Applications of Derivatives - AP Calculus BC

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Question

Evaluate:

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

Answer

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

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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Question

Evaluate:

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

Answer

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 , $0 \leq \sin^{2} x \leq 1$ ; 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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Question

Evaluate:

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

Answer

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

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,

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

So

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

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

But $\lim_{x\rightarrow \infty }a^{x} = \infty$ for any , 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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Question

Evaluate:

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

Answer

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

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,

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

so

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

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

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Question

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 , using a step size of .

Answer

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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Question

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

Answer

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

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Question

Calculate the following limit.

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

Answer

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 to get an answer of .

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Question

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 $\small \small h=\frac{1}{4}$ .

Answer

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 , 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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Question

Find the

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

Answer

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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Question

Evaluate the limit using L'Hopital's Rule.

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

Answer

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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Question

Evaluate the limit using L'Hopital's Rule.

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

Answer

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

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

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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Question

Evaluate the limit using L'Hopital's Rule.

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

Answer

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

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

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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Question

Calculate the following limit.

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

Answer

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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Question

Evaluate the following limit:

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

Answer

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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Question

Find the limit if it exists.

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

Hint: Apply L'Hospital's Rule.

Answer

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

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

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

As such the limit exists and is

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

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Question

Find the limit if it exists.

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

Hint: Apply L'Hospital's Rule.

Answer

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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Question

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

Answer

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 .

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Question

Find the limit if it exists

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

Hint: Use L'Hospital's rule

Answer

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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Question

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

Answer

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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Question

Evaluate the limit:

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

Answer

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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