The function $f:\mathbb{R}\setminus \{0\}\to \mathbb{R}$ given by $f(x)=\frac{1}{x}-\frac{2}{e^{2x}-1}$ can be made continuous at $x=0$ by defining $f(0)$ as:

To make $f(x)$ continuous at $x=0$, we must have $f(0)={\lim }_{x\to 0}f(x)$.
$$L=\underset{x\to 0}{\lim }\left(\frac{1}{x}-\frac{2}{e^{2x}-1}\right)$$
$$L=\underset{x\to 0}{\lim }\frac{e^{2x}-1-2x}{x(e^{2x}-1)}$$
Using L'Hôpital's Rule (form 0/0):
$$L=\underset{x\to 0}{\lim }\frac{2e^{2x}-2}{(e^{2x}-1)+2x{e}^{2x}}$$
Still 0/0, use L'Hôpital's again:
$$L=\underset{x\to 0}{\lim }\frac{4e^{2x}}{2e^{2x}+2e^{2x}+4x{e}^{2x}}$$
$$L=\frac{4}{2+2+0}=\frac{4}{4}=1$$
So $f(0)=1$.