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

Let $\displaystyle f(x) = $
$\displaystyle
(3a)sin(2x), x>0$
$\displaystyle xe^x+b, x\leq0$

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

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

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

Can someone show me where I went wrong?