Evaluating $\lim_{x\to 0}\frac{x}{2^x-1}$ without using L'Hôspital's rule So lately, when one my friends had asked for help, he showed me this task:

Evaluate the following limit $$\lim_{x\to 0}\frac{x}{2^x-1}$$

The problem is that one is not ought to use L'Hôspital's rule (which yields $\frac{1}{\ln 2}$), because derivatives weren't even introduced yet. 
I am sorry that I can't see a way to change this term algebraically and it would be a pleasure if you helped me.
 A: One way
Apply definition of derivative of $2^x$
$$\Longrightarrow \lim_{x\to 0} \frac{2^x-2^0}{x-0}$$
$$=\frac{d}{dx}(2^x)|_0$$
$$=2^0\log 2$$
$$\Longrightarrow \lim_{x\to 0}\frac{x}{2^x-1}=\frac{1}{\log 2}$$
Second way
Apply change of variable, let $2^x=y$
$$\Longrightarrow x=\frac{\log y}{\log 2}$$
Also, as $x\rightarrow 0, y\rightarrow 1$. So, required limit
$$=\lim_{y\to 1}\frac{\frac{\log y}{\log 2}}{y-1}$$
$$=\lim_{y\to 1}\frac{\log y}{\log 2 \times (y-1)}$$
Now, apply the well-known fact that $\forall y>0$
$$\frac{y-1}{y}\leq\log y\leq y-1$$
Now, divide both sides by $y-1$ and apply sandwich theorem as $y\rightarrow 1$.
A: Assuming that you know that $(2^x)'=\log(2)\times2^x$, then, in particular, you know that$$\lim_{x\to0}\frac{2^x-1}x=\log(2).$$Therefore,$$\lim_{x\to0}\frac x{2^x-1}=\frac1{\log(2)}.$$
A: As $a=e^{\ln a}$ for $a>0$
$$\lim_{x\to0}\dfrac{a^x-1}x=\ln a\cdot\lim_{x\to0}\dfrac{e^{x\ln a}-1}{x\ln a}=?$$
A: Is your friend allowed to use/familiar with derivatives to begin with?
We have
$$\displaystyle\lim_{x\to 0}{1\over {2^x - 2^0\over x-0}} = {1\over f’(0)}$$ where $f(x) = 2^x$, so the limit is $\frac 1{ln2}$. 
A: Since we are not allowed to use derivative we need to start from some well grounded point.
Usually the starting point is the following limit for sequences (wich can be proved by monotonicity theorem):
$$\lim_{n\to \infty}\left(1+\frac1n\right)^n =e$$
which can be easily extended to real functions
$$\lim_{y\to \infty}\left(1+\frac1y\right)^y =e$$
From the latter by $z=\frac1y \to 0$ we easily obtain that
$$\lim_{y\to \infty}\left(1+\frac1y\right)^y =\lim_{z\to 0}\left(1+z\right)^\frac1z=e \implies \lim_{z\to 0}\frac{\log(1+z)}{z}=1$$
and finally by $y=e^x-1 \to 0$ with $x\to 0$
$$\lim_{z\to 0}\frac{\log(1+z)}{z}=1 \implies \lim_{x\to 0}\frac x{e^x-1}=1$$
and then
$$\lim_{x\to 0}\frac{x}{2^x-1}=\lim_{x\to 0}\;\frac1{\log 2}\cdot\frac{x\log 2}{e^{x\log 2}-1}=\frac1{\log 2}\cdot\cdot 1 =\frac1{\log 2}$$
