Answer
= $1$
Work Step by Step
$f(x)$ = $\mathop {\lim }\limits_{n \to \infty }$ $(\frac{e^{x}+1}{e^{x}-1})^{\ln{x}}$
$\ln{f(x)}$ = $\mathop {\lim }\limits_{n \to \infty }$ $\ln(\frac{e^{x}+1}{e^{x}-1})^{\ln{x}}$
$\ln{f(x)}$ = $\mathop {\lim }\limits_{n \to \infty }$ $\\ln(x)ln(\frac{e^{x}+1}{e^{x}-1})$
$\ln{f(x)}$ = $\mathop {\lim }\limits_{n \to \infty }$ $\\ln(x)ln(\frac{e^{\frac{x}{2}}+e^{\frac{-x}{2}}}{e^{\frac{x}{2}}-e^{\frac{-x}{2}}})$
$\ln{f(x)}$ = $\mathop {\lim }\limits_{n \to \infty }$ $\\ln(x)ln(coth(\frac{x}{2}))$
$\ln{f(x)}$ = $\mathop {\lim }\limits_{n \to \infty }$ $\frac{ln(coth(\frac{x}{2}))}{(\ln{x})^{-1}}$
$\ln{f(x)}$ = $\mathop {\lim }\limits_{n \to \infty }$ $\frac{x(tanh(\frac{x}{2}))(csch^{2}(\frac{x}{2}))}{2(\ln(x))^{-2}}$
$\ln{f(x)}$ = $\mathop {\lim }\limits_{n \to \infty }$ $\frac{x(\ln(x))^{2}}{2cosh(\frac{x}{2})sinh(\frac{x}{2})}$
$\ln{f(x)}$ = $\mathop {\lim }\limits_{n \to \infty }$ $\frac{x(\ln(x))^{2}}{sinh(x)}$
$\ln{f(x)}$ = $\mathop {\lim }\limits_{n \to \infty }$ $\frac{2\ln(x)+(\ln(x))^{2}}{cosh(x)}$
$\ln{f(x)}$ = $\mathop {\lim }\limits_{n \to \infty }$ $\frac{\frac{2}{x}+\frac{2(\ln(x))}{x}}{sinh(x)}$
$\ln{f(x)}$ = $\mathop {\lim }\limits_{n \to \infty }$ $\frac{2+2(\ln{x})}{x(sinh(x))}$
$\ln{f(x)}$ = $\mathop {\lim }\limits_{n \to \infty }$ $\frac{\frac{2}{x}}{x(cosh(x))+sinh(x)}$
$\ln{f(x)}$ = $\mathop {\lim }\limits_{n \to \infty }$ $\frac{2}{x^{2}(cosh(x))+x(sinh(x))}$
$\ln{f(x)}$ = $0$
${f(x)}$ = $e^{0}$
${f(x)}$ = $1$
