Answer
$e^3$
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, $\ln f(x)=x \ln (\dfrac{x+2}{x-1})$
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 the limit, so we need to use L'Hospital's rule.
$e^{\lim\limits_{x \to \infty} (\dfrac{\ln \frac{x+2}{x-1}}{1/x})}=e^{\lim\limits_{x \to \infty} (\dfrac{\ln (x+2)-\ln (x-1)}{x^{-1}})}$
or,
$e^{\lim\limits_{x \to \infty} \dfrac{-3/(x+2)(x+1)}{-1/x^2}}=e^{\lim\limits_{x \to 0} \dfrac{3}{1+1/x-1/x^2}}$
Thus,
$e^{3/1+0+0}=e^3$
