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.

For the second problem the limit does not exist for both cases $0<a<1 $ and $a>1$ since 

1) case $0<a<1$ $$ \lim_{x\to 0^+} x(a^{1/x}-1) = 0,\quad \lim_{x\to 0^-} x(a^{1/x}-1) = -\infty $$
2) case $ a>1$ $$ \lim_{x\to 0^+} x(a^{1/x}-1) = \infty,\quad \lim_{x\to 0^-} x(a^{1/x}-1) = 0. $$

Now try to study these cases.
Note: I believe you studied the notion of left limit ($\lim_{x\to 0^-}$) and right limit ($\lim_{x\to 0^+}$).
