Two complicated limits: $\lim_{x\to 0}\frac{e^{ax}-e^{bx}}{\sin(cx)}$ and $\lim_{x\to 0} x(a^{\frac1x}-1)$ I need to solve these 2 limits ( without using L'Hospital's Rule) , but I can't figure out how to go about them:
Let $a \neq b$, $c \neq 0$.
$$\lim_{x\to 0}\frac{e^{ax}-e^{bx}}{\sin(cx)}$$
Also, let $a>0$, $a \neq 1$.
$$\lim_{x\to 0} x(a^{\frac1x}-1)$$
I don't necessarily need the result, more like understanding the process.

In short: factor the smaller exponential out. Use l'Hopital's rule on what remains. 
One iteration will do: exponentials stay, and $sin(cx)$ will become $c\cos (cx)$, which goes to $1$ instead.
