thnx captain and perfecthacker.

ok this is what i understood from what you said perfecthacker. here's my proof from both your help..

you used the sequential definition of continuity? which implies that:

continuity at 0 = limit at 0 exists and equals f(0)?

**PROOF**
sin(1/x) is continuous at $\displaystyle x_0 = 0$ if for any sequence $\displaystyle x_n$ $\displaystyle \epsilon$ R, such that $\displaystyle \lim_{n \to \infty}(x_n) = x_0$, we have $\displaystyle \lim_{n \to \infty}f(x_n) = f(x_0)=0$

now let $\displaystyle x_n = \frac{1}{(2n)\pi}$ and $\displaystyle b_n = \frac{1}{(2n + \frac{1}{2})\pi}$

so $\displaystyle \lim_{n \to \infty}(x_n) = \lim_{n \to \infty}(b_n) = 0$

if we consider, $\displaystyle \lim_{n \to \infty}f(x_n)$ we have:

$\displaystyle \lim_{n \to \infty}f(x_n)$ = $\displaystyle \lim_{n \to \infty}sin \frac{1}{\frac{1}{(2n)\pi}}$ = $\displaystyle \lim_{n \to \infty}sin(2\pi n)$ = 0

but, if we consider, $\displaystyle \lim_{n \to \infty}f(b_n)$ we have:

$\displaystyle \lim_{n \to \infty}f(b_n)$ = $\displaystyle \lim_{n \to \infty}sin \frac{1}{\frac{1}{(2n + 1/2)\pi}}$ = $\displaystyle \lim_{n \to \infty}sin((2n + 1/2)\pi)$ = 1

we see there are points arbitrarily close to 0 where sin(1/x) = 1.

so this means the limit at 0 doesn't exist

as the limit doesnt exist. the function is not continuous. this implies that it is not differentiable at 0. Q.E.D

is this completely correct? and is there anything else i need to add onto this proof? thanks