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AP Calculus BC : AP Calculus BC

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2 Diagnostic Tests 86 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #22 : L'hospital's Rule

Find the limit if it exists

$\lim_{x\rightarrow\infty }\frac{x^2+3x}{2+x^2}$

Hint: Use L'Hospital's rule

Possible Answers:

$\frac{3}{2}$

$\frac{1}{2}$

Correct answer:

Explanation:

Directly evaluating for $x=\infty$ yields the indeterminate form

$\frac{\infty ^2+3(\infty )}{2+ \infty}=\frac{\infty}{\infty}$

we are able to apply L'Hospital's rule which states that if the limit is in indeterminate form when evaluated, then

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

As such the limit in the problem becomes

$\lim_{x\rightarrow \infty}\frac{x^2+3x}{2+x^2}=\lim_{x\rightarrow \infty}\frac{\frac{\mathrm{d} }{\mathrm{d} x}[x^2+3x]}{\frac{\mathrm{d} }{\mathrm{d} x}[2+x^2]}=\lim_{x\rightarrow \infty}\frac{2x+3}{2x}$

$\lim_{x\rightarrow \infty}\frac{2x+3}{2x} = \lim_{x\rightarrow \infty}\frac{2x}{2x} + \frac{3}{2x}= \lim_{x\rightarrow \infty}1 + \frac{3}{2x}$

Evaluating for $x=\infty$ yields

$1+\frac{3}{2(\infty)} = 1+0 =1$

As such

$\lim_{x\rightarrow \infty}1 + \frac{3}{2x}=1$

and thus

$\lim_{x\rightarrow\infty }\frac{x^2+3x}{2+x^2}=1$

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Example Question #482 : Ap Calculus Bc

Evaluate $\lim_{x \to 0} \frac{\sin(x)}{x}$ using L'hopital's rule.

Possible Answers:

$\pi/3$

$\pi$

$\infty$

Correct answer:

Explanation:

This important limit from elementary limit theory is usually proven using trigonometric arguments, but it can be shown using L'hopital's rule too.

$\lim_{x \to 0} \frac{\sin(x)}{x} \longrightarrow \frac{0}{0}$

$=\lim_{x \to 0}\frac{(\sin(x))'}{(x)'} = \lim_{x \to 0}\frac{\cos(x)}{1} = \frac{1}{1} =1.$

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Example Question #483 : Ap Calculus Bc

Evaluate the limit:

$\lim_{x\rightarrow 3}\frac{x-3}{\sin(x-3)}$

Possible Answers:

$\infty$

The limit does not exist

Correct answer:

Explanation:

When evaluating the limit using normal methods, we find that the indeterminate form $\frac{0}{0}$ results. When this occurs, we must use L'Hopital's Rule, which states that for $\lim_{x\rightarrow a} \frac{f(x)}{g(x)}=\lim_{x\rightarrow a} \frac{f'(x)}{g'(x)}$ .

Taking the derivative of the top and bottom functions and evaluating the limit, we get

$\lim_{x\rightarrow 3}\frac{1}{\cos(x-3)}=1$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)$

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Example Question #213 : Derivatives

Evaluate the limit:

$\lim_{x\rightarrow 2}\frac{\ln(x-1)}{x-2}$

Possible Answers:

The limit does not exist

$-\infty$

Correct answer:

Explanation:

When evaluating the limit using normal methods, we find that we get the indeterminate form $\frac{0}{0}$ . When this occurs, we must use L'Hopital's Rule to evaluate the limit. The rule states that for the limits for which the indeterminate forms $\frac{0}{0}, \frac{\infty}{\infty}$ result,

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

Taking the derivatives for our limit, we get

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

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \ln(u)=\frac{1}{u}\frac{\mathrm{d} u}{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

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Example Question #482 : Ap Calculus Bc

Evaluate the limit:

$\lim_{x\rightarrow \pi}\frac{(x-\pi)^2}{\cos(\frac{3\pi}{2})}$

Possible Answers:

$-\infty$

$\infty$

The limit does not exist

Correct answer:

Explanation:

When evaluating the limit using normal methods (substitution), we get the indeterminate form $\frac{0}{0}$ . When this (or $\frac{\infty}{\infty}$ ) occurs, we must use L'Hopital's Rule to evaluate the limit. The rule states that when an indeterminate form results from the limit,

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

Taking the derivatives in our limit, we get

$\lim_{x\rightarrow \pi}\frac{2(x-\pi)}{\frac{3}{2}\sin(\frac{3x}{2})}=0$

The derivatives were found using the following rules:

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

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Example Question #481 : Ap Calculus Bc

Evaluate the limit

$\lim_{x\to1}\frac{7x^{9/2}-9x^{7/2}+2}{5x^{7/2}-7x^{5/2}+2}$ .

Possible Answers:

$\frac{9}{5}$

$\frac{9}{10}$

$\frac{27}{10}$

$\frac{18}{5}$

Correct answer:

$\frac{9}{5}$

Explanation:

Admittedly a more computationally intensive problem, we begin by trying to plug in x=1 into the limit directly, but we get the indeterminate form $\frac{0}{0}$ .

We then apply L'Hôpital's rule and try again:

$\lim_{x\to1}\frac{7x^{9/2}-9x^{7/2}+2}{5x^{7/2}-7x^{5/2}+2}$

$=\lim_{x\to1}\frac{7\left (\frac{9}{2} \right )x^{7/2}-9\left (\frac{7}{2} \right )x^{5/2}}{5\left (\frac{7}{2} \right )x^{5/2}-7\left (\frac{5}{2} \right )x^{3/2}}$ .

However, plugging in x=1 still yields the indeterminate form $\frac{0}{0}$ , so we are forced to apply L'Hôpital's rule a second time:

$=\lim_{x\to1}\frac{7\left (\frac{9}{2} \right )\left (\frac{7}{2} \right )x^{5/2}-9\left (\frac{7}{2} \right )\left (\frac{5}{2} \right )x^{3/2}}{5\left (\frac{7}{2} \right )\left (\frac{5}{2} \right )x^{3/2}-7\left (\frac{5}{2} \right )\left (\frac{3}{2} \right )x^{1/2}}$ .

This time, plugging in x=1 does not produce an indeterminate form, so we may evaluate the limit by setting x=1 as follows:

$=\frac{7\left (\frac{9}{2} \right )\left (\frac{7}{2} \right )-9\left (\frac{7}{2} \right )\left (\frac{5}{2} \right )}{5\left (\frac{7}{2} \right )\left (\frac{5}{2} \right )-7\left (\frac{5}{2} \right )\left (\frac{3}{2} \right )}$

$= \frac{(9)(7)(7-5)}{(5)(7)(5-3)}$

$=\frac{9}{5}$ .

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Example Question #22 : Euler's Method And L'hopital's Rule

Evaluate the limit

$\lim_{x\to 1}\frac{x^{5/2}-1}{x^{7/2}-1}$ .

Possible Answers:

$\frac{3}{5}$

$\frac{3}{7}$

$\frac{5}{7}$

$\frac{9}{7}$

$\frac{7}{5}$

Correct answer:

$\frac{5}{7}$

Explanation:

Simply plugging in x=1 into the expression yields the indeterminate form of $\frac{0}{0}$ , so we must resort to using L'Hôpital's rule. We take the derivative of the numerator and denominator, and then look at the limit again.

$\lim_{x\to1}\frac{x^{5/2}-1}{x^{7/2}-1}=\lim_{x\to1}\frac{\left (\frac{5}{2} \right )x^{3/2}}{\left (\frac{7}{2} \right )x^{5/2}}$

This time, when we plug in x=1 , we do not get an indeterminate form, so we can evaluate the limit by setting x=1 :

$=\frac{\left (\frac{5}{2} \right )1^{3/2}}{\left (\frac{7}{2} \right )1^{5/2}}=\frac{5}{7}$ .

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Example Question #23 : Euler's Method And L'hopital's Rule

$\begin{align*}&\text{The expression of a particular function is unknown;}\\&\text{however, we have an expression for its derivative.}\\&\text{Knowing that }f'(x)=4e^{(2x)}\text{ and }f(0)=9\\&\text{Approximate } f(4.8)\text{ using Euler's Method and }4\text{ steps.}\end{align*}$

Possible Answers:

7079.23

1769.81

13.8

66.71

28316.9

Correct answer:

7079.23

Explanation:

$\begin{align*}&\text{The general form of Euler's method, when a derivative function,}\\&\text{initial value, and step size are known, is:}\\&y_n=y_{n+1} +\Delta x f'(x_n,y_n)\\&\text{To calculate the step size find the distance between the final}\\&\text{and initial value and divide by the number of steps to be used:}\\&\Delta x = \frac{x_f-x_i}{Steps}\\&\text{For this problem, we are told: }f'(x)=4e^{(2x)};f(0)=9\\&\text{Knowing this, we may take the steps to estimate our function}\\&\text{value at our final x value:}\\&\Delta x = \frac{4.8-(0)}{4}=1.2\\&y_0=9;x_0=0\\&y_{1}=9+(1.2)(4e^{(2(0))})=13.8\\&y_{2}=13.8+(1.2)(4e^{(2(1.2))})=66.71\\&y_{3}=66.71+(1.2)(4e^{(2(2.4))})=649.96\\&y_{4}=649.96+(1.2)(4e^{(2(3.6))})=7079.23\\&\\&\end{align*}$

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Example Question #212 : Derivatives

$\begin{align*}&\text{The expression of a particular function is unknown;}\\&\text{however, we have an expression for its derivative.}\\&\text{Knowing that }f'(x)=-4cos(5x)\text{ and }f(-2)=-8\\&\text{Approximate } f(5.6)\text{ using Euler's Method and }2\text{ steps.}\end{align*}$

Possible Answers:

18.6

6.2

37.21

9.3

4.75

Correct answer:

18.6

Explanation:

$\begin{align*}&\text{The general form of Euler's method, when a derivative function,}\\&\text{initial value, and step size are known, is:}\\&y_n=y_{n+1} +\Delta x f'(x_n,y_n)\\&\text{To calculate the step size find the distance between the final}\\&\text{and initial value and divide by the number of steps to be used:}\\&\Delta x = \frac{x_f-x_i}{Steps}\\&\text{For this problem, we are told: }f'(x)=-4cos(5x);f(-2)=-8\\&\text{Knowing this, we may take the steps to estimate our function}\\&\text{value at our final x value:}\\&\Delta x = \frac{5.6-(-2)}{2}=3.8\\&y_0=-8;x_0=-2\\&y_{1}=-8+(3.8)(-4cos(5(-2)))=4.75\\&y_{2}=4.75+(3.8)(-4cos(5(1.8)))=18.6\\&\\&\end{align*}$

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Example Question #221 : Derivatives

$\begin{align*}&\text{The expression of a particular function is unknown;}\\&\text{however, we have an expression for its derivative.}\\&\text{Knowing that }f'(x)=8sin(x)\text{ and }f(-1)=5\\&\text{Approximate } f(7)\text{ using Euler's Method and }4\text{ steps.}\end{align*}$

Possible Answers:

-2.69

-8.08

-32.34

10.26

-8.46

Correct answer:

-8.08

Explanation:

$\begin{align*}&\text{The general form of Euler's method, when a derivative function,}\\&\text{initial value, and step size are known, is:}\\&y_n=y_{n+1} +\Delta x f'(x_n,y_n)\\&\text{To calculate the step size find the distance between the final}\\&\text{and initial value and divide by the number of steps to be used:}\\&\Delta x = \frac{x_f-x_i}{Steps}\\&\text{For this problem, we are told: }f'(x)=8sin(x);f(-1)=5\\&\text{Knowing this, we may take the steps to estimate our function}\\&\text{value at our final x value:}\\&\Delta x = \frac{7-(-1)}{4}=2\\&y_0=5;x_0=-1\\&y_{1}=5+(2)(8sin(-1))=-8.46\\&y_{2}=-8.46+(2)(8sin(1))=5\\&y_{3}=5+(2)(8sin(3))=7.26\\&y_{4}=7.26+(2)(8sin(5))=-8.08\\&\\&\end{align*}$

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AP Calculus BC : AP Calculus BC

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #22 : L'hospital's Rule

Example Question #482 : Ap Calculus Bc

Example Question #483 : Ap Calculus Bc

Example Question #213 : Derivatives

Example Question #482 : Ap Calculus Bc

Example Question #481 : Ap Calculus Bc

Example Question #22 : Euler's Method And L'hopital's Rule

Example Question #23 : Euler's Method And L'hopital's Rule

Example Question #212 : Derivatives

Example Question #221 : Derivatives

All AP Calculus BC Resources

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