.

We know that for any real number x different from 0, we have :

$x^2\ge 0$ .

Now we need to see that for any integer n we have:

$x^{2n}\ge 0$ . Adding in this case,

we have $1+x^2+x^4+\cdots+ x^{2n}> 1$

and therefore p(x)>1 , this implies by definiton of q(x) that:

q(x)< 1 .

We also have q(x)>0 .

This means that 0<q(x)<1 .

Therefore q(x) can never be 0 and this means that it does not have an asymptote.

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Example Question #11 : Limits And Asymptotes

Let and be positive integers. Determine the following limit:

$\lim_{x\rightarrow \infty}\frac{x^n}{x^m-1}$ ,

given that m<n .

Possible Answers:

$\infty$

Correct answer:

$\infty$

Explanation:

We note first we have a fractional function here that we can write:

$q(x)=\frac{f(x)}{g(x)}$

with,

f(x)=x^n , g(x)=x^m-1 .

We know that n>m, this means that

$q(x)\sim {x^{n-m}}$ and

n-m>0 . This means that

$\lim_{x\rightarrow \infty} x^{n-m}=\infty$

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Example Question #11 : Limits And Asymptotes

Find the following limit given that and are positive integers such that n<m :

$\lim_{x\rightarrow \infty }\frac{e^{nx}}{e^{mx}}$

Possible Answers:

$\infty$

$-\infty$

Correct answer:

Explanation:

We need to know that:

$\lim_{x\rightarrow \infty } e^{x}=\infty$ , $\lim_{x\rightarrow -\infty} e^{x}=0$

We also know that from the properties of the exponentials that:

$\frac{e^n}{e^m}=e^{n-m}$ .

Now we need to note that since n<m, we have n-m<0

and therefore:

$\lim_{x\rightarrow \infty }\frac{e^{nx}}{e^{mx}}=\lim_{x\rightarrow \infty }{e^{n(x-m)}}$ .

This means that :

$\lim_{x\rightarrow \infty }{e^{n(x-m)}}=0$ .

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Example Question #13 : Limits And Asymptotes

Find the following limit:

$\lim_{x\rightarrow \pi}\frac{sin^2{x}}{1+cosx}$

Possible Answers:

Correct answer:

Explanation:

Recall that for all real numbers we have:

sin^2x+cos^2x=1 , from which we deduce that:

sin^2x=1-cos^2x .

We know that:

1-cos^2x=(1-cosx)(1+cosx) .

Therefore:

$\lim_{x\rightarrow \pi}\frac{sin^2{x}}{1+cosx}=\lim_{x\rightarrow \pi}\frac{(1-cosx)(1+cosx))}{1+cosx}$ .

Simplifying the above expression gives:

$\lim_{x\rightarrow \pi}\frac{sin^2{x}}{1+cosx}=\lim_{x\rightarrow \pi} (1-cosx)=2$

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Example Question #12 : Limits And Asymptotes

Consider $y=\frac{1}{x}+2$ . What is the limit as approaches zero from the left?

Possible Answers:

$-\infty$

There is no limit.

$\infty$

Correct answer:

$-\infty$

Explanation:

For the function $y=\frac{1}{x}+2$ , there is an asymptote at x=0 .

The graph approaches to positive infinity as approaches zero from the right side of the graph, and negative infinity as approaches zero from the left.

Therefore, the limit of $y=\frac{1}{x}+2$ from the left approaches to $-\infty$ .

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Example Question #461 : Limits

Find the horizontal asymptote for $f(x) = \frac{3x^3 + 2x}{x^3+1}$ .

Possible Answers:

$y = \frac{3}{2}$

y = 2

y = 1

There is no horizontal asymptote / the asymptote is undefined.

y = 3

Correct answer:

y = 3

Explanation:

When finding horizontal asymptotes, there are 3 conditions / rules to follow.

1) if the leading terms of the numerator and denominator are of the same degree, then the HA is equal to the ratio of the coefficients of the leading terms

2) if the leading term of the denominator is of a higher degree than the leading term of the numerator, then the HA is y = 0

3) if the leading term of the numerator is of a higher degree than the leading term of the denominator, then there is no horizontal asymptote (or it is undefined)

This question falls into condition 1, and therefore the horizontal asymptote is y = 3 .

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Calculus 2 : Limits

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #4 : Limits And Asymptotes

Example Question #5 : Limits And Asymptotes

Example Question #3 : Limits And Asymptotes

Example Question #7 : Limits And Asymptotes

Example Question #3 : Limits And Asymptotes

Example Question #11 : Limits And Asymptotes

Example Question #11 : Limits And Asymptotes

Example Question #13 : Limits And Asymptotes

Example Question #12 : Limits And Asymptotes

Example Question #461 : Limits

All Calculus 2 Resources

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