Find the limit of $\dfrac{e^{x^2}}{xe^x}$ Please help me find this limit. I have already tried to use L'Hopitals rule and logarithmic differentiation but it turns into a quadratic.

I will assume you mean to evaluate the limit as $x\to\infty$.
Note, the expression is actually $\dfrac{e^{x^2-x}}{x}$. Now, we know for large $y$, the inequality $e^y\geq 1+y$ is true.
So noting that our expression is always non-negative, and putting $y=x^2-x$ in the above inequality, $\dfrac{e^{x^2-x}}{x}\geq \dfrac{1+x^2-x}{x}=\dfrac{1}{x}+x-1$.
As $x\to\infty$, the right side goes to $\infty$, so the limit is indeed $\infty$.
