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

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For f(x) to be continuous at x = 0, we must have $f(0)=\lim\limits_{x \rightarrow 0} f(x)$

$\Rightarrow f(0)=\lim\limits_{x \rightarrow 0}\left(\frac{1}{x}-\frac{2}{e^{2 x}-1}\right)$

$\Rightarrow f(0)=\lim\limits_{x \rightarrow 0} \frac{e^{2 x}-1-2 x}{x\left(e^{2 x}-1\right)}$

$\Rightarrow f(0)=\lim\limits_{x \rightarrow 0} \frac{\left(1+2 x+\frac{(2 x)^2}{2 !}+\frac{(2 x)^3}{3 !}+....\right)-1-2 x}{x\left(1+2 x+\frac{(2 x)^2}{2 !}+....-1\right)}$

$\Rightarrow f(0)=\lim\limits_{x \rightarrow 0} \frac{(2 x)^2\left\{\frac{1}{2 !}+\frac{2 x}{3 !}+....\right\}}{x\left\{2 x+\frac{(2 x)^2}{2 !}+....\right\}}$

$\Rightarrow f(0)=\lim\limits_{x \rightarrow 0} \frac{2\left\{\frac{1}{2 !}+\frac{2 x}{3 !}+....\right\}}{\left\{1+\frac{2 x}{2 !}+....\right\}}=2 \times \frac{1}{2}=1$