I am trying to compute the limits of $\lim_{n \rightarrow \infty} \int\limits_0^{\infty} \dfrac{1}{(1+\dfrac{x}{n})^n \sqrt[n]x}dx $ by using Monotone convergence theorem of integrals and switching the limits, I know that $(\dfrac{1}{1+\dfrac{x}{n}})^n $ converges to $e^{-x}$ but is unsure of what should I do with the root itself. Also can I claim that the whole expression converges to $e^x$ as n tends to infinity?
Thanks!
On
Let $f_n$ denote the $n$th integrand. As you noticed, $f_n(x)\to e^{-x}$ pointwise on $(0,\infty).$
On $(0,1),$ $f_n(x)\le 1/\sqrt x$ for all $n.$ On $[1,\infty),$ $f_n(x)\le 1/(1+x/n)^n$ for all $n.$ Since $(1+x/n)^n$ is pointwise increasing, we can say that $f_n(x)\le 1/(1+x/2)^2$ on $[1,\infty)$ for $n\ge 2.$
It follows that
$$\tag 1 f_n(x)\le \frac{1}{\sqrt x}\chi_{(0,1)} + \frac{1}{(1+x/2)^2}\chi_{[1,\infty)},$$
this for $x\in (0,\infty)$ and $n\ge 2.$ Since the function on the right of $(1)$ is in $L^1(0,\infty),$ the DCT impilies
$$\int_0^\infty f_n(x)\,dx \to \int_0^\infty e^{-x}\, dx =1.$$
$$I_n=\int_{0}^{+\infty}\frac{x^{-1/n}}{(1+x/n)^n}\,dx=\int_{0}^{n}\frac{x^{-1/n}}{(1+x/n)^n}\,dx+\int_{n}^{+\infty}\frac{x^{-1/n}}{(1+x/n)^n}\,dx$$ where: $$ 0\leq \int_{n}^{+\infty}\frac{x^{-1/n}}{(1+x/n)^n}\,dx\leq \frac{1}{2^n}\int_{n}^{+\infty}x^{-1/n}\,dx\leq\frac{1}{2^n}\int_{1}^{+\infty}x^{-1/n}\,dx = \frac{1}{2^n\left(1-\frac{1}{n}\right)}$$ and: $$ \int_{0}^{n}\frac{x^{-1/n}}{(1+x/n)^n}\,dx = n^{1-\frac{1}{n}}\int_{0}^{1}\frac{x^{-1/n}}{(1+x)^n}.$$ Since: $$\int_{0}^{1}\frac{x^{-1/n}}{(1+x)^n}\,dx\geq \int_{0}^{1}\frac{1}{(1+x)^n}\,dx = \frac{1-2^{1-n}}{n-1}$$ but over $[0,1]$ we have $1+x\geq\exp\left(\frac{x}{1+x/2}\right)$, it follows that $I_n\to 1$ as $n\to +\infty$, also without the dominated convergence theorem.