$\text{If }\underset{x\to 0}{\mathrm{Lt}}g\left(x\right)\left(\frac{e^{\frac{1}{x}}-e^{-\frac{1}{x}}}{e^{\frac{1}{x}}+e^{-\frac{1}{x}}}\right)$ exists, then g(x) =

$=g\left(x\right)\times \left(1-\frac{2}{{\mathrm{e}}^{2/\mathrm{x}}+1}\right)$

$=\mathrm{g}\left(\mathrm{x}\right)-\frac{2\mathrm{g}\left(\mathrm{x}\right)}{{\mathrm{e}}^{2/\mathrm{x}}+1}$

L.H.L

$=\underset{x\to 0^{-}}{lim} \left(g\left(x\right)-\frac{2g\left(x\right)}{e^{2/x}+1}\right)$

$=\mathrm{g}\left(0\right)-\frac{2\mathrm{g}\left(0\right)}{{\mathrm{e}}^{-\mathrm{\infty }}+1}\text{ as }\left(\mathrm{x}\to 0^{-};\frac{1}{\mathrm{x}}\to -\mathrm{\infty }\right)$

R.H.L

$\underset{x\to 0^{+}}{lim} f\left(x\right)=\underset{x\to 0^{+}}{lim} \left(g\left(x\right)-\frac{2g\left(x\right)}{e^{2/x}+1}\right)$

$=g\left(0\right)-\frac{2g\left(0\right)}{e^{\mathrm{\infty }}+1}\text{ as }\left(x\to 0^{+};\frac{1}{x}\to \mathrm{\infty }\right)$

Limit exists when $=\underset{\mathrm{x}\to 0}{\mathrm{Lt}} g\left(x\right)=0$

then option (B) correct