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

Solution :
$$\eqalign{ & {\text{Given, }}f\left( x \right) = \frac{1}{x} - \frac{2}{{{e^{2x}} - 1}} \cr & \Rightarrow f\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \frac{1}{x} - \frac{2}{{{e^{2x}} - 1}} \cr & \Rightarrow f\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {{e^{2x}} - 1} \right) - 2x}}{{x\left( {{e^{2x}} - 1} \right)}}\,\,\,\left[ {\frac{0}{0}\,\,{\text{from}}} \right] \cr} $$
$$\therefore $$ using, L'Hospital rule
$$\eqalign{ & \Rightarrow f\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \frac{{4{e^{2x}}}}{{2\left( {x{e^{2x}}\,2 + {e^{2x}}.1} \right) + {e^{2x}}.2}} \cr & \Rightarrow f\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \frac{{4{e^{2x}}}}{{4x{e^{2x}}\, + 2{e^{2x}} + 2{e^{2x}}}}\,\,\,\left[ {\frac{0}{0}\,\,{\text{from}}} \right] \cr & \Rightarrow f\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \frac{{4{e^{2x}}}}{{4\left( {x{e^{2x}}\, + {e^{2x}}} \right)}}\,\, = \frac{{4.{e^0}}}{{4\left( {0 + {e^0}} \right)}}\,\, = 1 \cr} $$