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Calculus 2 : Derivatives

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Example Questions

Example Question #65 : L'hospital's Rule

Evaluate the limit:

$\lim_{x \to 2} \left ( \frac{1}{\ln (x-1)} - \frac{1}{(x-2)}\right )$

Possible Answers:

$\frac{1}{2}$

$-\frac{1}{2}$

Correct answer:

$\frac{1}{2}$

Explanation:

If we evaluate the expression with the limit of x=2 , it returns the indeterminate form $\infty-\infty$ . We start by rewriting the expression

$\\ \frac{1}{\ln (x-1)} - \frac{1}{(x-2)}=\frac{1}{\ln (x-1)}\cdot \frac{x-2}{x-2} - \frac{1}{(x-2)}\cdot \frac{\ln(x-1)}{\ln(x-1)}=\frac{x-2-\ln(x-1)}{ln(x-1)\cdot (x-2)} \\ \lim_{x \to 2}\frac{x-2-\ln(x-1)}{ln(x-1)\cdot (x-2)} = \frac{0}{0}$

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

$\\ f(x) = x-2-\ln(x-1)\\ g(x)=ln(x-1)\cdot (x-2)$

So,

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

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

$\lim_{x \to 2}\frac{1 - \frac{1}{x-1}}{\frac{1}{x-1}\cdot (x-2)+\ln(x-2)} = \frac{0}{0}$

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

$\frac{1 - \frac{1}{x-1}}{\frac{1}{x-1}\cdot (x-2)+\ln(x-2)}\cdot \frac{x-1}{x-1}=\frac{x-1-1}{(x-2)-(x-2)\ln(x-2)}$

Where,

$\\f(x) = x-1-1 \\ g(x)=(x-2)-(x-2)\ln(x-2)$

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

$\lim_{x \to 2}\frac{1}{1+\ln(x-1)+\frac{x-1}{x-1}}= \frac{1}{2}$

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Example Question #601 : Derivatives

Evaluate the limit:

$\lim_{x \to -1} \frac{x^2+7x+6}{x^2-6x-7}$

Possible Answers:

$-\frac{5}{8}$

$\infty$

$- \infty$

$-\frac{9}{4}$

Correct answer:

$-\frac{5}{8}$

Explanation:

If we evaluate the expression with the limit of x = -1, 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,

$\\ f(x) = x^2+7x+6 \\ g(x)=x^2-6x-7$

Therefore,

$\\ f'(x)=2x+7 \\ g'(x)=2x-6$

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

$\lim_{x \to -1}\frac{2x+7}{2x-6} = -\frac{5}{8}$

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Example Question #602 : Derivatives

Evaluate the limit:

$\lim_{x \to \frac{2}{3}} \frac{x^2-\frac{11}{3}x+2}{x^2+\frac{25}{3}x-6}$

Possible Answers:

$-\infty$

$\infty$

$-\frac{7}{29}$

$-\frac{11}{25}$

Correct answer:

$-\frac{7}{29}$

Explanation:

If we evaluate the expression with the limit of $x=\frac{2}{3}$ , 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,

$\\ {f(x) =x^2-\frac{11}{3}x+2}\\\\ {g(x)=x^2+\frac{25}{3}x-6}$

Therefore,

$\\ {f'(x)=2x-\frac{11}{3}}\\\\ {}g'(x)=2x+\frac{25}{3}$

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

$\lim_{x \to \frac{2}{3}}\frac{2x-\frac{11}{3}}{2x+\frac{25}{3}} = -\frac{7}{29}$

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Example Question #603 : Derivatives

Evaluate the limit:

$\lim_{x \to 8} \frac{x^2-4x-32}{x^2-3x-40}$

Possible Answers:

$\frac{4}{5}$

$-\infty$

$\infty$

$\frac{12}{13}$

Correct answer:

$\frac{12}{13}$

Explanation:

If we evaluate the expression with the limit of x = 8, 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,

$\\ f(x) = x^2-4x-32 \\ g(x)=x^2-3x-40$

Therefore,

$\\ f'(x)=2x-4 \\ g'(x)=2x-3$

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

$\lim_{x \to 8}\frac{2x-4}{2x-3} = \frac{2\cdot 8-4}{2\cdot 8-3}= \frac{12}{13}$

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Example Question #604 : Derivatives

Evaluate the limit:

$\lim_{x \to \infty} \frac{e^{2x}}{x^{2}}$

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,

$\\ f(x) = e^{2x} \\ g(x)=x^2$

Therefore,

$\\f'(x)=2e^{2x} \\ g'(x)=2x$

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

$\lim_{x \to \infty}\frac{2e^{2x}}{2x} = \frac{\infty}{\infty}$

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

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

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Example Question #605 : Derivatives

Evaluate the limit:

$\lim_{x \to \infty} \frac{5x-7}{7x+1}$

Possible Answers:

$\infty$

$\frac{5}{7}$

$-\infty$

$\frac{7}{5}$

$\frac{-7}{1}$

Correct answer:

$\frac{5}{7}$

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,

$\\ f(x) = 5x-7 \\ g(x)=7x+1$

Therefore,

$\\ f'(x)=5 \\ g'(x)=7$

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

$\lim_{x \to \infty}\frac{5}{7} = \frac{5}{7}$

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Example Question #606 : Derivatives

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 #607 : Derivatives

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 #608 : Derivatives

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 #71 : L'hospital's Rule

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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Calculus 2 : Derivatives

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #65 : L'hospital's Rule

Example Question #601 : Derivatives

Example Question #602 : Derivatives

Example Question #603 : Derivatives

Example Question #604 : Derivatives

Example Question #605 : Derivatives

Example Question #606 : Derivatives

Example Question #607 : Derivatives

Example Question #608 : Derivatives

Example Question #71 : L'hospital's Rule

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