• Nov 21st 2009, 10:45 PM
rphagoo

let g(x) be such that l g(x) l <= M for all x in [-1,1]
let h(x) = (x^2)g(x) if x is not equal to 0
= 0 if x is equal to 0
show that h(x) is diffrentiable at x = 0 and find h'(0)
• Nov 21st 2009, 10:52 PM
Jose27
$0 \leq \vert \frac{h(x)}{x} \vert = \vert xg(x) \vert \leq M\vert x \vert$ so taking $x\rightarrow 0$ we have $h'(0)=0$
• Nov 21st 2009, 11:02 PM
rphagoo
can u break down a little more please
• Nov 21st 2009, 11:17 PM
Jose27
Remember that $h'(0)= \lim_{x\rightarrow 0} \frac{h(x)-h(0)}{x-0} =\lim_{x\rightarrow 0} \frac{h(x)}{x}$ and to evaluate the limit we use the squeeze theorem.
• Nov 21st 2009, 11:27 PM
rphagoo
thanks so much