define f(x) to be 0 at 0 and exp(-1/x^2) everwhere else.
i'm having trouble showing that f'(0) = 0.
can anyone help?
thanks!
(Note: since the function is even, $\displaystyle f_{+}'(0)$=0 is enough. But L'Hospital's rule goes here, so no need to distinguish.)
I'm sure there's a slicker way than this:
$\displaystyle {\rm lim}_{h\rightarrow0}f(h)/h={\rm lim}_{h\rightarrow0}{\rm e}^{-1/h^2}/h={\rm lim}_{h\rightarrow0}(1/h)/{\rm e}^{1/h^2}$
and apply L'Hospital's rule to get this equal to $\displaystyle {\rm lim}_{h\rightarrow0}h/{\rm e}^{1/h^2}={\rm lim}_{h\rightarrow0}h{\rm e}^{-1/h^2}=0$.
Just a quick point here. This function is smooth at 0, but not analytic at 0!
This is a classic. Only seen such misbehaving functions, when creating partitions of unity... Are you studying Differential Geometry by the way?
no, not studying differential geometry yet. i'm actually preparing for the math subject gre. i know i should resist these sort of problems for now and concentrate on gre type problems. but i just can't help myself. this problem was a harder problem in a standard calculus book.
by the way. technically, isn't it circular reasoning to use L'Hopital's rule since you are differentiating the function we are interested in? or is it ok since you've differentiated somewhere near zero using the standard differentiation formulas?
Technically, isn't it circular reasoning to use L'Hopital's rule since you are differentiating the function we are interested in? or is it ok since you've differentiated somewhere near zero using the standard differentiation formulas?
f(h)/h is inside a limiting process, so we are only interested in values of f(h) in the vicinity of zero.
Careful, not needed minus in the denominator.f'(0) = lim (x-->0) (1/x)*exp(-1/x^2)
Let
$\displaystyle f(x) = \left\{ \begin{array}{c}e^{-1/x^2} \mbox{ for }x\not =0\\ 0 \mbox{ for }x=0\end{array}\right.$
Now,
$\displaystyle f'(0) = \lim_{x\to 0}\frac{f(x)-f(0)}{x-0} = \lim_{x\to 0}\frac{e^{-1/x^2}}{x}$
Using the limit composite rule we can write,
$\displaystyle \lim_{x\to \infty} xe^{-x^2}$
But,
$\displaystyle e^x \geq x^2/2 \mbox{ for }x>0$
Thus,
$\displaystyle xe^{-x^2} \leq \frac{2}{x} \to 0$.