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AP Calculus BC : Euler's Method and L'Hopital's Rule

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Example Questions

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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Example Question #21 : Euler's Method And L'hopital's Rule

$\begin{align*}&\text{The expression of a particular function is unknown;}\\&\text{however, we have an expression for its derivative.}\\&\text{Knowing that }f'(x)=13tan(5x)\text{ and }f(2)=9\\&\text{Approximate } f(11.9)\text{ using Euler's Method and }3\text{ steps.}\end{align*}$

Possible Answers:

36.08

901.94

6.5

36.81

180.39

Correct answer:

180.39

Explanation:

$\begin{align*}&\text{The general form of Euler's method, when a derivative function,}\\&\text{initial value, and step size are known, is:}\\&y_n=y_{n+1} +\Delta x f'(x_n,y_n)\\&\text{To calculate the step size find the distance between the final}\\&\text{and initial value and divide by the number of steps to be used:}\\&\Delta x = \frac{x_f-x_i}{Steps}\\&\text{For this problem, we are told: }f'(x)=13tan(5x);f(2)=9\\&\text{Knowing this, we may take the steps to estimate our function}\\&\text{value at our final x value:}\\&\Delta x = \frac{11.9-(2)}{3}=3.3\\&y_0=9;x_0=2\\&y_{1}=9+(3.3)(13tan(5(2)))=36.81\\&y_{2}=36.81+(3.3)(13tan(5(5.3)))=244.67\\&y_{3}=244.67+(3.3)(13tan(5(8.6)))=180.39\\&\\&\end{align*}$

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

$\begin{align*}&\text{The expression of a particular function is unknown;}\\&\text{however, we have an expression for its derivative.}\\&\text{Knowing that }f'(x)=-19ln(5x)\text{ and }f(1)=-3\\&\text{Approximate } f(3.8)\text{ using Euler's Method and }2\text{ steps.}\end{align*}$

Possible Answers:

-559.55

-22.38

-58.94

-45.81

-111.91

Correct answer:

-111.91

Explanation:

$\begin{align*}&\text{The general form of Euler's method, when a derivative function,}\\&\text{initial value, and step size are known, is:}\\&y_n=y_{n+1} +\Delta x f'(x_n,y_n)\\&\text{To calculate the step size find the distance between the final}\\&\text{and initial value and divide by the number of steps to be used:}\\&\Delta x = \frac{x_f-x_i}{Steps}\\&\text{For this problem, we are told: }f'(x)=-19ln(5x);f(1)=-3\\&\text{Knowing this, we may take the steps to estimate our function}\\&\text{value at our final x value:}\\&\Delta x = \frac{3.8-(1)}{2}=1.4\\&y_0=-3;x_0=1\\&y_{1}=-3+(1.4)(-19ln(5(1)))=-45.81\\&y_{2}=-45.81+(1.4)(-19ln(5(2.4)))=-111.91\\&\\&\end{align*}$

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

$\begin{align*}&\text{The expression of a particular function is unknown;}\\&\text{however, we have an expression for its derivative.}\\&\text{Knowing that }f'(x)=18\cdot 5^x\text{ and }f(0)=10\\&\text{Approximate } f(6.6)\text{ using Euler's Method and }3\text{ steps.}\end{align*}$

Possible Answers:

48531

97061.9

49.6

738721

16177

Correct answer:

48531

Explanation:

$\begin{align*}&\text{The general form of Euler's method, when a derivative function,}\\&\text{initial value, and step size are known, is:}\\&y_n=y_{n+1} +\Delta x f'(x_n,y_n)\\&\text{To calculate the step size find the distance between the final}\\&\text{and initial value and divide by the number of steps to be used:}\\&\Delta x = \frac{x_f-x_i}{Steps}\\&\text{For this problem, we are told: }f'(x)=18\cdot 5^x;f(0)=10\\&\text{Knowing this, we may take the steps to estimate our function}\\&\text{value at our final x value:}\\&\Delta x = \frac{6.6-(0)}{3}=2.2\\&y_0=10;x_0=0\\&y_{1}=10+(2.2)(18\cdot 5^{(0)})=49.6\\&y_{2}=49.6+(2.2)(18\cdot 5^{(2.2)})=1415.53\\&y_{3}=1415.53+(2.2)(18\cdot 5^{(4.4)})=48531\\&\\&\end{align*}$

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Example Question #24 : Euler's Method And L'hopital's Rule

$\begin{align*}&\text{The expression of a particular function is unknown;}\\&\text{however, we have an expression for its derivative.}\\&\text{Knowing that }f'(x)=-17e^{(5x)}\text{ and }f(-2)=-20\\&\text{Approximate } f(1.8)\text{ using Euler's Method and }2\text{ steps.}\end{align*}$

Possible Answers:

-39.59

-158.37

-7.92

Correct answer:

-39.59

Explanation:

$\begin{align*}&\text{The general form of Euler's method, when a derivative function,}\\&\text{initial value, and step size are known, is:}\\&y_n=y_{n+1} +\Delta x f'(x_n,y_n)\\&\text{To calculate the step size find the distance between the final}\\&\text{and initial value and divide by the number of steps to be used:}\\&\Delta x = \frac{x_f-x_i}{Steps}\\&\text{For this problem, we are told: }f'(x)=-17e^{(5x)};f(-2)=-20\\&\text{Knowing this, we may take the steps to estimate our function}\\&\text{value at our final x value:}\\&\Delta x = \frac{1.8-(-2)}{2}=1.9\\&y_0=-20;x_0=-2\\&y_{1}=-20+(1.9)(-17e^{(5{(-2)})})=-20\\&y_{2}=-20+(1.9)(-17e^{(5{(-0.1)})})=-39.59\\&\\&\end{align*}$

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AP Calculus BC : Euler's Method and L'Hopital's Rule

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

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

Example Question #21 : Euler's Method And L'hopital's Rule

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

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

Example Question #24 : Euler's Method And L'hopital's Rule

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