Answer
The limit approaches $\infty$
Work Step by Step
The original limit is of the indeterminate form $\frac{\infty}{\infty}$. Hence, we use L'Hopital's: $\displaystyle\lim_{x\to\infty}\frac{e^{2x}}{x^2}=\lim_{x\to\infty}\frac{2e^{2x}}{2x}=\lim_{x\to\infty}\frac{4e^{2x}}{2}$. This final limit is unbounded, and the solution approaches $\fbox{$\infty$}$
