Answer
$\infty$
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, $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty}
\dfrac{e^{x^2}e^{-x}}{x}$
and
$\lim\limits_{x \to \infty} \dfrac{e^{x^2-x}}{x}=\dfrac{\infty}{\infty}$
This shows an indeterminate form of the limit, so we need to use L'Hospital's rule.
$\lim\limits_{x \to \infty} \dfrac{e^{x^2}-x(2x-1)}{1}=\dfrac{\infty}{1}=\infty$