For a measure theory class, I'm trying to evaluate:
$$\lim_{n\to\infty}\int^\infty_1\frac 1 {nx} e^{-x/n}\ \text d\lambda$$
Obviously I want to try and move the limit through the integral and conclude that the limit is $0$, so I need to either show this sequence of functions is increasing (monotone convergence), or dominate it (dominated convergence). I decided to try and dominate it, but my method feels a bit roundabout.
Fix $x$ and view $\frac 1 {nx} e^{-x/n}$ as a function in $n$. The derivative is:
$$e^{-x/n}\left(\frac 1 {n^3} - \frac 1 {xn^2} \right)$$
It's now easy to show that this has a single global maximum at $n=x$. Therefore I can conclude (I think) that $\frac 1 {e x^2}$ is greater than each of the integrands and thus apply LDC.
Hint: $$\lim_{n\to\infty}\int^\infty_1\frac 1 {nx} e^{-x/n}\ \text dx = \lim_{n\to\infty}\int^1_{0}\frac 1 {nu} e^{-1/nu}\ \text du = \lim_{n\to\infty}\frac 1n\int^n_{0}\frac 1 {v} e^{-1/v}\ \text dv $$
Now use the fact that if $f\sim g>0$ then $$ \int_0^\infty f = \infty \implies \int_0^n f \sim \int_0^n g $$