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Calculus 2 : L'Hospital's Rule

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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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Example Question #81 : New Concepts

$\lim_{x\rightarrow a} \frac{\sqrt{x}-\sqrt{a}}{x-a}$

Possible Answers:

$\frac{a}{4}$

$\frac{\sqrt{a}}{2}$

$\frac{\sqrt{a}}{2a}$

$\frac{1}{2a}$

$\frac{1}{2}$

Correct answer:

$\frac{\sqrt{a}}{2a}$

Explanation:

The first step to computing a limit is direct substitution:

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

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!

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

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

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

However, we still have to rationalize the denominator. Therefore:

$\frac{1}{2\sqrt{a}}\frac{\sqrt{a}}{\sqrt{a}}=\frac{\sqrt{a}}{2a}$ .

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Example Question #82 : New Concepts

Evaluate the following limit by L'hospital's Rule

$\lim_{x\rightarrow 0}\frac{x}{2Cos(x)-2e^x}$

Possible Answers:

$-\frac{1}{2}$

Undefined

$\infty$

Correct answer:

$-\frac{1}{2}$

Explanation:

Recall L'hospital's Rule for an indeterminate limit is as follows: $\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}$

Since $\lim_{x\rightarrow 0}\frac{x}{2Cos(x)-2e^x}$ is an indeterminate limit, one must use L'hospital's rule.

$\frac{d}{dx}(\frac{x}{2Cos(x)-2e^x})=\frac{1}{-2Sin(x)-2e^x}$

Therefore the question now becomes,

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

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Example Question #61 : L'hospital's Rule

Solve the limit:

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

Possible Answers:

$-\infty$

$\infty$

Correct answer:

Explanation:

Notice if we try to plug infinity into the limit we get $\frac{\infty }{\infty }$ so we apply L'Hospital's rule.

We then take the derivative of the top and the bottom and get

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

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Example Question #84 : New Concepts

Evaluate $\lim_{x \to \0}\frac{\tan x}{\ln (x+1)}$

Possible Answers:

$\infty$

$-\infty$

Correct answer:

Explanation:

Substituting 0 directly into x gives

$\frac{\tan(0)}{\ln(0+1)}\Rightarrow \frac{0}{\ln(1)}\Rightarrow \frac{0}{0}$ ,

which is an indeterminant form that allows the use of L'hospital's rule. Applying L'hospital's rule, we get

$\lim_{x \to \0}\frac{\frac{\mathrm{d} }{\mathrm{d} x} (\tan x)}{\frac{\mathrm{d} }{\mathrm{d} x}(\ln(x+1))}\Rightarrow \lim_{x\to \0}\frac{\sec^{2} x}{\frac{1}{x}}\Rightarrow \lim_{x \to \0}x\sec^{2}x$

Using the the pythagorean trig. identity, $\sec ^{2}x=1+\tan^{2}x$ , we rewrite the limit as

$\lim_{x \to \0}x(1+\tan^{2}x)$

Now we plug 0 in for x.

$0(1+0) \Rightarrow 0$ , giving us the answer.

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Example Question #85 : New Concepts

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 #86 : New Concepts

Evaluate the limit:

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

Possible Answers:

$- \infty$

$-\frac{9}{4}$

$\infty$

$-\frac{5}{8}$

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 #87 : New Concepts

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:

$-\frac{11}{25}$

$-\frac{7}{29}$

$\infty$

$-\infty$

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 #88 : New Concepts

Evaluate the limit:

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

Possible Answers:

$\frac{12}{13}$

$-\infty$

$\infty$

$\frac{4}{5}$

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 #89 : New Concepts

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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Calculus 2 : L'Hospital's Rule

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #71 : New Concepts

Example Question #81 : New Concepts

Example Question #82 : New Concepts

Example Question #61 : L'hospital's Rule

Example Question #84 : New Concepts

Example Question #85 : New Concepts

Example Question #86 : New Concepts

Example Question #87 : New Concepts

Example Question #88 : New Concepts

Example Question #89 : New Concepts

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