## Derivative piece function

Using $f'(x)=\lim_{x->c}\frac{f(x)-f(c)}{x-c}$ find the values of a, b, that makes f differentiable at x=0

Let $f(x) =$
$
(3a)sin(2x), x>0$

$xe^x+b, x\leq0$

So far I've done $f'(x)=\lim_{x->0^-}\frac{xe^x+b-b}{x} = \lim_{x->0^-}e^x = 1$

This seems right to me, but following through with it I get this:

$f'(x) = \lim_{x->0^+}\frac{(3a)sin(2x)-1}{x} = \lim_{x->0^+}\frac{(3a)sin(2x)}{x} - \lim_{x->0^+}\frac{1}{x} DNE$

Can someone show me where I went wrong?