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Calculus 2 : Derivatives

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Example Questions

Example Question #44 : New Concepts

Evaluate the following limit:

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

Possible Answers:

$-\frac{1}{6}$

The limit does not exist

$\infty$

$\frac{1}{6}$

Correct answer:

$-\frac{1}{6}$

Explanation:

When we evaluate the limit using normal methods, we arrive at the indeterminate form $\frac{0}{0}$ . When this occurs, to evaluate the limit, we must use L'Hopital's Rule, which states that

$\lim_{x\rightarrow n}\frac{ f(x)}{g(x)}=\lim_{x\rightarrow n} \frac{f'(x)}{g'(x)}$

So, we must find the derivative of the top and bottom functions:

f'(x)=-1, g'(x)=2x

The derivatives were found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Now, rewrite the limit and evaluate it:

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

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Example Question #23 : L'hospital's Rule

Evaluate the following limit:

$\lim_{x\rightarrow 2}\frac{x^2-4}{\sin(2-x)}$

Possible Answers:

$\infty$

Correct answer:

Explanation:

When we evaluate the limit using normal methods, we get the indeterminate form $\frac{0}{0}$ . When this happens, we must use L'Hopital's Rule, which states that $\lim_{x\rightarrow n}\frac{f(x)}{g(x)}=\lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}$

Now, we must find the derivatives of the numerator and denominator:

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

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)$

Next, rewrite the limit and evaluate it:

$\lim_{x\rightarrow 2}\frac{2x}{-\cos(2-x)}=-4$

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Example Question #24 : L'hospital's Rule

Use l'Hopital's rule to find the limit:

$\lim_{x \to 1} \frac{x^2-3x+2}{x-1}$

Possible Answers:

$\infty$

Correct answer:

Explanation:

The first thing we always have to do is to check that l'Hopital's rule is actually applicable when we want to use it.

$\frac{1^2-3\cdot1+2}{1-1} = \frac{0}{0}$

So it is applicable here.

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

$\lim_{x \to 1} \frac{x^2-3x+2}{x-1} = \lim_{x \to 1} \frac{2x-3}{1}$

and now we can safely plug in x=1 and get that the limit equals

$2\cdot 1 -3 = -1$ .

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Example Question #561 : Derivatives

Evaluate $\lim_{x \to 0^+} x\ln(x)$

Possible Answers:

Does not exist

$-\infty$

None of the other answers

Correct answer:

Explanation:

Plugging into the function $x\ln(x)$ head on yields the inteterminate form of zero times negative infinity, so we must rewrite the problem

$\lim_{x \to 0^+} x\ln(x)$ . Start

$=\lim_{x \to 0^+} \frac{\ln(x)}{\frac{1}{x}}$

In this expression, when approaches from the positive side, the limit "approaches $-\frac{\infty}{\infty}$ " So we can use L'Hospital's rule.

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

$=\lim_{x \to 0}-x$

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Example Question #562 : Derivatives

Evaluate the following limit:

$\lim_{x\rightarrow 10} \frac{\cos(\frac{\pi x}{4})}{x-10}$

Possible Answers:

$\frac{\pi}{4}$

The limit does not exist

$-\frac{\pi}{4}$

Correct answer:

$-\frac{\pi}{4}$

Explanation:

When we evaluate the limit using normal methods, we get the indeterminate form $\frac{0}{0}$ .

So, we must use L'Hopital's Rule to evaluate the limit, which states that

$\lim_{x\rightarrow n} \frac{f(x)}{g(x)}=\lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}$

Using the above, we get

$\lim_{x\rightarrow 10 } \frac{-\sin(\frac{\pi x}{4})(\frac{\pi}{4})}{1}=-\frac{\pi}{4}$

when we evaluate the limit using substitution.

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Example Question #31 : L'hospital's Rule

$\lim_{x\rightarrow 0} \; \frac{sin(4x)}{2x}$

Possible Answers:

$\frac{1}{2}$

$\frac{1}{4}$

Correct answer:

Explanation:

If you plug in the limit value, the function turns into $\frac{0}{0}$ . Therefore, we are allowed to use l'Hospital's Rule. We start by taking the derivative of both the numerator and the denominator until when you plug in the value of the limit, you do not get something in the form $\frac{0}{0}$ . Luckily, in this case, we only need to take the derivative once. By taking the derivatives separately, we get a new limit:

$\lim_{x\rightarrow 0} \frac{4cos(4x)}{2}=\frac{4cos(0)}{2}=\frac{4}{2}=2$ .

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Example Question #561 : Derivatives

Evaluate: $lim_{x->\infty} \frac{e^x}{x^2}$

Possible Answers:

e^x

$\infty$

Limit Does Not Exist

Correct answer:

$\infty$

Explanation:

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Example Question #33 : L'hospital's Rule

Find the limit using L'Hospital's Rule.

$\lim_{x \to \infty}xe^{-x}$

Possible Answers:

DNE

$\infty$

Correct answer:

Explanation:

We rewrite the limit as

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

Substituting $x=\infty$ yields the indeterminate form

$\frac{\infty}{\infty}$

L'Hospital's Rule states that when the limit is in indeterminate form, the limit becomes

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

For f(x)=x and g(x)=e^x we solve the limit

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

and substituting $x=\infty$ we find that

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

As such

$\lim_{x \to \infty}xe^{-x}=0$

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Example Question #34 : L'hospital's Rule

Find the limit using L'Hospital's Rule.

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

Possible Answers:

DNE

$\infty$

Correct answer:

Explanation:

Substituting $x=\infty$ yields the indeterminate form

$\frac{\infty}{\infty}$

L'Hospital's Rule states that when the limit is in indeterminate form, the limit becomes

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

For $f(x)=ln\,x$ and g(x)=x we solve the limit

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

and substituting $x=\infty$ we find that

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

As such

$\lim_{x \to \infty}\frac{ln\,x}{x}=0$

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Example Question #35 : L'hospital's Rule

Evaluate the following limit:

$\lim_{x \to 0} \frac{6^x-8^x}{x}$

Possible Answers:

$ln(\frac{3}{4})$

ln(12)

$\infty$

Correct answer:

$ln(\frac{3}{4})$

Explanation:

Simply substituting x=0 in the given limit will not work:

$\lim_{x \to 0} \frac{6^x-8^x}{x} = \frac{6^0-8^0}{0}$

$= \frac{1-1}{0}$

$= \frac{0}{0}$

Because direct substitution yields an indeterminate result, we must apply L'Hospital's rule to the limit:

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

if and only if $g'(x)\neq 0$ and both f'(x) and g'(x) exist at .

Here,

f'(x) = ln(6)6^x-ln(8)8^x, and g'(x)=1 .

Hence,

$\lim_{x \to 0}\frac{6^x-8^x}{x}=\lim_{x \to 0}\frac{ln(6)6^x-ln(8)8^x}{1}$

$= \lim_{x \to 0}(ln(6)6^x-ln(8)8^x)$

= ln(6)6^0-ln(8)8^0

= ln(6)-ln(8)

$= ln(\frac{3}{4})$

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Calculus 2 : Derivatives

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #44 : New Concepts

Example Question #23 : L'hospital's Rule

Example Question #24 : L'hospital's Rule

Example Question #561 : Derivatives

Example Question #562 : Derivatives

Example Question #31 : L'hospital's Rule

Example Question #561 : Derivatives

Evaluate: $lim_{x->\infty} \frac{e^x}{x^2}$

Example Question #33 : L'hospital's Rule

Example Question #34 : L'hospital's Rule

Example Question #35 : L'hospital's Rule

All Calculus 2 Resources

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