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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 #91 : New Concepts

Evaluate the limit:

$\lim_{x \rightarrow -3}\frac{x+3}{\ln(x+4)}$

Possible Answers:

$\frac{3}{4}$

$-\infty$

$\infty$

Correct answer:

Explanation:

If we evaluate the expression with the limit of x = -3, it returns the indeterminate form $\frac{0}{0}$ .

We can instead use L’Hospital’s Rule to evaluate, using the form:

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

Where,

$\\ f(x) = x+3 \\ g(x)=\ln(x+4)$

Therefore,

$\\f'(x)=1 \\ g'(x)=\frac{1}{x+4}$

If we rewrite the limit with L'Hospital's Rule,

$\lim_{x \to -3}\frac{1}{\frac{1}{x+4}} = \frac{1}{\frac{1}{-3+4}}=\frac{1}{\frac{1}{1}}=1$

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Example Question #92 : New Concepts

Evaluate the limit:

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

Possible Answers:

$\infty$

$-\infty$

Correct answer:

$\infty$

Explanation:

If we evaluate the expression with the limit of $x=\infty$ , it returns the indeterminate form $\frac{\infty}{\infty}$ .

We can instead use L’Hospital’s Rule to evaluate, using the form:

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

Where,

$\\ f(x) = 4^x \\ g(x)=x^2+2x-4$

So,

$\\ f'(x)=4^x\ln4 \\ g'(x)=2x+2$

If we rewrite the limit with L'Hospital's Rule,

$\lim_{x \to \infty}\frac{4^x\ln4}{2x+2} =\frac{\infty}{\infty}$

This is another indeterminate form, so we simply go through L'Hospital's Rule a second time.

$\lim_{x \to \infty}\frac{\ln4\cdot 4^x\cdot \ln4}{2} = \infty$

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Example Question #93 : New Concepts

Evaluate the limit:

$\lim_{x \to \infty}\frac{\ln x}{\sqrt{x}}$

Possible Answers:

$\infty$

$-\infty$

Correct answer:

Explanation:

If we evaluate the expression with the limit of $x=\infty$ , it returns the indeterminate form $\frac{\infty}{\infty}$ .

We can instead use L’Hospital’s Rule to evaluate, using the form:

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

Where,

$\\ f(x) = \ln x \\ g(x)=\sqrt{x}$

Therefore,

$\\ f'(x)= \frac{1}{x} \\\\ g'(x)=\frac{1}{2} \cdot x^{-\frac{1}{2}}$

If we rewrite the limit with L'Hospital's Rule,

$\lim_{x \to \infty}\frac{\frac{1}{x}}{\frac{1}{2} \cdot x^{-\frac{1}{2}}} =\frac{0}{0}$

This is another indeterminate form, and since $x\neq0$ , we can multiple the fraction by $\frac{x}{x}$ .

$\lim_{x \to \infty}\frac{\frac{1}{x}\cdot x}{\frac{1}{2} \cdot x^{-\frac{1}{2}}\cdot x} =\lim_{x \to \infty}\frac{1}{\frac{1}{2} \cdot x^{\frac{1}{2}}}=0$

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Example Question #94 : New Concepts

Use L'Hospital's rule to find $\lim_{x\rightarrow 0}\frac{x^2}{cosx-1}$ .

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.

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

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

$\lim_{x\rightarrow 0}\frac{2x}{-sin(x)}=\frac{0}{0}$

This means we can use L'Hospital's rule again. Taking the derivative of the top and bottom of the fraction a second time gives us

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

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Example Question #95 : New Concepts

Evaluate the limit:

$\lim_{x \to 0} \frac{\sin x}{x}$

Possible Answers:

$\infty$

$\pi$

$-\infty$

Correct answer:

Explanation:

If we evaluate the expression with the limit of $x=\infty$ , it returns the indeterminate form $\frac{\infty}{\infty}$ .

We can instead use L’Hospital’s Rule to evaluate, using the form:

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

Where,

$\\ f(x) = \sin x \\ g(x)=x$

Therefore,

$\\f'(x)=\cos x \\ g'(x)=1$

If we rewrite the limit with L'Hospital's Rule,

$\lim_{x \to 0}{\frac{\cos x}{1}} = \frac{1}{1}=1$

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Example Question #96 : New Concepts

Evaluate the limit:

$\lim_{x \to \infty}(2x-\ln 4x)$

Possible Answers:

$\infty$

$-\infty$

Correct answer:

Explanation:

If we evaluate the expression with the limit of $x=\infty$ , it returns the indeterminate form $\infty-\infty$ .

We can instead use L’Hospital’s Rule to evaluate, using the form:

$\lim_{x \to a} (f(x)-g(x)) = \lim_{x \to a} (f'(x)-g'(x))$

Where,

$\\ f(x) = 2x \\ g(x)=\ln4x$

So,

$\\ f'(x)=2 \\ g'(x)=\frac{1}{4x}\cdot 4=\frac{1}{x}$

If we rewrite the limit with L'Hospital's Rule,

$\lim_{x \to \infty} \left( 2-\frac{1}{x} \right ) = 2-0 = 2$

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Example Question #97 : New Concepts

Use L'Hospital's rule to find $\lim_{x\rightarrow \infty }\frac{3x+7}{x-4}$ .

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.

$\lim_{x\rightarrow \infty }\frac{3x+7}{x-4}=\frac{\infty}{\infty}$

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

$\lim_{x\rightarrow \infty }\frac{3}{1}=3$

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Example Question #98 : New Concepts

Evaluate the limit:

$\lim_{x \to \infty}x^\frac{1}{x}$

Possible Answers:

$-\infty$

$\infty$

Correct answer:

Explanation:

If we evaluate the expression with the limit of $x =\infty$ , it returns the indeterminate form $\infty^0$ .

$\\ \textup{Let}\ y = x^\frac{1}{x} \\ \textup{then} \ln y = \ln x^\frac{1}{x} = \frac{1}{x}\cdot \ln x = \frac{\ln x}{x} \\ \ln y = \frac{\ln x}{x} \\ y = e^{\frac{\ln x}{x}}$

$\lim_{x \to \infty} \frac{\ln x}{x} = \frac{\infty}{\infty}$

This also returns an indeterminate form of $\frac{\infty}{\infty}$ .

We can instead use L’Hospital’s Rule to evaluate, using the form:

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

Where,

$\\ f(x) =\ln x \\ g(x)=x$

So,

$\\f'(x)= x^{-1} \\ g'(x)=1$

If we rewrite the limit with L'Hospital's Rule,

$\\ \lim_{x \to \infty} \frac{x^{-1}}{1} = \lim_{x \to \infty}\frac{1}{x} = 0 \\ \textup{So,}\ \lim_{x \to \infty} \frac{\ln x}{x} = 0$

Then we can conclude

$\lim_{x \to \infty}x^\frac{1}{x} = \lim_{x \to \infty}e^{\frac{\ln x}{x}} = e^{0} = 1$

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Example Question #99 : New Concepts

Use L'Hospital's rule to find $\lim_{x\rightarrow 0}\frac{e^x-1}{x}$ .

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.

$\lim_{x\rightarrow 0}\frac{e^x-1}{x}=\frac{e^0-1}{0}=\frac{1-1}{0}=\frac{0}{0}$

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

$\lim_{x\rightarrow 0}\frac{e^x}{1}=\frac{e^0}{1}=\frac{1}{1}=1$

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Example Question #81 : L'hospital's Rule

Use L'Hospital's rule to find $\lim_{x\rightarrow 0}\frac{x^2-2x}{sinx}$ .

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.

$\lim_{x\rightarrow 0}\frac{x^2-2x}{sinx}=\frac{0^2-2(0)}{sin(0)}=\frac{0}{0}$

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

$\lim_{x\rightarrow 0}\frac{2x-2}{cosx}=\frac{2(0)-2}{cos(0)}=\frac{-2}{1}=-2$

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Calculus 2 : New Concepts

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #91 : New Concepts

Example Question #92 : New Concepts

Example Question #93 : New Concepts

Example Question #94 : New Concepts

Example Question #95 : New Concepts

Example Question #96 : New Concepts

Example Question #97 : New Concepts

Example Question #98 : New Concepts

Example Question #99 : New Concepts

Example Question #81 : L'hospital's Rule

All Calculus 2 Resources

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