Prove that $\mathop {\lim }\limits_{n \to \infty } \int_0^\infty {\frac{{dx}}{{{{(1 + \frac{x}{n})}^n}{x^{\frac{1}{n}}}}}} = 1$ using dominated convergence theorem (DCT). By DCT we need to show $\left| {\frac{1}{{{{(1 + \frac{x}{n})}^n}{x^{\frac{1}{n}}}}}} \right| \le \phi (x)$ for some Lebesgue integral function $\phi(x)$ on $(0,\infty)$.
My attempt solved part of the problem on $[1,\infty)$. When $x\geq1$, we have $\frac{1}{{{{(1 + \frac{x}{n})}^n}{x^{\frac{1}{n}}}}} \le \frac{1}{{{{(1 + \frac{x}{n})}^n}}}$. Then ${(1 + \frac{x}{n})^n} = 1 + C_n^1\frac{x}{n} + C_n^2\frac{{{x^2}}}{{{n^2}}} + ... \le 1 + C_n^1\frac{x}{n} + C_n^2\frac{{{x^2}}}{{{n^2}}} = 1 + x + \frac{{n(n - 1)}}{{2{n^2}}}{x^2}$.
Since $\frac{{n(n - 1)}}{{2{n^2}}} \ge \frac{1}{4}$ when $n\ge2$, we get ${(1 + \frac{x}{n})^n} \ge 1 + x + \frac{{{x^2}}}{4} \Rightarrow \frac{1}{{{{(1 + \frac{x}{n})}^n}}} \le \frac{1}{{1 + x + \frac{1}{4}{x^2}}}$ when $n\ge2$.
$\frac{1}{{1 + x + \frac{1}{4}{x^2}}}$ is Lebesgue integrable on $[1,\infty)$, then by dominated convergence theorem $\mathop {\lim }\limits_{n \to \infty } \int_1^\infty {\frac{{dx}}{{{{(1 + \frac{x}{n})}^n}{x^{\frac{1}{n}}}}}} = \int_1^\infty {\mathop {\lim }\limits_{n \to \infty } \frac{{dx}}{{{{(1 + \frac{x}{n})}^n}{x^{\frac{1}{n}}}}}} = \int_1^\infty { - {e^{ - x}}dx} $
But I cannot prove the limit and integration are exchangeable on $[0,1]$. Hope someone can help. Thank you.