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

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Example Questions

Example Question #1 : Euler's Method And L'hopital's Rule

Evaluate:

$\lim_{x\rightarrow \infty } \frac{x \sin^{2} x}{2^{x} - 1}$

Possible Answers:

$\frac{1}{2}$

The limit does not exist.

$\infty$

Correct answer:

Explanation:

Let's examine the limit

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

first.

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

$\frac{\mathrm{d} }{\mathrm{d} x}x = 1$

and

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

$= \frac{\mathrm{d} }{\mathrm{d} x}e^{ x \ln 2} - \frac{\mathrm{d} }{\mathrm{d} x} 1$

$=\ln 2 \cdot e^{ x \ln 2} - 0 = \ln 2 \cdot 2^{x}$ ,

so by L'Hospital's Rule,

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

Since $\lim_{x\rightarrow \infty } \ln 2 \cdot 2^{x} = \infty$ ,

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

Now, for each x > 0 , $0 \leq \sin^{2} x \leq 1$ ; therefore,

$0 \leq \frac{x \sin^{2} x}{2^{x} - 1} \leq \frac{x }{2^{x} - 1}$

By the Squeeze Theorem,

$0 \leq \lim_{x\rightarrow \infty } \frac{x \sin^{2} x}{2^{x} - 1} \leq \lim_{x\rightarrow \infty } \frac{x }{2^{x} - 1} = 0$

and

$\lim_{x\rightarrow \infty } \frac{x \sin^{2} x}{2^{x} - 1} = 0$

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Example Question #2 : Euler's Method And L'hopital's Rule

Evaluate:

$\lim_{x\rightarrow \infty } \frac{5^{x} - 1}{3^{x} - 1}$

Possible Answers:

The limit does not exist.

$\ln \frac{5}{3}$

$\frac{\ln 5}{\ln 3}$

$\infty$

$\frac{5}{3}$

Correct answer:

$\infty$

Explanation:

$\lim_{x\rightarrow \infty }\left ( 5^{x} - 1 \right ) = \lim_{x\rightarrow \infty }\left ( 3^{x} - 1 \right ) = \infty$

Therefore, by L'Hospital's Rule, we can find $\lim_{x\rightarrow \infty } \frac{5^{x} - 1}{3^{x} - 1}$ by taking the derivatives of the expressions in both the numerator and the denominator:

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( 5^{x} - 1 \right ) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( e^{x \ln 5} - 1 \right )$

$= \frac{\mathrm{d} }{\mathrm{d} x} e^{x \ln 5 } - \frac{\mathrm{d} }{\mathrm{d} x}1$

$=\ln 5\cdot e^{x \ln 5 } - 0 = \ln 5\cdot 5^{x }$

Similarly,

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( 3^{x} - 1 \right ) = \ln 3 \cdot 3 ^{x}$

$\lim_{x\rightarrow \infty } \frac{5^{x} - 1}{3^{x} - 1} = \lim_{x\rightarrow \infty } \frac{\ln 5 \cdot 5^{x} }{\ln 3 \cdot 3^{x} }$

$=\frac{\ln 5 }{\ln 3 } \cdot \lim_{x\rightarrow \infty } \frac{ 5^{x} }{ 3^{x} }= \frac{\ln 5 }{\ln 3 } \cdot \lim_{x\rightarrow \infty }\left ( \frac{ 5 }{ 3 } \right )^{x}$

But $\lim_{x\rightarrow \infty }a^{x} = \infty$ for any a > 1 , so

$\lim_{x\rightarrow \infty } \frac{5^{x} - 1}{3^{x} - 1}= \frac{\ln 5 }{\ln 3 } \cdot \lim_{x\rightarrow \infty }\left ( \frac{ 5 }{ 3 } \right )^{x} = \infty$

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Example Question #3 : Euler's Method And L'hopital's Rule

Evaluate:

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

Possible Answers:

$\log_{2}7$

$\infty$

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

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

The limit does not exist.

Correct answer:

$\log_{2}7$

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,

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

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

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

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Example Question #4 : Euler's Method And L'hopital's Rule

Evaluate:

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

Possible Answers:

$\ln 2$

$\frac{1}{\ln2}$

Correct answer:

Explanation:

$\lim_{x\rightarrow 0^{+} } x \sin x= 0 \cdot \sin 0 = 0$

and

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

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

$\frac{\mathrm{d} }{\mathrm{d} x}x \sin x = \frac{\mathrm{d} }{\mathrm{d} x} x \cdot \sin x + \frac{\mathrm{d} }{\mathrm{d} x}\sin x \cdot x$

$= 1 \cdot \sin x + \cos x \cdot x = \sin x + x \cos x$

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

$= \frac{\mathrm{d} }{\mathrm{d} x} e^{x \ln 2} - \frac{\mathrm{d} }{\mathrm{d} x} 1$

$= \ln 2\cdot e^{x \ln 2} - 0 = \ln 2\cdot 2^{x }$

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

$= \frac{\sin 0 + 0 \cos 0}{ \ln 2\cdot 2^{0 }} = \frac{ 0 + 0 }{ \ln 2\cdot 1} = 0$

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Example Question #1 : Euler's Method

Suppose we have the following differential equation with the initial condition:

$\frac{\partial p }{\partial x} = 0.5x(1-x), \ p(0) = 2$

Use Euler's method to approximate p(2) , using a step size of .

Possible Answers:

Correct answer:

Explanation:

We start at x = 0 and move to x=2, with a step size of 1. Essentially, we approximate the next step by using the formula:

$p(x + \Delta x) = p(x) + p'(x)\cdot (\Delta x)$ .

So applying Euler's method, we evaluate using derivative:

$p'(x) = 0.5x(1-x), \ p(0) = 2$

And two step sizes, at x = 1 and x = 2.

${}p(x=1) = p(0) + (1 - 0)p'(0) = 2 + 1\cdot p'(0) = 2 + 1 \cdot 0.5(0)(1-0) = 2 + 0 = 2$

${}p(x=2) = p(1) + (2-1)\cdot p'(1) = 2 + 1\cdot p'(1) = 2 + 0.5(1)(1-1) = 2 + 0 = 2$

And thus the evaluation of p at x = 2, using Euler's method, gives us p(2) = 2.

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Example Question #2 : Euler's Method

Approximate $\small e^{\frac{1}{2}}$ by using Euler's method on the differential equation

$\small y'=y$

with initial condition $\small y(0)=1$ (which has the solution $\small y(t)=e^t$ ) and time step $\small \small h=\frac{1}{4}$ .

Possible Answers:

$\small 1.5$

$\small 1.56$

$\small 1.649$

$\small 1.5625$

Correct answer:

$\small 1.5625$

Explanation:

Using Euler's method with $\small h=1/4$ means that we use two iterations to get the approximation. The general iterative formula is

$\small y_{i+1}=y_i+h\cdot f(t_i,y_i)$

where each $\small t_i$ is

$\small t_0=0$

$\small t_1=\frac{1}{4}$

$\small t_2=\frac{1}{2}$

$\small y_i$ is an approximation of $\small y(t_i)$ , and y'=f(t,y)=y , for this differential equation. So we have

$\small \small \small \small y_1=y_0+\frac{1}{4}f(t_0,y_0)=1+\frac{1}{4}\cdot 1=1.25$

$\small \small \small \small e^{1/2}= y(\frac{1}{2})\approx y_2=y_1+\frac{1}{4}f(t_1,y_1)=1.25+\frac{1}{4}\cdot 1.25=1.5625$

So our approximation of $\small e^{1/2}$ is

$\small e^{1/2}\approx 1.5625$

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Example Question #1 : L'hospital's Rule

Evaluate the limit using L'Hopital's Rule.

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

Possible Answers:

Undefined

$\infty$

Correct answer:

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.

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Example Question #1 : L'hospital's Rule

Evaluate the limit using L'Hopital's Rule.

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

Possible Answers:

$\infty$

Undefined

Correct answer:

Explanation:

$\lim_{x\rightarrow 0}\frac{4x}{\frac{1}{x}}=\lim_{x\rightarrow 0}4x^2$ .

Since the first set of derivatives eliminates an x term, we can plug in zero for the x term that remains. We do this because the limit approaches zero.

This gives us

$4\cdot 0^2=0$ .

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Example Question #2 : L'hospital's Rule

Evaluate the limit using L'Hopital's Rule.

$\lim_{x\rightarrow \infty }\frac{5x^3+1}{3x^2}$

Possible Answers:

$\frac{15}{6}$

$\frac{5}{3}$

$\infty$

Undefined

Correct answer:

$\infty$

Explanation:

$\lim_{x\rightarrow \infty }\frac{15x^2}{6x}$ .

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

$\lim_{x\rightarrow \infty }\frac{30x}{6}=5x$ .

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

$5\cdot \infty=\infty$ .

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Example Question #3 : L'hospital's Rule

Calculate the following limit.

$\lim_{x\rightarrow2} \frac{x^2-4}{2x-4}$

Possible Answers:

Correct answer:

Explanation:

To calculate the limit, often times we can just plug in the limit value into the expression. However, in this case if we were to do that we get $\frac00$ , which is undefined.

What we can do to fix this is use L'Hopital's rule, which says

$\lim_{x\rightarrow 2} \frac{f(x)}{g(x)}=\lim_{x\rightarrow 2} \frac{f'(x)}{g'(x)}$ .

So, L'Hopital's rule allows us to take the derivative of both the top and the bottom and still obtain the same limit.

$\lim_{x-\rightarrow 2} \frac{x^2-4}{2x-4}=\lim_{x\rightarrow 2} \frac{2x}{2}$ .

Plug in x=2 to get an answer of .

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

Study concepts, example questions & explanations for AP Calculus BC

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Example Questions

Example Question #1 : Euler's Method And L'hopital's Rule

Example Question #2 : Euler's Method And L'hopital's Rule

Example Question #3 : Euler's Method And L'hopital's Rule

Example Question #4 : Euler's Method And L'hopital's Rule

Example Question #1 : Euler's Method

Example Question #2 : Euler's Method

Example Question #1 : L'hospital's Rule

Example Question #1 : L'hospital's Rule

Example Question #2 : L'hospital's Rule

Example Question #3 : L'hospital's Rule

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