Show f(x) = sin(1/x), for x does not equal 0, is differentiable for nonzero real numbers.
I was wondering if this would be enough to show the the previous statement:
Let c<0 => c does not equal 0.
limit of f(x) as x approaches c from the left = lim sin(1/x) as x approaches c from the left = sin (1/c) since c does not equal 0.
The same works as x approaches c from the right.
Then, repeat this process with c>0.