Answer
$1$
Work Step by Step
Here, we have $\ln f(x)=(\dfrac{1}{x}) \ln (\dfrac{x^2+1}{x-1})$
But $e^{\lim\limits_{x \to \infty} (\dfrac{\ln (\dfrac{x^2+1}{x-1})}{x})}=\dfrac{\ln \infty}{\infty}=\dfrac{\infty}{\infty}$
This shows an indeterminate form of limit, thus we will apply L-Hospital's rule such as:
$\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$
$e^{\lim\limits_{x \to \infty} (\dfrac{\ln (\dfrac{x^2+1}{x-1})}{x})}=e^{\lim\limits_{x \to \infty} (\dfrac{\ln (x^2+1)-\ln (x+2)}{x})}$
or, $e^{\lim\limits_{x \to \infty} \dfrac{\frac{x^2+4x-1}{x^3+x^2+x+2}}{1}}=\dfrac{\infty}{\infty}$
This shows an indeterminate form of limit, thus we will again apply L-Hospital's rule such as:
$\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$
$e^{\lim\limits_{x \to \infty} \dfrac{\frac{2x+4}{6x+4}}{1}}=e^{ \frac{0}{1}}=1$
