limit with a parametrical integral $ \lim_{n \to \infty} n^2 \int_{0}^{\infty} \frac{sin(x)}{(1 + x)^n} dx $ I need help in calculating this strange limit.
$$
\lim_{n \to \infty} n^2 \int_{0}^{\infty} \frac{sin(x)}{(1 + x)^n} dx
$$
 A: We can substitute $x=\frac{t}{n}$:
$$\lim_{n \to \infty} n^2 \int_{0}^{\infty} \frac{\sin(x)}{(1 + x)^n} dx=\lim_{n \to \infty} n^2 \int_{0}^{\infty} \frac{\sin(t/n)}{(1 + \frac tn)^n}\frac{1}{n}dt$$
Now in the denominator we get $(1+\frac{t}{n})^n$, and we're taking the limit $n\to\infty$; this becomes $e^t$, so that we get (also take the $\frac 1n$ out of the integral):
$$\lim_{n \to \infty} n \int_{0}^{\infty} \frac{\sin(t/n)}{e^t}dt$$
Now also see that $\lim_{n\to\infty}n\sin(\frac{t}{n})=t$ so that we get
$$\lim_{n \to \infty}\int_{0}^{\infty} \frac{t}{e^t}dt$$
Now the $n$'s are gone so we can write
$$\int_{0}^{\infty} \frac{t}{e^t}dt$$
Can you take it from here?
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
\lim_{n \to \infty}\bracks{n^{2}\int_{0}^{\infty}
{\sin\pars{x} \over \pars{1 + x}^{n}}\,\dd x} & =
\lim_{n \to \infty}\bracks{n^{2}\int_{0}^{\infty}
\sin\pars{x}\exp\pars{-n\ln\pars{1 + x}}\,\dd x}
\\[5mm] & =
\lim_{n \to \infty}\bracks{n^{2}\int_{0}^{\infty}
x\,\exp\pars{-nx}\,\dd x} =
\lim_{n \to \infty}\int_{0}^{\infty}
x\,\exp\pars{-x}\,\dd x
\\[5mm] & = \bbx{1}
\end{align}

See Laplace's Method.

