# Math Help - prove this is differentiable everywhere

1. ## prove this is differentiable everywhere

f(x) = p(1/x) * e^(-1/x^2) , x/=0
0 , x=0

p(t) is a nonzero polynomial in t

prove this is differentiable everywhere, i cant prove this limit::

limit( (p(1/h)*e^(-1/h^2))/h, h goes to 0)

2. Originally Posted by szpengchao
f(x) = p(1/x) * e^(-1/x^2) , x/=0
0 , x=0

p(t) is a nonzero polynomial in t

prove this is differentiable everywhere, i cant prove this limit::

limit( (p(1/h)*e^(-1/h^2))/h, h goes to 0)
Do you mean $\lim_{h\to{0}}p\bigg(\frac{1}{h}\bigg)\cdot\frac{e ^{\frac{-1}{h^2}}}{h}$?

Alternatively you could consider the fact that since e^x is analytic everywhere it is thus differentiable everywhere and work at it from the power series point of view

3. ## yes

yes. what i did is : show f(x) is differentiable when x/=0,
now i m trying to show that limit =0, therefore, the whole function is differentiable.

please show how to do that limit..

4. Originally Posted by szpengchao
yes. what i did is : show f(x) is differentiable when x/=0,
now i m trying to show that limit =0, therefore, the whole function is differentiable.

please show how to do that limit..
Did you try using Maclaurin series on the limit...Besides that there is nothing I can readily think of...I will report back if I do

5. ## ok

ok....i m waiting for u

6. $\lim_{h\to{0}}p \left ( \frac{1}{h} \right ) \cdot \frac{e^{\frac{-1}{h^2}}}{h}$

You know that p is a non-zero polynomial, so this limit is of the form
$\sum_n a_n \lim_{h \to 0} \frac{e^{\frac{-1}{h^2}}}{h^n}$

So we need to find
$\lim_{h \to 0} \frac{e^{\frac{-1}{h^2}}}{h^n}$
for some arbitrary positive integer n.

Let
$x = \frac{1}{h}$

Then
$h^n = \frac{1}{x^n}$

So
$\lim_{h \to 0} \frac{e^{\frac{-1}{h^2}}}{h^n} = \lim_{x \to \infty} x^n e^{-x^2}$

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

which is in a slightly more "doable" form.

-Dan

7. Originally Posted by Mathstud28
Do you mean $\lim_{h\to{0}}p\bigg(\frac{1}{h}\bigg)\cdot\frac{e ^{\frac{-1}{h^2}}}{h}$?

Alternatively you could consider the fact that since e^x is analytic everywhere it is thus differentiable everywhere and work at it from the power series point of view
Rewriting this as $\bigg(\frac{1}{h^{n+1}}\bigg)\cdot{e^{\frac{-1}{h^2}}}$

we get $\lim_{h\to{0}}\frac{1-\frac{1}{h^2}+\frac{(\frac{1}{h^2})}{2!}-\frac{(\frac{1}{h^2})^3}{3!}-...\pm\frac{(\frac{1}{h^2})^{n-1}}{(n-1)!}\mp\frac{(\frac{1}{h^2})^n}{n!}}{h^{n+1}}$

can you go from there

8. Originally Posted by Mathstud28
Rewriting this as $\bigg({\color{red}\frac{1}{h^{n+1}}}\bigg)\cdot{e^ {\frac{-1}{h^2}}}$
huh ?

we get $\lim_{n\to{\infty}}\frac{1-\frac{1}{h^2}+\frac{(\frac{1}{h^2})}{2!}-\frac{(\frac{1}{h^2})^3}{3!}-...}{h^{n+1}}$

can you go from there
Ok, and where did the limit of h go ??
Plus, you, can you go from there ?

*After l'Hôpital's rule, power series, here comes the newly trend : MacLaurin series !*

(am kidding)

9. Originally Posted by Moo
huh ?

Ok, and where did the limit of h go ??
Plus, you, can you go from there ?

*After l'Hôpital's rule, power series, here comes the newly trend : MacLaurin series !*

(am kidding)
I misread...it should be let $p\bigg(\frac{1}{h}\bigg)=\sum_{n=0}^{c}a_n\bigg(\f rac{1}{h}\bigg)^{n}=a_0+a_1\bigg(\frac{1}{h}\bigg) +a_2\bigg(\frac{1}{h}\bigg)^2+...a_n\bigg(\frac{1} {h}\bigg)^n$

10. Originally Posted by Moo
huh ?

Ok, and where did the limit of h go ??
Plus, you, can you go from there ?

*After l'Hôpital's rule, power series, here comes the newly trend : MacLaurin series !*

(am kidding)
No I have always liked power series more it was just I was not adept enough at LaTeX yet to use them...now I am...and yes you could go from there...technically the answer would be ∞ due to an infinite amount of ∞ + an infinite amount of 0's...but as you pointed out I read the question wrong so that limit does not apply