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- Dec 14th 2006, 11:07 PMsbsitehelp with some problems?
thanks ?

- Dec 15th 2006, 08:58 AMThePerfectHacker
We need to find the limits,

$\displaystyle \lim_{x\to 0} x\sin \frac{1}{x}$

We can write,

$\displaystyle -1\leq \sin \frac{1}{x} \leq 1$

Multiply by "x" (positive or negative equality is preseved),

$\displaystyle -x\leq x\sin \frac{1}{x} \leq x$.

This inequality is true at some open interval containing zero except possibly at zero itself. Thus, by the squeeze theorem,

$\displaystyle x\to 0,-x\to 0$.

Thus, the limit of that is also zero.

That means the function is continous.

Similarly,

$\displaystyle \lim_{x\to 0}x^2\cos (1/x)=0$

Thus, it is continous. - Dec 15th 2006, 09:06 AMThePerfectHacker
Now we need to determine the derivative of $\displaystyle f,g$ at $\displaystyle x=0$.

The derivative (if it exists) for $\displaystyle f$ is given by,

$\displaystyle \lim_{\Delta x\to 0}\frac{(0+\Delta x)\sin (1/(0+\Delta x)) - f(0)}{\Delta x}$

Which is equivalent to after simplification,

$\displaystyle \lim_{\Delta x\to 0}\sin \frac{1}{\Delta x}$.

This is a famous limits, which does not exist.

However, if we used the second function after some manipulation, we would eventually reach after doing the same steps above,

$\displaystyle \lim_{\Delta x\to 0}\Delta x\cos \frac{1}{\Delta x}=0$

By using the Squeeze theorem trick employed before.

Thus, $\displaystyle f$ is not differenciable at $\displaystyle x=0$. While $\displaystyle g$ is differenciable at $\displaystyle x=0$. That immediatly answers the (c) the first function's derivative is not defined at the point, thus it is not continous there. - Dec 15th 2006, 09:11 AMThePerfectHacker
The only thing remains is to show that the derivative of $\displaystyle g$ is continous at $\displaystyle x=0$.

If $\displaystyle x\not =0$ then the derivative of $\displaystyle g$ exists and is,

$\displaystyle g'=(x^2)'\cos \frac{1}{x}+x^2\left( \cos \frac{1}{x} \right)'$

$\displaystyle g'=2x\cos \frac{1}{x}+x^2\left(\frac{1}{x^2} \right) \sin \frac{1}{x}$

$\displaystyle g'=2x\cos \frac{1}{x}+\sin \frac{1}{x}$.

I do not think the limit $\displaystyle x\to 0$ exists. Because the first summand exists (by squeeze theorem) but the second one does not.

Thus, the derivative is not continous. - Dec 15th 2006, 01:47 PMCaptainBlack
$\displaystyle

f(x)=\left\{ {e^{-1/x^2} \ \ ,x \ne0 \atop 0 \ \ \ \ \ \ \ ,x=0} \right.

$

This is a well known function which has the curious property that all its

derivatives at $\displaystyle x=0$ are $\displaystyle 0$. Showing this depends on showing that:

$\displaystyle

\lim_{x \to \infty} x^n e^{-x^2}=0

$

for all $\displaystyle n>0$, which I will leave as an excercise to the reader,

it can be shown by repeated application of L'Hopital's rule. Alternativly you may

be in the position of knowing that $\displaystyle e^{-x}$ decreases faster than any power of

$\displaystyle x$, which is also sufficient to justify the above limit.

$\displaystyle

f'(x)=\frac{2}{x^3}e^{-1/x^2},\ x \ne 0

$

and:

$\displaystyle

f'(0)=\lim_{x \to 0} \frac{2}{x^3}e^{-1/x^2}

$

Put $\displaystyle u=1/x$, then:

$\displaystyle

f'(0)=\lim_{u \to \infty} 2\, u^3\,e^{-u^2}=0

$

The second derivative is similar, but a bit more fiddly.

RonL - Dec 15th 2006, 02:03 PMCaptainBlack
Problem 20, show:

$\displaystyle

x-(x^3/6) < \sin(x) < x,\ \ \ 0<x<\pi

$

This is a result of taking the Taylor expansion of $\displaystyle \sin(x)$ up to the term in $\displaystyle x^2$, and keeping the remainder term in:

$\displaystyle

\sin(x)=\sin(0)+\cos(0)x-\sin(0)x^2/2!-\cos(\zeta)x^3/3!,\ \ \ \zeta \in (0,\pi)

$

$\displaystyle

\sin(x)=x-\cos(\zeta)x^3/3!,\ \ \ \zeta \in (0,\pi)

$

Now the required inequality follows by using the upper and lower bounds on $\displaystyle \cos(\zeta)$ for $\displaystyle \zeta \in [0,\pi]$. - Dec 16th 2006, 02:54 AMSoroban
Hello, sbsite!

Here's #11 . . . but check my work,*please.*

Quote:

11) Find $\displaystyle \frac{dy}{dx}\bigg|_{x=0}$ for: .$\displaystyle y \;=\;\left[\ln(3x+2)\right]^{\sin^{-1}(2x+0.5)}$

Take logs: .$\displaystyle \ln(y) \;=\;\ln\bigg\{\left[\ln(3x+2)\right]^{\sin^{-1}(2x+0.5)}\bigg\}\;=\;\sin^{-1}(2x+0.5)\cdot\ln\left[\ln(3x+3)\right]

$

Differentiate implicitly:

. . $\displaystyle \frac{1}{y}\left(\frac{dy}{dx}\right)\:=\:\sin^{-1}(2x+0.5)\cdot\frac{1}{\ln(3x+2)}\cdot\frac{1}{3x +2}\cdot3$$\displaystyle + \frac{2}{\sqrt{1 - (2x+0.5)^2}}\cdot\ln\left[\ln(3x+2)\right] $

We have: .$\displaystyle \frac{dy}{dx}\;=\;y\bigg\{\frac{3\sin^{-1}(2x+0.5)}{(3x+2)\ln(3x+2)} + \frac{2\ln\left[\ln(3x+2)\right]}{\sqrt{1 - (2x+0.5)^2}}\bigg\}

$

When $\displaystyle x = 0\!:\;y \:=\:\left(\ln 2\right)^{\sin^{-1}(0.5)} \:=\:\left(\ln 2\right)^{\frac{\pi}{6}} $

Hence: .$\displaystyle \frac{dy}{dx}\:=\:(\ln 2)^{\frac{\pi}{6}}\left[\frac{3\cdot\sin^{-1}(0.5)}{2\cdot\ln2} + \frac{2\ln(\ln 2)}{\sqrt{1 - (0.5)^2}}\right] $

. . . . . . . . $\displaystyle =\:(\ln 2)^{\frac{\pi}{6}}\left[\frac{3\cdot\frac{\pi}{6}}{2\ln 2} + \frac{2\ln(\ln 2)}{\sqrt{1-0.25}}\right]$

. . . . . . . . $\displaystyle = \:(\ln2)^{\frac{\pi}{6}}\left[\frac{\pi}{4\ln2} + \frac{2\ln(\ln2)}{\sqrt{0.75}}\right] $