Show that $\lim\limits_{n \to \infty} \int\limits_0^\infty \frac{x^\frac{1}{n}}{(1+\frac{x}{n})^n} = 1$ Show that $\displaystyle\lim_{n \to \infty} \int\limits_0^\infty \frac{x^\frac{1}{n}}{\left( 1+\frac{x}{n} \right)^n}\ dx = 1$.
I guess that I'm supposed to use the dominated convergence theorem because it is easy to calculate the limit inside of the integral, but I can't find a function that bounds $\frac{x^\frac{1}{n}}{(1+\frac{x}{n})^n}$.
 A: The function $$f(x) = \begin{cases}1 & x\leq 1 \\ 2 \exp(-x) & x>1\end{cases}$$ works fine.
A: This argument is not rigorous, but maybe someone else can make it rigorous.
Since 
$$\frac{\sqrt[n]{x}}{(1+\frac{x}{n})^n}\sim \sqrt[n]{x}e^{-x}$$
\begin{align}
l&=\lim_{n\to\infty} \int_0^{\infty}\frac{\sqrt[n]{x}}{(1+\frac{x}{n})^n}\,dx\\
&=\lim_{n\to\infty} \int_0^{\infty} \sqrt[n]{x}e^{-x}\,dx\\
&=\lim_{n\to\infty} \Gamma\left(1+\frac{1}{n}\right)\\
&=\Gamma\left(\lim_{n\to\infty}1+\frac{1}{n}\right)\\
&=\Gamma(1)\\
&=1
\end{align}
Where the interchange of the limit and function composition was justified because $\Gamma(n)$ and $\left(1+\frac{1}{n}\right)$ are continuous on $]0,\infty[$.
A: Let us assume $n\geq 3$.
The problem can be tackled by squeezing. As already remarked by Messney,
$$\int_{0}^{+\infty}\frac{x^{1/n}}{\left(1+\frac{x}{n}\right)^n}\,dx \geq \int_{0}^{+\infty}x^{1/n}e^{-x}\,dx = \Gamma\left(1+\frac{1}{n}\right)\geq 1-\frac{\gamma}{n}. \tag{1}$$
On the other hand
$$ \int_{0}^{+\infty}\frac{x^{1/n}}{\left(1+\frac{x}{n}\right)^n}\,dx= n^{1+1/n}\int_{0}^{+\infty}\frac{x^{1/n}\,dx}{(1+x)^n}\stackrel{\text{AM-GM}}{\leq}n^{1/n}\int_{0}^{+\infty}\frac{(n-1)+x}{(1+x)^n}\,dx\tag{2}$$
hence the given integral is bounded between $1-\frac{\gamma}{n}$ and $1+\frac{C\log n}{n}$.
