# Evaluate <munder> <mo movablelimits="true" form="prefix">lim <mrow class="MJX-TeXAtom-ORD

Evaluate $\underset{x\to \mathrm{\infty }}{lim}{e}^{-{x}^{2}}{\int }_{x}^{x+\mathrm{ln}\left(x\right)/x}{e}^{-{t}^{2}}dt$
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soosenhc
First, we'll use a limit law, saying the limit of a product is a product of the limits.
$\underset{x\to \mathrm{\infty }}{lim}{e}^{-{x}^{2}}\cdot \underset{x\to \mathrm{\infty }}{lim}{\int }_{x}^{x+\frac{\mathrm{ln}\left(x\right)}{x}}{e}^{-{t}^{2}}dt$
Let's evaluate the first limit.
$\underset{x\to \mathrm{\infty }}{lim}{e}^{-{x}^{2}}=\underset{x\to \mathrm{\infty }}{lim}\frac{1}{{e}^{{x}^{2}}}$
As x approaches $\mathrm{\infty }$, ${e}^{{x}^{2}}$ also approaches $\mathrm{\infty }$, therefore $\frac{1}{{e}^{{x}^{2}}}$ approaches 0, therefore one of the limits calculated is 0.
However, we're not done. If the limit of the integral is $\mathrm{\infty }$, we would get an indeterminate form. By monotonicity,
$|{\int }_{x}^{x+\frac{\mathrm{ln}\left(x\right)}{x}}{e}^{-{t}^{2}}dt|\le \left(x+\frac{\mathrm{ln}\left(x\right)}{x}-x\right){e}^{-{x}^{2}}=\frac{\mathrm{ln}\left(x\right)}{x}{e}^{-{x}^{2}}$
, which if we evaluate as x approaches $\mathrm{\infty }$, we get $0\cdot 0=0$, so we didn't have to use L'Hôpital's rule.
Hope this helps!