Answer
$1$
Work Step by Step
L'Hospital's rule states that $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{A'(x)}{B'(x)}$
Here, $\ln f(x)=(1/x) \ln (\dfrac{x^2+1}{x-1})$
Now, $e^{\lim\limits_{x \to \infty} (\dfrac{\ln (\dfrac{x^2+1}{x-1})}{x})}=\dfrac{\ln \infty}{\infty}=\dfrac{\infty}{\infty}$
This shows an indeterminate form of the limit, so we need to use L'Hospital's rule.
$e^{\lim\limits_{x \to \infty} (\dfrac{\ln (\dfrac{x^2+1}{x-1})}{x})}=e^{\lim\limits_{x \to \infty} (\dfrac{\ln (x^2+1)-\ln (x+2)}{x})}$
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 the limit, so we need to use L'Hospital's rule.
$e^{\lim\limits_{x \to \infty} \dfrac{\frac{2x+4}{6x+4}}{1}}=e^{0}=1$
