<s>[INST] Evaluate $\lim_{x\to 0} \frac 1x -\frac{2}{e^{2x}-1}$ Factorizing the expression
$$\lim_{x\to 0} \frac{e^{2x}-1-2x}{x(e^{2x}-1)}$$
$$=\lim_{x\to 0} \frac{e^{2x}-(1+2x)}{(x)(e^x-1)(e^x+1)}$$
How do I proceed?
 [/INST] Another approach using the Hopital's rule:
$$\lim_{x \to 0}\left(- \frac{2}{e^{2 x} - 1} + \frac{1}{x}\right)= \lim_{x \to 0} \frac{- 2 x + e^{2 x} - 1}{x \left(e^{2 x} - 1\right)}$$
$$\lim_{x \to 0} \frac{- 2 x + e^{2 x} - 1}{x \left(e^{2 x} - 1\right)} \stackrel{\,\,(0/0) \\\text{Hopital}}{=} \lim_{x \to 0} \frac{2 e^{2 x} - 2}{2 x e^{2 x} + e^{2 x} - 1}$$
$$\left(2 \lim_{x \to 0} \frac{e^{2 x} - 1}{2 x e^{2 x} + e^{2 x} - 1}\right)\stackrel{\,\,(0/0) \\\text{Hopital}}{=} 2 \lim_{x \to 0} \frac{2 e^{2 x}}{4 x e^{2 x} + 4 e^{2 x}}=\lim_{x \to 0} \frac{4e^{2x}}{4e^{2x}\left(x + 1\right)}=1$$
</s>