# help with some problems?

• Dec 15th 2006, 12:07 AM
sbsite
help with some problems?
thanks ?
• Dec 15th 2006, 09:58 AM
ThePerfectHacker
We need to find the limits,
$\lim_{x\to 0} x\sin \frac{1}{x}$
We can write,
$-1\leq \sin \frac{1}{x} \leq 1$
Multiply by "x" (positive or negative equality is preseved),
$-x\leq x\sin \frac{1}{x} \leq x$.
This inequality is true at some open interval containing zero except possibly at zero itself. Thus, by the squeeze theorem,
$x\to 0,-x\to 0$.
Thus, the limit of that is also zero.
That means the function is continous.

Similarly,
$\lim_{x\to 0}x^2\cos (1/x)=0$
Thus, it is continous.
• Dec 15th 2006, 10:06 AM
ThePerfectHacker
Now we need to determine the derivative of $f,g$ at $x=0$.

The derivative (if it exists) for $f$ is given by,
$\lim_{\Delta x\to 0}\frac{(0+\Delta x)\sin (1/(0+\Delta x)) - f(0)}{\Delta x}$
Which is equivalent to after simplification,
$\lim_{\Delta x\to 0}\sin \frac{1}{\Delta x}$.
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,
$\lim_{\Delta x\to 0}\Delta x\cos \frac{1}{\Delta x}=0$
By using the Squeeze theorem trick employed before.

Thus, $f$ is not differenciable at $x=0$. While $g$ is differenciable at $x=0$. 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, 10:11 AM
ThePerfectHacker
The only thing remains is to show that the derivative of $g$ is continous at $x=0$.

If $x\not =0$ then the derivative of $g$ exists and is,
$g'=(x^2)'\cos \frac{1}{x}+x^2\left( \cos \frac{1}{x} \right)'$
$g'=2x\cos \frac{1}{x}+x^2\left(\frac{1}{x^2} \right) \sin \frac{1}{x}$
$g'=2x\cos \frac{1}{x}+\sin \frac{1}{x}$.
I do not think the limit $x\to 0$ exists. Because the first summand exists (by squeeze theorem) but the second one does not.

Thus, the derivative is not continous.
• Dec 15th 2006, 02:47 PM
CaptainBlack
Quote:

Originally Posted by sbsite
thanks ?

$
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 $x=0$ are $0$. Showing this depends on showing that:

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

for all $n>0$, 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 $e^{-x}$ decreases faster than any power of
$x$, which is also sufficient to justify the above limit.

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

and:

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

Put $u=1/x$, then:

$
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, 03:03 PM
CaptainBlack
Quote:

Originally Posted by sbsite
thanks ?

Problem 20, show:

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

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

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

$
\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 $\cos(\zeta)$ for $\zeta \in [0,\pi]$.
• Dec 16th 2006, 03:54 AM
Soroban
Hello, sbsite!

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

Quote:

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

Take logs: . $\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:

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

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

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

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

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