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?
 A: HINT
We can use that by Taylor’s expansion as $u\to 0$
$$e^u=1+u+\frac12u^2+O(u^3)$$
A: $$\lim_{x\to 0} \frac{e^{2x}-1-2x}{x(e^{2x}-1)}$$
$$=\lim_{x\to 0} \frac{e^{2x}-(1+2x)}{(x)(e^x-1)(e^x+1)}$$
$$=\lim_{x\to 0} \frac{1}{e^x+1} \lim_{x\to 0} \frac{2 e^{2x} - 2 }{e^{x}-1+ x e^{x}}(\text{using L'hospital rule}) $$
$$=\frac{1}{2} \lim_{x\to 0} \frac{4 e^{2x}}{2 e^{x} + x e^{x}} (\text{ again using L'hospital rule}) $$
$$=1 $$
A: Since $\lim_{x\to0}\frac{2x}{e^{2x}-1}=1$, your limit is $\lim_{x\to0}\frac{e^{2x}-1-2x}{2x^2}=2\lim_{y\to0}\frac{e^y-1-y}{y^2}$. The limit in terms of $y$ is famously $\tfrac12$ by the series defining $\exp y$, making the final answer $1$.
A: $$\frac{e^{2x}-1-2x}{x(e^{2x}-1)}=\frac{2x}{e^{2x}-1}\cdot\frac{e^{2x}-1-2x}{2x^2}\rightarrow\frac{2e^{2x}-2}{4x}\rightarrow1.$$
A: Another approach using the Hopital's rule:
$$\lim_{x \to 0}\left(- \frac{2}{e^{2 x} - 1} + \frac{1}{x}\right)= \lim_{x \to 0} \frac{- 2 x + e^{2 x} - 1}{x \left(e^{2 x} - 1\right)}$$
$$\lim_{x \to 0} \frac{- 2 x + e^{2 x} - 1}{x \left(e^{2 x} - 1\right)} \stackrel{\,\,(0/0) \\\text{Hopital}}{=} \lim_{x \to 0} \frac{2 e^{2 x} - 2}{2 x e^{2 x} + e^{2 x} - 1}$$
$$\left(2 \lim_{x \to 0} \frac{e^{2 x} - 1}{2 x e^{2 x} + e^{2 x} - 1}\right)\stackrel{\,\,(0/0) \\\text{Hopital}}{=} 2 \lim_{x \to 0} \frac{2 e^{2 x}}{4 x e^{2 x} + 4 e^{2 x}}=\lim_{x \to 0} \frac{4e^{2x}}{4e^{2x}\left(x + 1\right)}=1$$
