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

Jim H
Featured 4 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 $1$.

## Integrate (cos^3 x +cos ^5 x)/(sin^2 x+sin^4 x)?

Eric Sia
Featured 2 weeks ago

$\int \frac{{\cos}^{3} x + {\cos}^{5} x}{{\sin}^{2} x + {\sin}^{4} x} \mathrm{dx} = \sin x - 2 \csc x - 6 {\tan}^{- 1} \left(\sin x\right) + C$

#### Explanation:

Let

$I = \int \frac{{\cos}^{3} x + {\cos}^{5} x}{{\sin}^{2} x + {\sin}^{4} x} \mathrm{dx}$

Factor out a $\cos x$:

$I = \int \left(\frac{{\cos}^{2} x + {\cos}^{4} x}{{\sin}^{2} x + {\sin}^{4} x}\right) \cdot \cos x \mathrm{dx}$

Rewrite in terms of $\sin x$:

$I = \int \left(\frac{2 - 3 {\sin}^{2} x + {\sin}^{4} x}{{\sin}^{2} x + {\sin}^{4} x}\right) \cdot \cos x \mathrm{dx}$

For ease of reading, apply the substitution $\sin x = u$:

$I = \int \frac{2 - 3 {u}^{2} + {u}^{4}}{{u}^{2} + {u}^{4}} \mathrm{du}$

Apply long division:

$I = \int \left(1 + \frac{2 - 4 {u}^{2}}{{u}^{2} \left(1 + {u}^{2}\right)}\right) \mathrm{du}$

Apply partial fraction decomposition:

$I = \int \left(1 + \frac{2}{u} ^ 2 - \frac{6}{1 + {u}^{2}}\right) \mathrm{du}$

Integrate term by term:

$I = u - \frac{2}{u} - 6 {\tan}^{- 1} \left(u\right) + C$

Reverse the substitution:

$I = \sin x - 2 \csc x - 6 {\tan}^{- 1} \left(\sin x\right) + C$

## What is the slope of the polar curve f(theta) = theta^2-theta - cos^3theta + tan^2theta at theta = pi/4?

mason m
Featured 1 week ago

The slope is: $\frac{{\pi}^{2} + 4 \pi + 8 \sqrt{2} + 64}{{\pi}^{2} + 12 \pi + 16 \sqrt{2} + 32} \approx 0.9564910$

#### Explanation:

The slope will be given by $\frac{\mathrm{dy}}{\mathrm{dx}}$ of the curve evaluated at $\theta = \frac{\pi}{4}$.

To get the equation in terms of $x$ and $y$, we should use the polar-to-rectangular identities: $\left\{\begin{matrix}x = r \cos \theta \\ y = r \sin \theta\end{matrix}\right.$

So here, where $r = f \left(\theta\right)$, we have:

$x = \cos \theta \left({\theta}^{2} - \theta - {\cos}^{3} \theta + {\tan}^{2} \theta\right)$

$\textcolor{w h i t e}{x} = {\theta}^{2} \cos \theta - \theta \cos \theta - {\cos}^{4} \theta + \tan \theta \sin \theta$

$y = \sin \theta \left({\theta}^{2} - \theta - {\cos}^{3} \theta + {\tan}^{2} \theta\right)$

$\textcolor{w h i t e}{y} = {\theta}^{2} \sin \theta - \theta \sin \theta - {\cos}^{3} \theta \sin \theta + \sin \theta {\tan}^{2} \theta$

To find $\frac{\mathrm{dy}}{\mathrm{dx}}$, we first need to find $\frac{\mathrm{dx}}{d \theta}$ and $\frac{\mathrm{dy}}{d \theta}$. This will take a fair amount of product and chain rule, so be careful.

$\frac{\mathrm{dx}}{d \theta} = 2 \theta \cos \theta - {\theta}^{2} \sin \theta - \cos \theta + \theta \sin \theta + 4 {\cos}^{3} \theta \sin \theta + {\sec}^{2} \theta \sin \theta + \tan \theta \cos \theta$

$\textcolor{w h i t e}{\frac{\mathrm{dx}}{d \theta}} = 2 \theta \cos \theta - {\theta}^{2} \sin \theta - \cos \theta + \theta \sin \theta + 4 {\cos}^{3} \theta \sin \theta + \tan \theta \sec \theta + \sin \theta$

$\frac{\mathrm{dy}}{d \theta} = 2 \theta \sin \theta + {\theta}^{2} \cos \theta - \sin \theta - \theta \cos \theta + 3 {\cos}^{2} \theta {\sin}^{2} \theta - {\cos}^{4} \theta + \cos \theta {\tan}^{2} \theta + 2 \sin \theta \tan \theta {\sec}^{2} \theta$

$\textcolor{w h i t e}{\frac{\mathrm{dy}}{d \theta}} = 2 \theta \sin \theta + {\theta}^{2} \cos \theta - \sin \theta - \theta \cos \theta + 3 {\cos}^{2} \theta {\sin}^{2} \theta - {\cos}^{4} \theta + \sin \theta \tan \theta + 2 {\tan}^{2} \theta \sec \theta$

Now we can find $\frac{\mathrm{dy}}{\mathrm{dx}}$ using $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy} / d \theta}{\mathrm{dx} / d \theta}$.

Just so we don't have to write out a giant fraction, let's evaluate $\frac{\mathrm{dx}}{d \theta}$ and $\frac{\mathrm{dy}}{d \theta}$ at $\theta = \frac{\pi}{4}$ separately.

$\frac{\mathrm{dx}}{d \theta} = 2 \frac{\pi}{4} \frac{1}{\sqrt{2}} + {\pi}^{2} / 16 \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} + \frac{\pi}{4} \frac{1}{\sqrt{2}} + 4 \frac{1}{2 \sqrt{2}} \frac{1}{\sqrt{2}} + 2 \frac{1}{\sqrt{2}} + 1 \frac{1}{\sqrt{2}}$

$\textcolor{w h i t e}{\frac{\mathrm{dx}}{d \theta}} = \frac{\pi}{2 \sqrt{2}} + {\pi}^{2} / \left(16 \sqrt{2}\right) + \frac{\pi}{4 \sqrt{2}} + 1 + \sqrt{2}$

$\textcolor{w h i t e}{\frac{\mathrm{dx}}{d \theta}} = \frac{1}{16 \sqrt{2}} \left({\pi}^{2} + 12 \pi + 16 \sqrt{2} + 32\right)$

$\frac{\mathrm{dy}}{d \theta} = 2 \frac{\pi}{4} \frac{1}{\sqrt{2}} + {\pi}^{2} / 16 \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} - \frac{\pi}{4} \frac{1}{\sqrt{2}} + 3 \frac{1}{2} \frac{1}{2} - \frac{1}{4} + \frac{1}{\sqrt{2}} 1 + 2 \cdot 1 \sqrt{2}$

$\textcolor{w h i t e}{\frac{\mathrm{dy}}{d \theta}} = \frac{\pi}{2 \sqrt{2}} + {\pi}^{2} / \left(16 \sqrt{2}\right) - \frac{\pi}{4 \sqrt{2}} + \frac{1}{2} + 2 \sqrt{2}$

$\textcolor{w h i t e}{\frac{\mathrm{dy}}{d \theta}} = \frac{1}{16 \sqrt{2}} \left({\pi}^{2} + 4 \pi + 8 \sqrt{2} + 64\right)$

Thus, at $\theta = \frac{\pi}{4}$, the slope is:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy} / d \theta}{\mathrm{dx} / d \theta} = \frac{{\pi}^{2} + 4 \pi + 8 \sqrt{2} + 64}{{\pi}^{2} + 12 \pi + 16 \sqrt{2} + 32} \approx 0.9564910$

## How do you find the critical points of k'(x)=x^3-3x^2-18x+40?

Michael F.
Featured 6 days ago

To find the critical points we have to examine three items as follows.
Besides we need the first (given) derivative and the second derivative of the function k.

Remarks:

${k}^{'} \left(x\right)$ means the term of the first derivative of k, ${k}^{' '} \left(x\right)$ is the term of the second derivative of k.

In other "words":

$\frac{d}{\mathrm{dx}} k \left(x\right) = {k}^{'} \left(x\right)$ and ${d}^{2} / {\mathrm{dx}}^{2} k \left(x\right) = {k}^{' '} \left(x\right)$.

i) Necessary condition: ${k}^{'} \left(x\right) = 0$

Using polynomial long division (taking some time to find $x = 2$ that fulfils ${k}^{'} \left(x\right) = 0$) we get

$\left({x}^{3} - 3 {x}^{2} - 18 x + 40\right) : \left(x - 2\right) = {x}^{2} - x - 20$
$- \left({x}^{3} - 2 {x}^{2}\right)$
$= - {x}^{2} - 18 x$
$- \left(- {x}^{2} + 2 x\right)$
$= - 20 x + 40$
$- \left(- 20 x + 40\right)$
$= 0$

Now using the pq-formula for quadratic equations:

${x}^{2} - x - 20 = 0$
$\implies x = \frac{1}{2} \pm \sqrt{\frac{1}{4} + 20} = \frac{1}{2} \pm \sqrt{\frac{81}{4}} = \frac{1}{2} \pm \frac{9}{2}$
$\implies x = - 4 \vee x = 5$.

ii) Necessary and sufficient condition: ${k}^{'} \left(x\right) = 0 \wedge {k}^{' '} \left(x\right) \ne 0$.
${k}^{' '} \left(- 4\right) = 54 > 0$, so minimum at $x = - 4$,
${k}^{' '} \left(2\right) = - 18 < 0$, so maximum at $x = 2$,
${k}^{' '} \left(5\right) = 27 > 0$, so minimum at $x = 5$.

iii) y - values and points

Provided that $k \left(x\right) = {x}^{4} / 4 - {x}^{3} - 9 {x}^{2} + 40 x$ we get

$k \left(- 4\right) = - 176$, so the first mimimum point is M_1 (- 4 ; - 176) ,

$k \left(2\right) = 40$, so maximum point is M (2 ; 40) ,

$k \left(5\right) = \frac{25}{4}$, so the second mimimum point is M_2 (5 ; 25/4) .

For other equations of the function k there are other values and therefore other points.

I hope it helps.

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

Steve M
Featured 4 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:

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

## 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 4 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.

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