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AP Calculus BC : Derivatives

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Example Questions

Example Question #1 : 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:

$\small 1.649$

$\small 1.5$

$\small 1.56$

$\small 1.5625$

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 #2 : 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 #5 : Euler's Method And L'hopital'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 #3 : Euler's Method And L'hopital's Rule

Evaluate the limit using L'Hopital's Rule.

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

Possible Answers:

$\frac{5}{3}$

Undefined

$\frac{15}{6}$

$\infty$

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

Calculate the following limit.

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

Possible Answers:

$-\infty$

$\infty$

$e^{2}$

Correct answer:

$\infty$

Explanation:

If we plugged in $-\infty$ directly, we would get an indeterminate value of $\frac{\infty}{\infty}$ .

We can use L'Hopital's rule to fix this. We take the derivate of the top and bottom and reevaluate the same limit.

$\lim_{x->-\infty}\frac{e^{-2x}}{x^2}=-\frac{e^{-2x}}{x}$ .

We still can't evaluate the limit of the new expression, so we do it one more time.

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

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Example Question #2 : L'hospital's Rule

Find the

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

Possible Answers:

$\small 1$

Does Not Exist

$\small e$

$\small 0$

$\small -1$

Correct answer:

$\small 1$

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 $\small \frac{-1}{-1}=1$ .

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Example Question #11 : L'hospital's Rule

Evaluate the following limit:

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

Possible Answers:

$-\infty$

Correct answer:

Explanation:

When you try to solve the limit using normal methods, you find that the limit approaches zero in the numerator and denominator, resulting in an indeterminate form "0/0".

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

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

when an indeterminate form occurs when evaluting the limit.

Next, simply find f'(x) and g'(x) for this limit:

$2t, -\cos(2-t)$

The derivatives were found using the following rules:

$\frac{d}{dx}(x^n)=nx^{n-1}$ , $\frac{d}{dx}\sin(u)=\cos(u)\frac{du}{dx}$

Next, using L'Hopital's Rule, evaluate the limit using f'(x) and g'(x):

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

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Example Question #11 : L'hospital's Rule

Find the limit if it exists.

$\lim_{x\rightarrow 0}\frac{2x^2-3x}{3x^2+3x}$

Hint: Apply L'Hospital's Rule.

Possible Answers:

$\frac{2}{3}$

DNE

Correct answer:

Explanation:

Through direct substitution, we see that the limit becomes

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

which is in indeterminate form.

As such we can use l'Hospital's Rule, which states that if the limit

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

is in indeterminate form, then the limit is equivalent to

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

Taking the derivatives we use the power rule which states

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

Using the power rule the limit becomes

$\lim_{x\rightarrow 0}\frac{2x^2-3x}{3x^2+3x} = \lim_{x\rightarrow 0}\frac{2\cdot2x^{2-1}-1\cdot3x^{1-1}}{2\cdot3x^{2-1}+1\cdot3x^{1-1}}$

$= \lim_{x\rightarrow 0}\frac{4x-3}{6x+3}$

=-1

As such the limit exists and is

$\lim_{x\rightarrow 0}\frac{2x^2-3x}{3x^2+3x}=-1$

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Example Question #22 : Limits

Find the limit if it exists.

$\lim_{x\rightarrow 0}\frac{\sin(\pi x)}{\sin(x)}$

Hint: Apply L'Hospital's Rule.

Possible Answers:

$\pi$

DNE

Correct answer:

$\pi$

Explanation:

Through direct substitution, we see that the limit becomes

$\lim_{x\rightarrow 0}\frac{\sin(\pi\cdot0)}{\sin(0)}=\lim_{x\rightarrow 0}\frac{0}{0}$

which is in indeterminate form.

As such we can use l'Hospital's Rule, which states that if the limit

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

is in indeterminate form, then the limit is equivalent to

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

Taking the derivatives we use the trigonometric rule which states

$\frac{\mathrm{d} }{\mathrm{d} x} \sin(Bx)=B\cos(Bx)$

where is a constant.

Using l'Hospital's Rule we obtain

$\lim_{x\rightarrow 0}\frac{\sin(\pi x)}{\sin(x)}=\lim_{x\rightarrow 0}\frac{\pi \cos(\pi x)}{\cos(x)}$

And through direct substitution we find

$\lim_{x\rightarrow 0}\frac{\pi \cos(\pi x)}{\cos(x)}=\frac{\pi \cos(\pi \cdot0)}{\cos(0)}$

$=\frac{\pi \cdot1}{1}$

$=\pi$

As such the limit exists and is

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

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AP Calculus BC : Derivatives

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #1 : Euler's Method

Example Question #2 : L'hospital's Rule

Example Question #5 : 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 : L'hospital's Rule

Example Question #2 : L'hospital's Rule

Example Question #11 : L'hospital's Rule

Example Question #11 : L'hospital's Rule

Example Question #22 : Limits

All AP Calculus BC Resources

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