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Calculus 2 : New Concepts

Study concepts, example questions & explanations for Calculus 2

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

Example Questions

Example Question #611 : Derivatives

Use L'Hospital's rule to find $\lim_{x\rightarrow 3}\frac{x^2-9}{x-3}$ .

Possible Answers:

$\infty$

Correct answer:

Explanation:

L'Hospital's rule state that if $\lim\frac{f(x)}{g(x)}=$ , $\frac{\infty }{\infty }$ , $\frac{-\infty }{-\infty }$ or $\frac{0}{0 }$ , then $\lim\frac{f(x)}{g(x)}=\lim\frac{f'(x)}{g'(x)}$

To solve this problem, we must first see if L'Hospital's rule applies, by substitution.

$\lim_{x\rightarrow 3}\frac{x^2-9}{x-3}=\frac{3^2-9}{3-3}=\frac{9-9}{3-3}=\frac{0}{0}$

Since $\lim\frac{f(x)}{g(x)}=$ $\frac{0}{0 }$ , we can use L'Hospital's rule. Take the derivative of the top and bottom of the fraction, gives us

$\lim_{x\rightarrow 3}\frac{2x}{1}=\frac{2(3)}{1}=6$

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Example Question #612 : Derivatives

Use L'Hospital's rule to find $\lim_{x\rightarrow 0}\frac{x}{sinx}$ .

Possible Answers:

$\infty$

Correct answer:

Explanation:

L'Hospital's rule state that if $\lim\frac{f(x)}{g(x)}=$ , $\frac{\infty }{\infty }$ , $\frac{-\infty }{-\infty }$ or $\frac{0}{0 }$ , then $\lim\frac{f(x)}{g(x)}=\lim\frac{f'(x)}{g'(x)}$

To solve this problem, we must first see if L'Hospital's rule applies, by substitution.

$\lim_{x\rightarrow 0}\frac{x}{sinx}=\frac{0}{sin(0)}=\frac{0}{0}$

Since $\lim\frac{f(x)}{g(x)}=$ $\frac{0}{0 }$ , we can use L'Hospital's rule. Take the derivative of the top and bottom of the fraction, gives us

$\lim_{x\rightarrow 0}\frac{1}{cosx}=\frac{1}{cos(0)}=\frac{1}{1}=1$

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Example Question #619 : Derivatives

Use L'Hospital's rule to find $\lim_{x\rightarrow \infty }\frac{x}{sinx}$ .

Possible Answers:

$\infty$

Correct answer:

Explanation:

L'Hospital's rule state that if $\lim\frac{f(x)}{g(x)}=$ , $\frac{\infty }{\infty }$ , $\frac{-\infty }{-\infty }$ or $\frac{0}{0 }$ , then $\lim\frac{f(x)}{g(x)}=\lim\frac{f'(x)}{g'(x)}$

To solve this problem, we must first see if L'Hospital's rule applies, by substitution.

$\lim_{x\rightarrow \infty }\frac{x}{sinx}=\frac{\infty}{sin\infty}=\frac{\infty}{\infty}$

Since $\lim\frac{f(x)}{g(x)}=$ $\frac{\infty }{\infty }$ , we can use L'Hospital's rule. Take the derivative of the top and bottom of the fraction, gives us

$\lim_{x\rightarrow \infty }\frac{1}{cosx}=\frac{1}{cos\infty}=\frac{1}{\infty}=0$

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Example Question #81 : L'hospital's Rule

Use L'Hospital's rule to find $\lim_{x\rightarrow \infty}\frac{e^x}{5x-3}$ .

Possible Answers:

$\infty$

Correct answer:

$\infty$

Explanation:

L'Hospital's rule state that if $\lim\frac{f(x)}{g(x)}=$ , $\frac{\infty }{\infty }$ , $\frac{-\infty }{-\infty }$ or $\frac{0}{0 }$ , then $\lim\frac{f(x)}{g(x)}=\lim\frac{f'(x)}{g'(x)}$

To solve this problem, we must first see if L'Hospital's rule applies, by substitution.

$\lim_{x\rightarrow \infty}\frac{e^x}{5x-3}=\frac{e^\infty}{5(\infty)-3}=\frac{\infty}{\infty}$

Since $\lim\frac{f(x)}{g(x)}=$ $\frac{\infty }{\infty }$ , we can use L'Hospital's rule. Take the derivative of the top and bottom of the fraction, gives us

$\lim_{x\rightarrow \infty}\frac{e^x}{5}=\frac{e^\infty}{5}=\frac{\infty}{5}=\infty$

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Example Question #101 : New Concepts

Find the following limit: $\lim_{x\to\infty}\frac{x^2}{e^{2x}}$

Possible Answers:

$\infty$

There is no such limit

Correct answer:

Explanation:

This problem is solved by utilizing the L'Hospital's rule:

$\lim_{x\to\infty} \frac{x^2}{e^{2x}}=\lim_{x\to\infty} \frac{\frac{d}{dx}(x^2)}{\frac{d}{dx}(e^{2x})}= \lim_{x\to\infty} \frac{2x}{2e^{2x}}=\lim_{x\to\infty} \frac{\frac{d}{dx}(2x)}{\frac{d}{dx}(2e^{2x})}= \lim_{x\to\infty} \frac{1}{2e^{2x}}=0$

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Example Question #83 : L'hospital's Rule

Use L'Hospital's rule to find $\lim_{x\rightarrow \infty}\frac{x}{lnx}$ .

Possible Answers:

$\infty$

Correct answer:

$\infty$

Explanation:

L'Hospital's rule state that if $\lim\frac{f(x)}{g(x)}=$ , $\frac{\infty }{\infty }$ , $\frac{-\infty }{-\infty }$ or $\frac{0}{0 }$ , then $\lim\frac{f(x)}{g(x)}=\lim\frac{f'(x)}{g'(x)}$

To solve this problem, we must first see if L'Hospital's rule applies, by substitution.

$\lim_{x\rightarrow \infty}\frac{x}{lnx}=\frac{\infty}{ln(\infty)}=\frac{\infty}{\infty}$

Since $\lim\frac{f(x)}{g(x)}=$ $\frac{\infty }{\infty }$ , we can use L'Hospital's rule. Take the derivative of the top and bottom of the fraction, gives us

$\lim_{x\rightarrow \infty}\frac{1}{1/x}=\frac{1}{1/\infty}=\frac{1}{0}=\infty$

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Example Question #102 : New Concepts

Use L'Hospital's rule to find $\lim_{x\rightarrow -3}\frac{sin(x^2-9)}{x+3}$ .

Possible Answers:

$\infty$

Correct answer:

Explanation:

L'Hospital's rule state that if $\lim\frac{f(x)}{g(x)}=$ , $\frac{\infty }{\infty }$ , $\frac{-\infty }{-\infty }$ or $\frac{0}{0 }$ , then $\lim\frac{f(x)}{g(x)}=\lim\frac{f'(x)}{g'(x)}$

To solve this problem, we must first see if L'Hospital's rule applies, by substitution.

$\lim_{x\rightarrow -3}\frac{sin(x^2-9)}{x+3}=\frac{sin((-3)^2-9)}{-3+3}=\frac{sin(9-9)}{-3+3}$

$=\frac{sin(0)}{0}=\frac{0}{0}$

Since $\lim\frac{f(x)}{g(x)}=$ $\frac{0}{0 }$ , we can use L'Hospital's rule. Take the derivative of the top and bottom of the fraction, gives us

$\lim_{x\rightarrow -3}\frac{(2x)cos(x^2-9)}{1}=\frac{(2(-3))cos((-3)^2-9)}{1}=\frac{-6cos(9-9)}{1}$

$=\frac{-6cos(0)}{1}=\frac{-6(1)}{1}=-6$

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Example Question #85 : L'hospital's Rule

Use L'Hospital's rule to find $\lim_{x\rightarrow \infty}\frac{x^2-x}{x+1}$ .

Possible Answers:

$\infty$

Correct answer:

$\infty$

Explanation:

L'Hospital's rule state that if $\lim\frac{f(x)}{g(x)}=$ , $\frac{\infty }{\infty }$ , $\frac{-\infty }{-\infty }$ or $\frac{0}{0 }$ , then $\lim\frac{f(x)}{g(x)}=\lim\frac{f'(x)}{g'(x)}$

To solve this problem, we must first see if L'Hospital's rule applies, by substitution.

$\lim_{x\rightarrow \infty}\frac{x^2-x}{x+1}=\frac{\infty^2-\infty}{\infty+1}=\frac{\infty}{\infty}$

Since $\lim\frac{f(x)}{g(x)}=$ $\frac{\infty }{\infty }$ , we can use L'Hospital's rule. Take the derivative of the top and bottom of the fraction, gives us

$\lim_{x\rightarrow \infty}\frac{2x-1}{1}=\frac{2*\infty-1}{1}=\infty$

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Example Question #86 : L'hospital's Rule

Use L'Hospital's rule to find $\lim_{x\rightarrow 4}\frac{ln(x-3)}{x^2-16}$ .

Possible Answers:

$\frac{1}{4}$

$\frac{1}{8}$

$\infty$

Correct answer:

$\frac{1}{8}$

Explanation:

L'Hospital's rule state that if $\lim\frac{f(x)}{g(x)}=$ , $\frac{\infty }{\infty }$ , $\frac{-\infty }{-\infty }$ or $\frac{0}{0 }$ , then $\lim\frac{f(x)}{g(x)}=\lim\frac{f'(x)}{g'(x)}$

To solve this problem, we must first see if L'Hospital's rule applies, by substitution.

$\lim_{x\rightarrow 4}\frac{ln(x-3)}{x^2-16}=\frac{ln(4-3)}{4^2-16}=\frac{ln(1)}{16-16}=\frac{0}{0}$

Since $\lim\frac{f(x)}{g(x)}=$ $\frac{0}{0 }$ , we can use L'Hospital's rule. Take the derivative of the top and bottom of the fraction, gives us

$\lim_{x\rightarrow 4}\frac{1/(x-3)}{2x}=\frac{1/(4-3)}{2*4}=\frac{1/(1)}{8}=\frac{1}{8}$

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Calculus 2 : New Concepts

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #611 : Derivatives

Example Question #612 : Derivatives

Example Question #619 : Derivatives

Example Question #81 : L'hospital's Rule

Example Question #101 : New Concepts

Example Question #83 : L'hospital's Rule

Example Question #102 : New Concepts

Example Question #85 : L'hospital's Rule

Example Question #86 : L'hospital's Rule

All Calculus 2 Resources

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