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.
On
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.
On
hint: Break it into $2$ parts:$$\dfrac{1}{c}\cdot \dfrac{e^{ax}-e^{bx}}{x}\cdot \dfrac{cx}{\sin(cx)}$$
On
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 ($\lim_{x\to 0^+}$).
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.