Euler's Method and L'Hopital's Rule

Help Questions

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

Questions 1 - 10

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

Explanation

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

Evaluate the limit using L'Hopital's Rule.

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

Undefined

$\infty$

Explanation

L'Hopital's Rule is used to evaluate complicated limits. The rule has you take the derivative of both the numerator and denominator individually to simplify the function. In the given function we take the derivatives the first time and get

$\lim_{x\rightarrow \infty }\frac{2x}{e^x}$ .

This still cannot be evaluated properly, so we will take the derivative of both the top and bottom individually again. This time we get

$\lim_{x\rightarrow \infty }\frac{2}{e^x}$ .

Now we have only one x, so we can evaluate when x is infinity. Plug in infinity for x and we get

$\frac{2}{e^\infty }$ and $e^\infty=\infty$ .

So we can simplify the function by remembering that any number divided by infinity gives you zero.

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

Explanation

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

Find the

$\small \small \small \lim_{x\rightarrow 0}\frac{e^{x^{2}}-ln(1+x)-1}{cosx-sinx-1}$ .

$\small 1$

$\small 0$

$\small e$

$\small -1$

Does Not Exist

Explanation

Subbing in zero into $\small \frac{(e^{x^{2}}-ln(1+x)-1)}{cosx-sinx-1}$ will give you $\small \frac{0}{0}$ , so we can try to use L'hopital's Rule to solve.

First, let's find the derivative of the numerator.

$\small e^{x^{2}}$ is in the form $\small a^{u}$ , which has the derivative $\small a^{u}u'lna$ , so its derivative is $\small 2xe^{x^{2}}$ .

$\small ln(1+x)$ is in the form $\small lnu$ , which has the derivative $\small u'/u$ , so its derivative is $\small 1/(1+x)$ .

The derivative of $\small 1$ is $\small 0$ so the derivative of the numerator is $\small 2xe^{x^{2}}-1/(1+x)$ .

In the denominator, the derivative of $\small cosx$ is $\small -sinx$ , and the derivative of $\small sinx$ is $\small cosx$ . Thus, the entire denominator's derivative is $\small -(sinx+cosx)$ .

Now we take the

$\small \lim_{x\rightarrow 0}\frac{2xe^{x^{2}}-1/(x+1)}{-(sinx+cosx)}$ , which gives us $\small \frac{-1}{-1}=1$ .

Find the limit if it exists

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

Hint: Use L'Hospital's rule

$\frac{1}{2}$

$\frac{3}{2}$

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

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

$\begin{align*}&\text{The expression of a particular function is unknown;}\&\text{however, we have an expression for its derivative.}\&\text{Knowing that }f'(x)=-2cos(6x)\text{ and }f(-1)=-11\&\text{Approximate } f(6.2)\text{ using Euler's Method and }4\text{ steps.}\end{align*}$

Explanation

$\begin{align*}&\text{The general form of Euler's method, when a derivative function,}\&\text{initial value, and step size are known, is:}\&y_n=y_{n+1} +\Delta x f'(x_n,y_n)\&\text{To calculate the step size find the distance between the final}\&\text{and initial value and divide by the number of steps to be used:}\&\Delta x = \frac{x_f-x_i}{Steps}\&\text{For this problem, we are told: }f'(x)=-2cos(6x);f(-1)=-11\&\text{Knowing this, we may take the steps to estimate our function}\&\text{value at our final x value:}\&\Delta x = \frac{6.2-(-1)}{4}=1.8\&y_0=-11;x_0=-1\&y_{1}=-11+(1.8)(-2cos(6{(-1)}))=-14.46\&y_{2}=-14.46+(1.8)(-2cos(6{(0.8)}))=-14.77\&y_{3}=-14.77+(1.8)(-2cos(6{(2.6)}))=-11.19\&y_{4}=-11.19+(1.8)(-2cos(6{(4.4)}))=-12.27\&\&\end{align*}$

Evaluate:

$\lim_{x\rightarrow 0 } \frac{7^{x} - 1}{2^{x} - 1}$

$\log_{2}7$

$\ln \frac{7}{2}$

$\frac{1}{2}\ln 7$

$\infty$

The limit does not exist.

Explanation

$\lim_{x\rightarrow 0 }\left ( 7^{x} - 1 \right ) = 7^{0} - 1 = 1- 1 = 0$

and

$\lim_{x\rightarrow 0 }\left ( 2^{x} - 1 \right ) = 2^{0} - 1 = 1- 1 = 0$

Therefore, by L'Hospital's Rule, we can find $\lim_{x\rightarrow 0 } \frac{7^{x} - 1}{2^{x} - 1}$ by taking the derivatives of the expressions in both the numerator and the denominator:

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( 7^{x} - 1 \right ) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( e^{x \ln 7} - 1 \right )$

$= \frac{\mathrm{d} }{\mathrm{d} x} e^{x \ln 7 } - \frac{\mathrm{d} }{\mathrm{d} x}1$

$=\ln 7\cdot e^{x \ln 7 } - 0 = \ln 7\cdot 7^{x }$

Similarly,

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( 2^{x} - 1 \right ) = \ln 2 \cdot 2 ^{x}$

$\lim_{x\rightarrow 0 } \frac{7^{x} - 1}{2^{x} - 1} = \lim_{x\rightarrow 0 } \frac{ \ln 7\cdot 7^{x }}{ \ln 2\cdot 2^{x }}$

$= \frac{ \ln 7 \cdot 7^{0 }}{ \ln 2\cdot 2^{0 }} = \frac{ \ln 7\cdot 1}{ \ln 2\cdot 1} = \frac{ \ln 7}{ \ln 2 }= \log_{2}7$

Find the limit: $\lim_{x \to 3} \frac{x^2+3x-18}{x-3}$

$\textup{Undefined}$

Explanation

By substituting the value of , we will find that this will give us the indeterminate form $\frac{0}{0}$ . This means that we can use L'Hopital's rule to solve this problem.

$\lim_{x \to 3} \frac{x^2+3x-18}{x-3}=\frac{3^2+3(3)-18}{x-3} = \frac{18-18}{3-3} = \frac{0}{0}$

L'Hopital states that we can take the limit of the fraction of the derivative of the numerator over the derivative of the denominator. L'Hopital's rule can be repeated as long as we have an indeterminate form after every substitution.

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

Take the derivative of the numerator.

$\frac{d}{dx}(x^2+3x-18) = 2x+3$

Take the derivative of the numerator.

$\frac{d}{dx}(x-3) =1$

Rewrite the limit and use substitution.

$\lim_{x \to 3}\frac{2x+3}{1} = 2(3)+3 =9$

The limit is .

Evaluate the limit

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

$\frac{5}{7}$

$\frac{3}{7}$

$\frac{3}{5}$

$\frac{9}{7}$

$\frac{7}{5}$

Explanation

Simply plugging in 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 , we do not get an indeterminate form, so we can evaluate the limit by setting :

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

Evaluate the limit:

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

The limit does not exist

$\infty$

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:

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

Page 1 of 4