# Math Help - help with some problems?

1. ## help with some problems?

thanks ?

2. 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.

3. 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.

4. 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.

5. 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

6. 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 $\zeta \in [0,\pi]$.

7. Hello, sbsite!

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

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

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

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