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