The question
Compute the limit:
$$\lim_{n\to\infty} \int_0^{\infty}\frac{1}{\displaystyle\left(1+\frac{x}{n}\right)^nx^{\frac{1}{n}}}\ \mathrm{d}\lambda$$
My attempt
It is clear to see that point-wise the integrand has limit $e^{-x}$ however I am struggling to call upon the proper limit theorem. Since the integrand is decreasing. Also I cannot seem to bound it by an integrable function as neither $\frac{1}{x+x^2}, \frac{1}{x}$ or $\frac{1}{x^2}$ are integrable.
Going forward
I think this must use the dominated convergence theorem but I cannot seem to find a suitable function. I was considering $f_2(x)$ and then using a substitution $u = x^{\frac{1}{2}}$ but this becomes awfully complicated. Any help would be great.
Best Answer
Instead of trying to find an upper bound given by a formula, think about the problem points. Here your integral has issues at $0$ and at infinity. Also, you don't have to bound for all $n$, just eventually.
At zero the integrand is basically $x^{-1/n}$, which is integrable for $n\geq2$. So, say on $[0,1]$ we can take $1_{[0,1]}\,x^{-1/2}$.
At infinity you can use that, for $x\geq1$, $$ \Big(1+\frac xn\Big)^{-n}\,x^{-1/n}\leq \Big(1+\frac x2\Big)^{-2}. $$ In all, the function $$ g(x)=1_{[0,1]}\,x^{-1/2}+1_{[1,\infty)}\,\Big(1+\frac x2\Big)^{-2} $$ is an upper bound for all $n\geq2$, and integrable. So Dominated Convergence applies.