 # Evaluate: <munder> <mo movablelimits="true" form="prefix">lim <mrow class="MJX-TeXAt polivijuye 2022-06-27 Answered
Evaluate:
$\underset{n\to \mathrm{\infty }}{lim}{\int }_{0}^{\mathrm{\infty }}{\left(1+\frac{x}{n}\right)}^{-n}\mathrm{sin}\frac{x}{n}\phantom{\rule{thinmathspace}{0ex}}dx.$
I have tried to use the (Lebesgue) dominated convergence theorem to evaluate the same. At first I noticed that:

Since $g\left(x\right)=1$ is Lebesgue integrable for each $x\in \left[0,\mathrm{\infty }\right).$ So by DCT, the given limit equals:
${\int }_{0}^{\mathrm{\infty }}\underset{n\to \mathrm{\infty }}{lim}{\left(1+\frac{x}{n}\right)}^{-n}\mathrm{sin}\frac{x}{n}\phantom{\rule{thinmathspace}{0ex}}dx={\int }_{0}^{\mathrm{\infty }}{e}^{-x}\cdot 0\phantom{\rule{thinmathspace}{0ex}}dx=0.$
This question was asked in our end term exam this semester and so I don't know the correct answer to this question. Therefore, just to check whether I have evaluated the limit correctly I am posting the same here on MSE. If I have gone wrong somewhere, please point out and give some insights. Thanks in advance.
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Using
$\left(1+z{\right)}^{n}\ge 1+\left(\genfrac{}{}{0}{}{n}{2}\right){z}^{2}$
for $z>0$, one has
$\left(1+\frac{x}{n}{\right)}^{n}\ge \frac{n\left(n-1\right)}{2}\frac{{x}^{2}}{{n}^{2}}=1+\frac{1}{2}\left(1-\frac{1}{n}\right){x}^{2}\ge 1+\frac{1}{4}{x}^{2}$
for $n\ge 2$. Then define
${f}_{n}\left(x\right)=\left(1+\frac{x}{n}{\right)}^{-n}\mathrm{sin}\left(\frac{x}{n}\right)$
to estimate ${f}_{n}$ as
$|{f}_{n}\left(x\right)|\le \frac{4}{4+{x}^{2}}$
for $x\ge 1$ and then one can use LDC.