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Calculus 2 : Derivatives

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Example Questions

Example Question #72 : New Concepts

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 #53 : L'hospital's Rule

Find the limit

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

Possible Answers:

$\ln e$

e^2

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 #53 : L'hospital's Rule

Use L'Hospital's Rule to evaluate

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

Possible Answers:

$\pi/2$

$\pi/3$

The limit does not exist.

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 #71 : New Concepts

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 #72 : New Concepts

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

Possible Answers:

2a^2

3a^2

a^2

$\infty$

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 #592 : Derivatives

Evaluate the following limit by L'hospital's Rule

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

Possible Answers:

$\infty$

Undefined

$-\frac{1}{2}$

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 #591 : Derivatives

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 #594 : Derivatives

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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Calculus 2 : Derivatives

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #72 : New Concepts

Example Question #53 : L'hospital's Rule

Example Question #51 : L'hospital's Rule

Example Question #53 : L'hospital's Rule

Example Question #71 : New Concepts

Example Question #72 : New Concepts

Example Question #81 : New Concepts

Example Question #592 : Derivatives

Example Question #591 : Derivatives

Example Question #594 : Derivatives

All Calculus 2 Resources

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