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3

## How do you use the Squeeze Theorem to find lim Sin(x)/x as x approaches zero?

Jim H
Featured 2 days ago

For a non-rigorous proof, please see below.

#### Explanation:

For a positive central angle of $x$ radians ($0 < x < \frac{\pi}{2}$) (not degrees)

Source:

The geometric idea is that

$\text{Area of "Delta KOA < "Area of " "Sector KOA" < "Area of } \Delta L O A$

$\text{Area of } \Delta K O A = \frac{1}{2} \left(1\right) \left(\sin x\right) \setminus \setminus \setminus$ ($\frac{1}{2} \text{base"*"height}$)

$\text{Area of " "Sector KOA} = \frac{1}{2} {\left(1\right)}^{2} x \setminus \setminus \setminus$ ($x$ is in radians)

$\text{Area of } \Delta L O A = \frac{1}{2} \tan x \setminus \setminus \setminus$ ($A L = \tan x$)

So we have:

$\sin \frac{x}{2} < \frac{x}{2} < \tan \frac{x}{2}$

For small positive $x$, we have $\in x > 0$ so we can multiply through by $\frac{2}{\sin} x$, to get

$1 < \frac{x}{\sin} x < \frac{1}{\cos} x$

So

$\cos x < \sin \frac{x}{x} < 1$ for $0 < x < \frac{\pi}{2}$.

${\lim}_{x \rightarrow {0}^{+}} \cos x = 1$ and ${\lim}_{x \rightarrow {0}^{+}} 1 = 1$

so ${\lim}_{x \rightarrow {0}^{+}} \sin \frac{x}{x} = 1$

We also have, for these small $x$, $\sin \left(- x\right) = - \sin x$, so $\frac{- x}{\sin} \left(- x\right) = \frac{x}{\sin} x$ and $\cos \left(- x\right) = \cos x$, so

$\cos x < \sin \frac{x}{x} < 1$ for $- \frac{\pi}{2} < x < 0$.

${\lim}_{x \rightarrow {0}^{-}} \cos x = 1$ and ${\lim}_{x \rightarrow {0}^{-}} 1 = 1$

so ${\lim}_{x \rightarrow {0}^{-}} \sin \frac{x}{x} = 1$

Since both one sided limits are $1$, the limit is $1$.

3

## Find the volume of the solid of revolution obtained by rotating the curve x=3cos^3theta , y=3sin^3theta about the x axis?

Frederico Guizini S.
Featured 3 days ago

4

## find a point on the graph of the given function where the tangent or normal line satisfies the indicated condition, and then write an equation for this line.?

Steve M
Featured 3 days ago

There are three possible coordinates:

 ( 1, 6 ) ; ( -1-1/(3sqrt(2)), 13/6+(2sqrt(2))/3 ) ; ( -1+1/(3sqrt(2)), 13/6-(2sqrt(2))/3 )

If we choose $\left(1 , 6\right)$ then the normal equation is:

$8 y + x - 49 = 0$

#### Explanation:

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. (If needed, then the normal is perpendicular to the tangent so the product of their gradients is −1).

We have:

$f \left(x\right) = 3 {x}^{2} + 2 x + 1$

Differentiating wrt $x$ we get

$f ' \left(x\right) = 6 x + 2$

Suppose the point we seek has coordinates $P \left(a , b\right)$, then as $P$ lies on our given curve we have:

$b = 3 {a}^{2} + 2 a + 1$ ..... [A]

The slope of the tangent line at $P$ is given by:

${m}_{T} = f ' \left(a\right) = 6 a + 2$

So, the slope of the normal line is given by:

${m}_{N} = - \frac{1}{6 a + 2}$

As the normal also passes through $\left(9 , 5\right)$ the equation of the normal using the point/slope form $y - {y}_{1} = m \left(x - {x}_{1}\right)$ is;

$y - 5 = - \frac{1}{6 a + 2} \left(x - 9\right)$ ..... [B]

This normal equation also passes through the point $P$ on the curve, and so:

$b - 5 = - \frac{1}{6 a + 2} \left(a - 9\right)$

Substituting [A] gives us:

$3 {a}^{2} + 2 a + 1 - 5 = - \frac{a - 9}{6 a + 2}$
$\therefore \left(6 a + 2\right) \left(3 {a}^{2} + 2 a - 4\right) = - a + 9$
$\therefore 18 {a}^{3} + 12 {a}^{2} - 24 a + 6 {a}^{2} + 4 a - 8 = - a + 9$
$\therefore 18 {a}^{3} + 18 {a}^{2} - 19 a - 17 = 0$

The challenge is to now solve this cubic for $a$. Fortunately, by observation, we see that $a = 1$ is a solution, hence by the factor theorem, $\left(x - 1\right)$ is a factor and thus we can can perform algebraic long division to form linear and quadratic factors, giving:

$\left(a - 1\right) \left(18 {a}^{2} + 36 a + 17\right) = 0$

Then using the quadratic formula, we gain the additional roots, and so we have:

$a = 1 , - 1 \pm \frac{1}{3 \sqrt{2}}$

Using [A] with these values of $a$ we get the corresponding values of $b$:

$a = \left\{\begin{matrix}1 \\ - 1 - \frac{1}{3 \sqrt{2}} \\ - 1 + \frac{1}{3 \sqrt{2}}\end{matrix}\right. \implies b = \left\{\begin{matrix}6 \\ \frac{13}{6} + \frac{2 \sqrt{2}}{3} \\ \frac{13}{6} - \frac{2 \sqrt{2}}{3}\end{matrix}\right.$

Hence there are three possible coordinates:

$\left(1 , 6\right)$
$\left(- 1 - \frac{1}{3 \sqrt{2}} , \frac{13}{6} + \frac{2 \sqrt{2}}{3}\right)$
$\left(- 1 + \frac{1}{3 \sqrt{2}} , \frac{13}{6} - \frac{2 \sqrt{2}}{3}\right)$

If we choose $\left(1 , 6\right)$ then using [B] the normal equation is:

$y - 5 = - \frac{1}{6 + 2} \left(x - 9\right)$
$\therefore 8 y - 40 = - x + 9$
$\therefore 8 y + x - 49 = 0$

We can confirm this graphically:

2

## Find the inverse Laplace of the (1)F(s)=1/(s²+4s+13)² and (2)F(s)=1/(s²+4)(s+1)² functions by using convolution theorem?

Steve M
Featured 3 days ago

 f_1(t) = ℒ^(-1){ 1/(s^2+4s+13)^2 }
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{54} \setminus {e}^{- 2 t} \left\{\sin \left(3 t\right) - 3 t \cos \left(3 t\right)\right\}$

 f_2(t) = ℒ^(-1){ 1/( (s^2+4)(s+1)^2 }
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = {e}^{- t} / 50 \left\{10 t + 4 - 3 \sin 2 t - 4 \cos 2 t\right\}$

#### Explanation:

We have:

A: ${F}_{1} \left(s\right) = \frac{1}{{s}^{2} + 4 s + 13} ^ 2$
B: ${F}_{2} \left(s\right) = \frac{1}{\left({s}^{2} + 4\right) {\left(s + 1\right)}^{2}}$

The Laplace Convolution Theorem tells us that if we define the convolution of two function $f \left(t\right)$ and $g \left(t\right)$ by:

$\left(f \star g\right) \left(t\right) = {\int}_{0}^{t} \setminus f \left(t - x\right) \setminus g \left(x\right) \setminus \mathrm{dx}$

Then:

 ℒ \ { (f star g)(t)} = F(s)G(s) iff ℒ^(-1) \ { F(s)G(s) }= (f star g)(t)

We will need the following standard Laplace transform and inverses:

 {: (ul(f(t)=ℒ^(-1){F(s)}), ul(F(s)=ℒ{f(t)}), ul("Notes")), (f(t), F(s),), (t^n, (n!)/(s^(n+1)),n in NN), (e^(at), 1/(s-a), a " constant"), (t^n e^(at), (n!)/(s-1)^(n+1), a " constant," nin NN), (sinat, a/(s^2+a^2), a " constant"), (e^(at)sinbt, b/((s-a)^2+b^2),a","b " constant") :}

======================================================

Part [A]:

We seek:

 f_1(t) = ℒ^(-1){ 1/(s^2+4s+13)^2 } = ℒ^(-1){ F(s)G(s) }

Where:

$F \left(s\right) = G \left(s\right) = \frac{1}{{s}^{2} + 4 s + 13}$

Completing the square, we can write as follows:

$F \left(s\right) = \frac{1}{{\left(s + 2\right)}^{2} - {2}^{2} + 13}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{{\left(s + 2\right)}^{2} + 9}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{{\left(s + 2\right)}^{2} + {3}^{2}}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{3} \setminus \frac{3}{{\left(s + 2\right)}^{2} + {3}^{2}}$

Then using the above table of transforms, we get:

$f \left(t\right) = g \left(t\right) = \frac{1}{3} {e}^{- 2 t} \sin 3 t$

So applying the convolution theorem, we have:

${f}_{1} \left(t\right) = \left(f \star g\right) \left(t\right) = {\int}_{0}^{t} \setminus f \left(t - x\right) \setminus g \left(x\right) \setminus \mathrm{dx}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {\int}_{0}^{t} \setminus \frac{1}{3} {e}^{- 2 \left(t - x\right)} \sin 3 \left(t - x\right) \setminus \frac{1}{3} {e}^{- 2 x} \sin \left(3 x\right) \setminus \mathrm{dx}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{9} \setminus {\int}_{0}^{t} \setminus {e}^{- 2 t + 2 x - 2 x} \setminus \sin \left(3 t - 3 x\right) \setminus \sin \left(3 x\right) \setminus \mathrm{dx}$

Now we apply the trigonometric identity:

$2 \sin A \sin B = \cos \left(A - B\right) - \cos \left(A + B\right)$

Then we have:

${f}_{1} = \frac{1}{9} {e}^{- 2 t} \setminus {\int}_{0}^{t} \setminus \frac{\cos \left(3 t - 6 x\right) - \cos \left(3 t\right)}{2} \setminus \mathrm{dx}$

$\setminus \setminus = \frac{1}{18} {e}^{- 2 t} {\left[\sin \frac{6 x - 3 t}{6} - x \cos \left(3 t\right)\right]}_{0}^{t}$

$\setminus \setminus = \frac{1}{18} {e}^{- 2 t} \left\{\left(\sin \frac{6 t - 3 t}{6} - t \cos \left(3 t\right)\right) - \left(\sin \frac{- 3 t}{6} - 0\right)\right\}$

$\setminus \setminus = \frac{1}{18} {e}^{- 2 t} \left\{\sin \frac{3 t}{6} - t \cos \left(3 t\right) - \sin \frac{- 3 t}{6}\right\}$

$\setminus \setminus = \frac{1}{18} {e}^{- 2 t} \left\{\sin \frac{3 t}{6} - t \cos \left(3 t\right) + \sin \frac{3 t}{6}\right\}$

$\setminus \setminus = \frac{1}{18} {e}^{- 2 t} \left\{\sin \frac{3 t}{3} - t \cos \left(3 t\right)\right\}$

$\setminus \setminus = \frac{1}{18} \cdot \frac{1}{3} \setminus {e}^{- 2 t} \left\{\sin \left(3 t\right) - 3 t \cos \left(3 t\right)\right\}$

$\setminus \setminus = \frac{1}{54} \setminus {e}^{- 2 t} \left\{\sin \left(3 t\right) - 3 t \cos \left(3 t\right)\right\}$

======================================================

Part [B]:

We seek:

 f_2(t) = ℒ^(-1){ 1/( (s^2+4)(s+1)^2 ) } = ℒ^(-1){ F(s)G(s) }

Where:

$F \left(s\right) = \frac{1}{s + 1} ^ 2$; and $G \left(s\right) = \frac{1}{{s}^{2} + 4}$

Then using the above table of transforms, we get:

$F \left(s\right) = \frac{1}{s + 1} ^ 2 \setminus \setminus \setminus \implies f \left(t\right) = t {e}^{- t}$
$G \left(s\right) = \frac{1}{2} \setminus \frac{2}{{s}^{2} + {2}^{2}} \implies g \left(t\right) = \frac{1}{2} \sin \left(2 t\right)$

So applying the convolution theorem, we have:

${f}_{2} \left(t\right) = \left(f \star g\right) \left(t\right) = {\int}_{0}^{t} \setminus f \left(t - x\right) \setminus g \left(x\right) \setminus \mathrm{dx}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = {\int}_{0}^{t} \setminus \left(t - x\right) {e}^{- \left(t - x\right)} \setminus \frac{1}{2} \setminus \sin \left(2 x\right) \setminus \mathrm{dx}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{2} \setminus {\int}_{0}^{t} \setminus \left(t - x\right) {e}^{x - t} \setminus \sin \left(2 x\right) \setminus \mathrm{dx}$

Due to the length of the solution thus far, I will omit the derivation of the integral, and quote the result:

${f}_{2} = \frac{1}{2} {\left[- {e}^{x - t} \left\{\frac{\left(5 x - 5 t + 3\right) \sin 2 x}{25} + \frac{\left(- 10 x + 10 t + 4\right) \cos 2 x}{25}\right\}\right]}_{0}^{t}$

$\setminus \setminus = {\left[- {e}^{x - t} / 50 \left\{\left(5 x - 5 t + 3\right) \sin 2 x + \left(- 10 x + 10 t + 4\right) \cos 2 x\right\}\right]}_{0}^{t}$

$\setminus \setminus = - {e}^{0} / 50 \left\{3 \sin 2 t + 4 \cos 2 t\right\} + {e}^{- t} / 50 \left\{\left(- 5 t + 3\right) \sin 0 + \left(10 t + 4\right) \cos 0\right\}$

$\setminus \setminus = {e}^{- t} / 50 \left\{10 t + 4 - 3 \sin 2 t - 4 \cos 2 t\right\}$

2

## Calculate the volume of the area by rotating the given region about the y axis. Use shell thrm. ?? x=2y, x=y^2 , 0 <= y <= 2

Frederico Guizini S.
Featured 3 days ago

2

## Riemann integration of sinx in [0,π/2]?

Steve M
Featured 3 days ago

${\int}_{0}^{\frac{\pi}{2}} \setminus \sin x \setminus \mathrm{dx} = 1$

#### Explanation:

By definition of an integral, then

${\int}_{a}^{b} \setminus f \left(x\right) \setminus \mathrm{dx}$

represents the area under the curve $y = f \left(x\right)$ between $x = a$ and $x = b$. We can estimate this area under the curve using thin rectangles. The more rectangles we use, the better the approximation gets, and calculus deals with the infinite limit of a finite series of infinitesimally thin rectangles.

That is

${\int}_{a}^{b} \setminus f \left(x\right) \setminus \mathrm{dx} = {\lim}_{n \rightarrow \infty} \frac{b - a}{n} {\sum}_{i = 1}^{n} \setminus f \left(a + i \frac{b - a}{n}\right)$

And we partition the interval $\left[a , b\right]$ equally spaced using:

$\Delta = \left\{a + 0 \left(\frac{b - a}{n}\right) , a + 1 \left(\frac{b - a}{n}\right) , \ldots , a + n \left(\frac{b - a}{n}\right)\right\}$
$\setminus \setminus \setminus = \left\{a , a + 1 \left(\frac{b - a}{n}\right) , a + 2 \left(\frac{b - a}{n}\right) , \ldots , b\right\}$

Here we have $f \left(x\right) = \sin x$ and so we partition the interval $\left[0 , \frac{\pi}{2}\right]$ using:

$\Delta = \left\{0 , 0 + 1 \frac{\frac{\pi}{2}}{n} , 0 + 2 \frac{\frac{\pi}{2}}{n} , 0 + 3 \frac{\frac{\pi}{2}}{n} , \ldots , \frac{\pi}{2}\right\}$

And so:

$I = {\int}_{0}^{\frac{\pi}{2}} \setminus \sin x \setminus \mathrm{dx}$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{\frac{\pi}{2}}{n} {\sum}_{i = 1}^{n} \setminus f \left(0 + i \cdot \frac{\frac{\pi}{2}}{n}\right)$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{\pi}{2 n} {\sum}_{i = 1}^{n} \setminus \sin \left(\frac{i \pi}{2 n}\right)$ ..... [A}

If we were dealing with polynomials, we would now utilise standard summation formulas for powers, but as we are dealing with a sine summation those results will not help.

In order to evaluate the sine sum, We consider the following:

$1 + z + {z}^{2} + \ldots + {z}^{n} = \frac{{z}^{n + 1} - 1}{z - 1} \setminus \setminus$ for z $\ne 1$

in combination with Euler's formula by taking

$z = {e}^{i \theta} = \cos \theta + i \sin \theta$

Then applying De Moivre's theorem:

${e}^{i n \theta} = \cos n \theta + i \sin n \theta$

and equating real and imaginary parts, we eventually find that:

 sum_(j=1)^n sin(jtheta) = (cos(theta/2)-cos(n+1/2)theta) / (2sin(theta/2)
 " " = (cos(theta/2)-cos(ntheta+theta/2)) / (2sin(theta/2)
 " " = (cos(theta/2)-(cos ntheta cos(theta/2) - sinnthetasin(theta/2))) / (2sin(theta/2)

$\text{ } = \frac{\cos \left(\frac{\theta}{2}\right) - \cos n \theta \cos \left(\frac{\theta}{2}\right) + \sin n \theta \sin \left(\frac{\theta}{2}\right)}{2 \sin \left(\frac{\theta}{2}\right)}$

So, if we put $\theta = \frac{\pi}{2 n}$ and use this result in the earlier summation [A], we get:

$I = {\lim}_{n \rightarrow \infty} \frac{\pi}{2 n} \left\{\frac{\cos \left(\frac{\pi}{4 n}\right) - \cos \left(\frac{\pi}{2}\right) \cos \left(\frac{\pi}{4 n}\right) + \sin \left(\frac{\pi}{2}\right) \sin \left(\frac{\pi}{4 n}\right)}{2 \sin \left(\frac{\pi}{4 n}\right)}\right\}$

$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{\pi}{2 n} \left\{\frac{\cos \left(\frac{\pi}{4 n}\right) + \sin \left(\frac{\pi}{4 n}\right)}{2 \sin \left(\frac{\pi}{4 n}\right)}\right\}$

$\setminus \setminus = \frac{\pi}{4} \setminus {\lim}_{n \rightarrow \infty} \frac{1}{n} \left(\cot \left(\frac{\pi}{4 n}\right) + 1\right)$

$\setminus \setminus = \frac{\pi}{4} \left\{{\lim}_{n \rightarrow \infty} \frac{1}{n} \left(\frac{1}{\tan} \left(\frac{\pi}{4 n}\right)\right) + {\lim}_{n \rightarrow \infty} \frac{1}{n}\right\}$

$\setminus \setminus = \frac{\pi}{4} \left\{{\lim}_{n \rightarrow \infty} \frac{1}{n} \left(\frac{\frac{\pi}{4 n}}{\tan} \left(\frac{\pi}{4 n}\right)\right) \cdot \frac{4 n}{\pi} + 0\right\}$

$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{\frac{\pi}{4 n}}{\tan} \left(\frac{\pi}{4 n}\right)$

And a standard calculus limit is:

${\lim}_{x \rightarrow 0} \frac{x}{\tan} x = 1$

Utilising this result with $x = \frac{\pi}{4 n}$ we get:

$I = 1$

Using Calculus

If we use Calculus and our knowledge of integration to establish the answer, for comparison, we get:

${\int}_{0}^{\frac{\pi}{2}} \setminus \sin x \setminus \mathrm{dx} = {\left[- \cos\right]}_{0}^{\frac{\pi}{2}}$
$\text{ } = - \left(\cos \left(\frac{\pi}{2}\right) - \cos 0\right)$
$\text{ } = - \left(0 - 1\right)$
$\text{ } = 1$

2

## How do you show that the function g(x)=sqrt(x^2-9)/(x^2-2) is continuous on its domain and what is the domain?

Frederico Guizini S.
Featured 3 days ago

1

## How do you find dy/dx by implicit differentiation given xy^2-3x^2y+x=1?

Featured yesterday

The answer is $= \frac{\left(6 x y - {y}^{2} - 1\right)}{\left(2 y x - 3 {x}^{2}\right)}$

#### Explanation:

We need

$\left(u v\right) ' = u ' v + u v '$

The function is

$x {y}^{2} - 3 {x}^{2} y + x - 1 = 0$

We differentiate with respect to x

$\frac{d}{\mathrm{dx}} \left(x {y}^{2}\right) - \frac{d}{\mathrm{dx}} \left(3 {x}^{2} y\right) + \frac{d}{\mathrm{dx}} \left(x\right) - \frac{d}{\mathrm{dx}} \left(1\right) = 0$

${y}^{2} + 2 y x \frac{\mathrm{dy}}{\mathrm{dx}} - 6 x y - 3 {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + 1 = 0$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(2 y x - 3 {x}^{2}\right) = 6 x y - {y}^{2} - 1$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(6 x y - {y}^{2} - 1\right)}{\left(2 y x - 3 {x}^{2}\right)}$

3

## Suppose f and g are differentiable and increasing functions. How do I show that if f(x)<=0 and g(x)>0 , then h(x)=f(x)/g(x) is increasing, differentiable and negative function?

Zack M.
Featured 2 days ago

We are given;

$f \left(x\right) \le 0$
$g \left(x\right) > 0$
$h \left(x\right) = f \frac{x}{g} \left(x\right)$

The first states that $f \left(x\right)$ is negative or zero, the second that $g \left(x\right)$ is strictly positive, and the third establishes a new function, $h \left(x\right)$ which is a ratio of the first two. We are also told that both are differentiable and increasing.

If a function is increasing, then we can infer that its derivative function must be positive. This can be written as;

$f ' \left(x\right) > 0$
$g ' \left(x\right) > 0$

In order to show that $h \left(x\right)$ is increasing, we must show that $h ' \left(x\right)$ is also positive. We can find $h ' \left(x\right)$ using the quotient rule.

$h ' \left(x\right) = \frac{f ' \left(x\right) g \left(x\right) - f \left(x\right) g ' \left(x\right)}{{g}^{2} \left(x\right)}$

With the information we have, only one term is potentially negative, $f \left(x\right)$, and $f \left(x\right)$ appears only once, in the subtracted term in the numerator. When subtracting a negative, the sign becomes positive, so the numerator is therefore positive. The denominator must also be positive, as $g \left(x\right)$ is strictly positive, so $h ' \left(x\right)$ is positive. Notice that $f \left(x\right)$ can also be 0, in which case, $h ' \left(x\right)$ is still positive. $h \left(x\right)$ is therefore increasing.

Of course, we already made the assumption that $h \left(x\right)$ is differentiable. Returning to $h ' \left(x\right)$, the denominator term is strictly positive, so we don't have to worry about any 0's there. Furthermore, we are told that $f \left(x\right)$ and $g \left(x\right)$ are differentiable, so they must also be continuous, and $f ' \left(x\right)$ and $g ' \left(x\right)$ exist. Therefore, $h ' \left(x\right)$ exists, so $h \left(x\right)$ is differentiable.

Lastly, $h \left(x\right)$ is indeed negative for all points where it is not 0.
The numerator $f \left(x\right)$ is negative or 0, and the denominator, $g \left(x\right)$, is strictly positive, and a ratio of a positive and a negative is always negative.

2

## How do you find the average rate of change of g(x)=1/(x-2) over [x, x+h] ?

George C.
Featured yesterday

$- \frac{1}{\left(x - 2\right) \left(x - 2 + h\right)}$

#### Explanation:

The average rate of change is the slope of the secant between:

$\left(x , g \left(x\right)\right)$ and $\left(x + h , g \left(x + h\right)\right)$

The slope $m$ of a line between two points $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ is given by the formula:

$m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$

So the slope of the secant is:

$\frac{g \left(x + h\right) - g \left(x\right)}{\left(x + h\right) - x} = \frac{g \left(x + h\right) - g \left(x\right)}{h}$

So given:

$g \left(x\right) = \frac{1}{x - 2}$

its average rate of change over $\left[x . x + h\right]$ is:

$\frac{\frac{1}{\textcolor{b l u e}{\left(x + h\right)} - 2} - \frac{1}{\textcolor{b l u e}{x} - 2}}{h} = \frac{\left(\textcolor{red}{\cancel{\textcolor{b l a c k}{x}}} - \textcolor{b r o w n}{\cancel{\textcolor{b l a c k}{2}}}\right) - \left(\textcolor{red}{\cancel{\textcolor{b l a c k}{x}}} + h - \textcolor{b r o w n}{\cancel{\textcolor{b l a c k}{2}}}\right)}{h \left(x - 2\right) \left(x + h - 2\right)}$

$\textcolor{w h i t e}{\frac{\frac{1}{\left(\left(x + h\right)\right) - 2} - \frac{1}{\left(x\right) - 2}}{h}} = \frac{- \textcolor{red}{\cancel{\textcolor{b l a c k}{h}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{h}}} \left(x - 2\right) \left(x + h - 2\right)}$

$\textcolor{w h i t e}{\frac{\frac{1}{\left(\left(x + h\right)\right) - 2} - \frac{1}{\left(x\right) - 2}}{h}} = - \frac{1}{\left(x - 2\right) \left(x - 2 + h\right)}$

In particular, we find:

${\lim}_{h \to 0} \frac{g \left(x + h\right) - g \left(x\right)}{h} = - \frac{1}{x - 2} ^ 2$

This is the slope of $g \left(x\right)$ at $x$