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Calculus 2 : New Concepts

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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Example Question #61 : New Concepts

Evaluate the following limit, if possible:

$\lim_{x\to 0}\frac{|x|}{x^2+x}$

Possible Answers:

$\infty$

The limit does not exist

Correct answer:

The limit does not exist

Explanation:

The limit does not exist. When we first plug in the limit value, 0, we get 0/0 and indeterminate form. We may try to use L'Hopital's rule. However, we quickly discover that

f(x)=|x|

is not differentiable at x=0, so we cannot use L'Hopital's rule. We then consider the limits from the left and right. We determine that

$\lim_{x\to 0^-}\frac{|x|}{x^2+x}=-1$

and

$\lim_{x\to 0^+}\frac{|x|}{x^2+x}=1$ .

Thus the two sided limit does not exist.

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Example Question #62 : New Concepts

Evaluate the following limit, if it exists:

$\lim_{x\to\frac{\pi}{2}} \frac{\sin^2{x}-1}{\cos{x}}$

Possible Answers:

$\frac{\pi}{2}$

$\pi$

Correct answer:

Explanation:

To calculate the limit we first plug the limit value into the numerator and denominator of the expression. When we do this we get $\frac{0}{0}$ , which is undefined. We now use L'Hopital's rule which says that if f(x) and g(x) are differentiable and

f(c)=g(c)=0 ,

then

$\lim_{x\to c} \frac{f(x)}{g(x)} = \lim_{x\to c} \frac{f'(x)}{g'(x)}$ .

We are trying to evaluate the limit

$\lim_{x\to\frac{\pi}{2}} \frac{\sin^2{x}-1}{\cos{x}}$ ,

so we have

$f(x)=\sin^2{x}-1$

and

$g(x)=\cos(x)$ .

We differentiate both of these functions and find

$f'(x)=2\sin{x}\cos{x}$

and

$g'(x)=-\sin{x}$ .

Using L'Hopital's rule we see

$\lim_{x\to\frac{\pi}{2}} \frac{\sin^2{x}-1}{\cos{x}} = \lim_{x\to\frac{\pi}{2}} \frac{2\sin{x}\cos{x}}{-\sin{x}}$ .

Now when we plug the limit value into the expression we get $\frac{0}{-1}$ , so

$\lim_{x\to\frac{\pi}{2}} \frac{\sin^2{x}-1}{\cos{x}} = 0$ .

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Example Question #63 : New Concepts

Evaluate the following limit, if possible:

$\lim_{x\to\infty} \frac{3x^3+x}{2x^2+3}$ .

Possible Answers:

$\infty$

The limit does not exist.

$\frac{3}{2}$

Correct answer:

$\infty$

Explanation:

If we plugged in the limit value, $\infty$ , directly we would get the indeterminate value $\frac{\infty}{\infty}$ . We now use L'Hopital's rule which says that if f(x) and g(x) are differentiable and

$f(x)=g(x)=\pm\infty$ ,

then

$\lim_{x\to c} \frac{f(x)}{g(x)} = \lim_{x\to c} \frac{f'(x)}{g'(x)}$ .

The limit we wish to evaluate is

$\lim_{x\to\infty} \frac{3x^3+x}{2x^2+3}$ ,

so we have

f(x)=3x^3+x

and

g(x)=2x^2+3 .

We evaluate the derivatives of these two functions and find

f'(x)=9x^2+1

and

g'(x)=4x .

Thus, using L'Hopital's we find

$\lim_{x\to\infty} \frac{3x^3+x}{2x^2+3} = \lim_{x\to\infty} \frac{9x^2+1}{4x}$ .

However, when we plug the limit value into this second expression we still end up with the indeterminate value $\frac{\infty}{\infty}$ . So we use L'Hopital's again.

f''(x)=18x

and

g''(x)=4 .

Using L'Hopital's rule again we see

$\lim_{x\to\infty} \frac{3x^3+x}{2x^2+1} = \lim_{x\to\infty} \frac{9x^2+3}{4x}= \lim_{x\to\infty} \frac{18x}{4}$ .

Plugging in the limit value now we see that the limit evaluates to $\infty.$

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Example Question #64 : New Concepts

Evaluate the following limit, if possible:

$\lim_{x\to 0} \frac{\cos{x}-1}{e^x-1}$

Possible Answers:

$\pi$

The limit does not exist

$\infty$

Correct answer:

Explanation:

f(c)=g(c)=0 ,

then

$\lim_{x\to c} \frac{f(x)}{g(x)} = \lim_{x\to c} \frac{f'(x)}{g'(x)}$ .

We are trying to evaluate the limit

$\lim_{x\to 0} \frac{\cos{x}-1}{e^x-1}$

so we have

$f(x)=\cos{x}-1$

and

g(x)=e^x-1 .

We calculate the derivatives of these functions and get

$f'(x)=-\sin{x}$

and

g'(x)=e^x .

Using L'Hopital's rule we find

$\lim_{x\to 0} \frac{\cos{x}-1}{e^x-1}= \lim_{x\to 0} \frac{-\sin{x}}{e^x}$ .

Now when we plug in the limit value we get $\frac{0}{1}$ , so the limit evaluates to 0.

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Example Question #65 : New Concepts

Evaluate the following limit, if possible:

$\lim_{x\to 0} \frac{\log(x+1)}{2x}$ .

Possible Answers:

The limit does not exist

$\frac{1}{2}$

Correct answer:

$\frac{1}{2}$

Explanation:

f(c)=g(c)=0 ,

then

$\lim_{x\to c} \frac{f(x)}{g(x)} = \lim_{x\to c} \frac{f'(x)}{g'(x)}$ .

In this case we are calculating

$\lim_{x\to 0} \frac{\log(x+1)}{2x}$

$f(x)=\log$x+1$$

and

g(x)=2x .

We calculate the derivatives and find that

$f'(x)=\frac{1}{x+1}$

and

g'(x)=2 .

Thus

$\lim_{x\to 0} \frac{\log(x+1)}{2x} = \lim_{x\to 0} \frac{1}{2(x+1)} = \frac{1}{2}$ .

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Example Question #66 : New Concepts

Evaluate the following limit, if possible:

$\lim_{x\to \infty} \frac{-4x^2+7x+3}{x^2-5}$

Possible Answers:

The limit does not exist

Correct answer:

Explanation:

If we plugged in the limit value, $\infty$ , directly we would get the indeterminate value $\frac{-\infty}{\infty}$ . We now use L'Hopital's rule which says that if f(x) and g(x) are differentiable and

$f(x)=g(x)=\pm\infty$ ,

then

$\lim_{x\to c} \frac{f(x)}{g(x)} = \lim_{x\to c} \frac{f'(x)}{g'(x)}$ .

The limit we wish to evaluate is

$\lim_{x\to \infty} \frac{-4x^2+7x+3}{x^2-5}$ ,

so we have

f(x)=-4x^2+7x+3

and

g(x)=x^2-5 .

We differentiate both functions and find

f'(x)=-8x+7

and

g'(x)=2x .

Now using L'Hopital's rule we find

$\lim_{x\to \infty} \frac{-4x^2+7x+3}{x^2-5} = \lim_{x\to \infty} \frac{-8x+7}{2x}$ .

When we plug the limit value in to this expression we still get the indeterminate value $\frac{-\infty}{\infty}$ . Thus we must use L'Hopital's rule again.

f''(x)=-8

and

g''(x)=2 .

Using L'Hopital's rule again we find

$\lim_{x\to \infty} \frac{-4x^2+7x+3}{x^2-5} = \lim_{x\to \infty} \frac{-8x+7}{2x}=\lim_{x\to \infty} \frac{-8}{2}=-4$ .

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Example Question #67 : New Concepts

Evaluate the following limit, if possible:

$\lim_{x\to 3} \frac{2x^2-\frac{4}{3}x-14}{\sin{(\pi x)}}$

Possible Answers:

$-\frac{6}{\pi}$

The limit does not exist

$\frac{16}{\pi}$

$-\frac{32}{3\pi}$

Correct answer:

$-\frac{32}{3\pi}$

Explanation:

f(c)=g(c)=0 ,

then

$\lim_{x\to c} \frac{f(x)}{g(x)} = \lim_{x\to c} \frac{f'(x)}{g'(x)}$ .

We are evaluating the limit

$\lim_{x\to 3} \frac{2x^2-\frac{4}{3}x-14}{\sin{(\pi x)}}$

so we have

$f(x)=2x^2-\frac{4}{3}x-14$

and

$g(x)=\sin{(\pi x)}$ .

We differentiate these functions and find

$f'(x)=4x-\frac{4}{3}$

and

$g'(x)=\pi\cos{(\pi x)}$ .

Using L'Hopital's rule we see

$\lim_{x\to 3} \frac{2x^2-\frac{4}{3}x-14}{\sin{(\pi x)}} = \lim_{x\to 3} \frac{4x-\frac{4}{3}}{\pi\cos{(\pi x)}} =\frac{\frac{32}{3}}{-\pi}=-\frac{32}{3\pi}$ .

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Example Question #68 : New Concepts

Evaluate the following limit, if possible:

$\lim_{x\to 0} \frac{\log{(x+1)}-\sin{x}}{3x}$ .

Possible Answers:

$\infty$

The limit does not exist

$\frac{1}{3}$

Correct answer:

Explanation:

f(c)=g(c)=0 ,

then

$\lim_{x\to c} \frac{f(x)}{g(x)} = \lim_{x\to c} \frac{f'(x)}{g'(x)}$ .

We are evaluating the limit

$\lim_{x\to 0} \frac{\log{(x+1)}-\sin{x}}{3x}$ .

In this case we have

$f(x)=\log{(x+1)}-\sin{x}$

and

g(x)=3x .

We differentiate both functions and get

$f'(x)=\frac{1}{x+1}-\cos{x}$

and

g'(x)=3 .

Using L'Hopital's rule we find

$\lim_{x\to 0} \frac{\log{(x+1)}-\sin{x}}{3x} = \lim_{x\to 0} \frac{\frac{1}{x+1}-\cos{x}}{3}=\frac{1-1}{3}=0$ .

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Example Question #69 : New Concepts

Evaluate:

$\lim_{n \to \infty} \frac{n^3+e^{2n}}{3n^3+1}$

Possible Answers:

$-\frac{1}{3}$

$\infty$

$\frac{1}{3}$

$-\infty$

Correct answer:

$\infty$

Explanation:

To evaluate the limit, we must compare the numerator and denominator. The numerator and denominator both have a term containing a third degree term. However, dividing their coefficients is not enough to determine the limit. There is an exponential term in the numerator, which grows faster than any polynomial terms. It dominates the function, and the limit goes to $\infty$ .

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Example Question #1711 : Calculus Ii

Evaluate:

$\lim_{x\rightarrow 2}\frac{x^2+2x-8}{x^3+2x^2-16}$

Possible Answers:

$\infty$

$\frac{3}{10}$

$\frac{3}{5}$

Correct answer:

$\frac{3}{10}$

Explanation:

When we evaluate the limit using normal methods (substitution), we get the indeterminate form $\frac{0}{0}$ . When this occurs, we must use L'Hopital's Rule to evaluate the limit. The rule states that

$\lim_{x\rightarrow a}\frac{ f(x)}{g(x)}=\lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}$

Using the rule for our limit, we get

$\lim_{x\rightarrow 2}\frac{2x+2}{3x^2+4x}=\frac{6}{20}=\frac{3}{10}$

We used the following rule to find the derivative:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

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Calculus 2 : New Concepts

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #61 : New Concepts

Example Question #62 : New Concepts

Example Question #63 : New Concepts

Example Question #64 : New Concepts

Example Question #65 : New Concepts

Example Question #66 : New Concepts

Example Question #67 : New Concepts

Example Question #68 : New Concepts

Example Question #69 : New Concepts

Example Question #1711 : Calculus Ii

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