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.

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

$$\lim_{x\to 0}\frac{x}{2^x-1}=\lim_{x\to 0}\;\frac1{\log 2}\cdot\frac{x\log 2}{e^{x\log 2}-1}=\frac1{\log 2}\cdot\cdot 1 =\frac1{\log 2}$$