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.

Write $x$, $e^{ax} = 1+ax +O(x^2)$, $e^{bx} = 1+bx+O(x^2)$,  and $\sin cx = cx+O(x^3)$.  Then, 
$$\begin{align}
\frac{e^{ax}-e^{bx}}{\sin cx} & = \frac{\left(1+ax\right)-\left(1+bx\right)+O(x^2)}{cx+O(x^3)} \\\\
& = \frac{(a-b)+O(x)}{c+O(x^2)} \\\\
&= \frac{a-b}{c} \left(1+O(x)\right)
\end{align}$$
which goes to $\frac{a-b}{c}$ as $x \to 0$.

For the second limit, we analyze the right-side limit only.  
Note that we can write 
$$\begin{align}
\lim_{x \to 0^+} x\left(a^{\frac{1}{x}}-1\right) & = \lim_{x \to 0^+} xa^{\frac{1}{x}}-\lim_{x \to 0^+} x \\\\
& = \lim_{x \to 0^+} xa^{\frac{1}{x}} \\\\
& = \lim_{x \to 0^+} e^{\log \left(xa^{\frac{1}{x}}\right)} \\\\
& = \lim_{x \to 0^+} e^{\log (x)+\frac{1}{x}\log(a)} \\\\
& = \begin{cases}\infty&,a>1\\\\0&,0 < a \le  1 \end{cases}
\end{align}$$
This last equality is justified since heuristically, $x^{-1} \to \infty$ faster than $\log x \to -\infty$.
The case for the left-side limit is left as an exercise.
