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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 #71 : New Concepts

Evaluate:

$\lim_{x\rightarrow 0}\frac{ \sin(x)}{x^3+x^2+x}$

Possible Answers:

$\infty$

$-\infty$

The limit does not exist

Correct answer:

Explanation:

When evaluating the limit using normal methods, we receive the indeterminate form $\frac{0}{0}$ . When this occurs, we must use L'Hopital's Rule to solve 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, we get

$\lim_{x\rightarrow 0}\frac{\cos(x)}{3x^2+2x+1}=1$

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Example Question #1713 : Calculus Ii

Evaluate the limit:

$\lim_{x\rightarrow 2^+}\frac{\ln\left | x-2\right |}{\ln\left | x^2-4\right |}$

Possible Answers:

$\infty$

$-\infty$

The limit does not exist

Correct answer:

Explanation:

When evaluating the limit using normal methods, we get the indeterminate form $\frac{\infty}{\infty}$ . 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

$\lim_{x\rightarrow 2^+}\frac{\frac{1}{x-2}}{\frac{2x}{x^2-4}}=\lim_{x\rightarrow 2^+}(\frac{1}{(x-2)}\cdot \frac{(x+2)(x-2)}{2x})=1$

We used the following rules to find the derivatives:

$\frac{\mathrm{d} }{\mathrm{d} x} \ln(u)=\frac{1}{u}\frac{\mathrm{d} u}{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

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Example Question #1714 : Calculus Ii

Evaluate $\lim_{n\rightarrow \infty }\frac{n^3+2n}{2n^2+5}$

Possible Answers:

$\infty$

$\frac{3}{2}$

Correct answer:

$\infty$

Explanation:

Evaluating the limit to begin with gets us $\frac{\infty}{\infty}$ , which is undefined. We can solve this problem using L'Hospital's rule. Taking the derivative of the numerator and denominator with respect to n, we get $\frac{3n^2+2}{4n}$ . The limit is still undefined. Another application of the rule gets us $\frac{6n}{4}$ , which evaluated at $n=\infty$ is in fact $\infty$ .

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Example Question #71 : New Concepts

Evaluate $\lim_{n\rightarrow \infty }\frac{2n^2+3n}{5n^2+n+1}$

Possible Answers:

$\frac{2}{5}$

Correct answer:

$\frac{2}{5}$

Explanation:

In evaluating the limit to begin with, you get $\frac{\infty }{\infty}$ , which is undefined. Applying L'Hospitals Rule, we take the derivative of both the numerator and denominator with respect to n. The first derivative gets us $\frac{4n+3}{10n+1}$ , which is still improper. Another application of the rule will get us $\frac{2}{5}$ , the correct solution.

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Example Question #1716 : Calculus Ii

Evaluate $\lim_{x\rightarrow \infty}\frac{2x^2}{7x^3}$

Possible Answers:

$\infty$

$\frac{4}{42}$

Correct answer:

Explanation:

Evaluating the limit to begin with, we get $\frac{\infty}{\infty}$ , which is undefined. Using L'Hospital's rule to solve, we take the derivative of the numerator and denominator of the expression. In doing so, we get $\frac{4x}{21x^2}$ . Evaluating the new limit, we still get $\frac{\infty}{\infty}$ . Another application of L'Hospitals rule gives us $\frac{4}{42x}$ . We can now solve the limit, which is $\frac{4}{\infty}=0$

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Example Question #72 : New Concepts

Find the limit

$\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$

Possible Answers:

e^2

$\ln e$

Correct answer:

Explanation:

Untitled

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Example Question #51 : L'hospital's Rule

Use L'Hospital's rule to evaluate

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

Possible Answers:

$\infty$

The limit does not exist.

Correct answer:

Explanation:

To use L'hospital's rule, evaluate the limit of the numerator of the fraction and the denominator separately. If the result is $\infty / \infty$ , $-\infty / \infty$ , or 0/0 , take the derivative of the numerator and the denominator separately, and try to evaluate the limit again.

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

$= \lim_{x \to \infty}\frac{6x^2}{4x^3+10x^4}$ (L'hospital's rule) $\longrightarrow \frac{\infty}{\infty}$

$= \lim_{x \to \infty} \frac{12x}{12x^2+40x^3}$ (L'hospital's rule again) $\longrightarrow \frac{\infty}{\infty}$

$= \lim_{x \to \infty}\frac{12}{24x+120x^2}$ (L'hospital's rule again) $\longrightarrow \frac{12}{\infty}$

= 0

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Example Question #1719 : Calculus Ii

Use L'Hospital's Rule to evaluate

$\lim_{x \to 0} \frac{\tan(t^2)}{t^2}$

Possible Answers:

The limit does not exist.

$\pi/3$

$\pi/2$

Correct answer:

Explanation:

$\lim_{x \to 0} \frac{\tan(x^2)}{x^2}$ $\longrightarrow \frac{0}{0}$

$=\lim_{x \to 0} \frac{2x\sec^{2}(x^2)}{2x}$ (L'hospital's Rule) $\longrightarrow \frac{0}{0}$

$=\lim_{x \to 0} \frac{(2x)(2\sec(x^2)(\sec(x^2)\tan(x^2))+\sec^2(x^2)(2)}{2}$ (L'hospital's Rule. Here the derivative of the numerator involves the Product Rule)

$= \frac{(0)(0)+(1)(2)}{2}=1$

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Example Question #1720 : Calculus Ii

Use L'hospital's Rule to evaluate

$\lim_{x \to 0} \frac{2^{-x}-1}{3^{-x}-1}$ .

Possible Answers:

None of the other answers

1/2

$\infty$

Correct answer:

None of the other answers

Explanation:

$\lim_{x \to 0} \frac{2^{-x}-1}{3^{-x}-1}$ $\longrightarrow \frac{2^0-1}{3^0-1}\longrightarrow \frac{0}{0}$

$=\lim_{x \to 0} \frac{-\ln(2)(2^{-x})-0}{-\ln(3)(3^{-x})-0}$ (L'hospital's Rule)

$=\frac{\ln(2)}{\ln(3)}$ .

So the answer is $\frac{\ln(2)}{\ln(3)}$ .

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Example Question #71 : New Concepts

$\lim_{x\rightarrow a} \frac{x^3-a^3}{x-a}$

Possible Answers:

2a^2

a^2

$\infty$

3a^2

Correct answer:

3a^2

Explanation:

The first step to computing a limit is direct substitution:

$\lim_{x\rightarrow a} \frac{x^3-a^3}{x-a}=\frac{a^3-a^3}{a-a}=\frac{0}{0}.$

Now, we see that this is in the form of L'Hopital's Rule. For those problems, we take a derivative of both the numerator and denominator separately. Remember, is simply a constant!

$\frac{(x^3-a^3)'}{(x-a)'}=\frac{3x^2}{1}.$

Now, we can go back and plug in the original limit value:

$\lim_{x\rightarrow a}\frac{3x^2}{1}=3a^2.$

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Calculus 2 : New Concepts

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #71 : New Concepts

Example Question #1713 : Calculus Ii

Example Question #1714 : Calculus Ii

Example Question #71 : New Concepts

Example Question #1716 : Calculus Ii

Example Question #72 : New Concepts

Example Question #51 : L'hospital's Rule

Example Question #1719 : Calculus Ii

Example Question #1720 : Calculus Ii

Example Question #71 : New Concepts

All Calculus 2 Resources

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