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