Evaluate
$$\lim_{n \to \infty} \int_0^\infty \frac{1+ \frac{x}{\sqrt{n}} e^{-x/n}}{(x+1)^2} \hspace{0.1cm} dx$$
As this is a limit of integrals, surely we will be using a (monotone/dominated/bounded) convergence theorem. Monotone convergence cannot be utilized here as it is not true that $f_n \nearrow f$ where $f_n$ is the above integrand (in fact, the sequence of $f_n$ is decreasing!). But we do note that each $f_n$ is bounded. Observe:
\begin{align}
|f_n| &= \frac{ \left| 1+ \frac{x}{\sqrt{n}} e^{-x/n} \right|}{\left|(x+1)^2\right|} \\
&\leq \frac{ 1+ \frac{|x|}{\sqrt{n}} e^{-x/n}}{(x+1)^2} \\
&= \frac{ 1+ \frac{|x|}{\sqrt{n}}}{(x+1)^2 e^{x/n}} \\
&= \frac{ 1+ \frac{|x|}{\sqrt{n}}}{(x+1)^2} \\
\end{align}
And now I fall short. I observe that for each $n$, $\sup\{f_n(x): x \in (0, \infty)\} = f_n(0) = 1$; that is to say, $f_n$ takes on its largest value at $x=0$ and decreases thereafter. How can I take this observation and manipulate $f_n$ to get this upper bound?
Best Answer
Here is another solution:
Note that $f(t) := te^{-t^2}$ is non-negative and bounded for $t \geq 0$. We can even show that $C = \frac{1}{\sqrt{2e}}$ is the sharpest bound, although this explicit value is not important in our argument. Then
$$ \int_{0}^{\infty} \frac{\frac{x}{\sqrt{n}}e^{-x/n}}{(1+x)^2} \, \mathrm{d}x = \int_{0}^{\infty} \frac{\sqrt{x}f(\sqrt{x/n})}{(1+x)^2} \, \mathrm{d}x. $$
Since the integrand is dominated by the integrable function $\frac{C\sqrt{x}}{(1+x)^2}$, it follows that
$$ \lim_{n\to\infty} \int_{0}^{\infty} \frac{\frac{x}{\sqrt{n}}e^{-x/n}}{(1+x)^2} \, \mathrm{d}x = \int_{0}^{\infty} \lim_{n\to\infty} \frac{\sqrt{x}f(\sqrt{x/n})}{(1+x)^2} \, \mathrm{d}x = \int_{0}^{\infty} \frac{\sqrt{x}f(0)}{(1+x)^2} \, \mathrm{d}x = 0. $$