Dominated convergence theorem exercise application

Question

I have to prove that $$\lim_{n \to \infty} \int_0^\infty \mathrm (1+x/n)^{-n}(x^{-1/n})\mathrm{d}x = 1$$

I've been told to use Dominated convergence theorem but I can't find a function $|f_n(x)| \le g(x)$. Any tips to do that? Or should I try with another theorem?

Answer 1

It is not difficult to see that, if $n\geq2 $, that $$\left(1+\frac{x}{n}\right)^{-n}x^{-1/n}\geq\left(1+\frac{x}{n+1}\right)^{-(n+1)}x^{-1/(n+1)} $$ so $$\int_{0}^{\infty}\left(1+\frac{x}{n}\right)^{-n}x^{-1/n}dx\leq\int_{0}^{\infty}\left(1+\frac{x}{2}\right)^{-2}x^{-1/2}dx=\frac{\pi}{\sqrt{2}} $$ then we can apply the DCT, and so $$\lim_{n\rightarrow\infty}\int_{0}^{\infty}\left(1+\frac{x}{n}\right)^{-n}x^{-1/n}dx=\int_{0}^{\infty}\lim_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^{-n}x^{-1/n}dx $$ $$=\int_{0}^{\infty}e^{-x}dx=1.$$