# Compute $\lim_{n\rightarrow \infty} n\int_0^1(1+x)^n e^{-3nx} \mathrm{d} \mu (x)$

Compute $\lim_{n\rightarrow \infty} n\int_0^1(1+x)^n e^{-3nx} \mathrm{d} \mu (x)$

Well, I think I know how to solve this one. I have to use the Lebesgue dominated convergence theorem, so that the limit gets inside the integral, the limit goes to $0$ and so the integral is $0$. Thing is, I have to find a dominating function. By drawing some sketches of the first 3 functions, I think $\frac{1}{x}$ is a dominationg one, but the only way to prove it that I thought of, was to take the function $g(x) = n(1+x)^n e^{-3nx}-\frac{1}{x}$ and take the first derivative, trying to find when it becomes $0$, and then find the monotonicity of $g$ to prove that $g\geq 0$, but it becomes really messy and I can't finish this. Any tips about this problem (and tips in general about similar problems with the Lebesgue dominated convergence theorem) are appreciated!

By the change of variable $t=nx$, this is the integral on the whole real line of the function $$g_n:t\mapsto\left(1+\frac{t}n\right)^n\mathrm e^{-3t}\,\mathbf 1_{0\lt t\lt n}.$$ Since $1+\frac{t}n\leqslant\mathrm e^{t/n}$ for every $t$, $g_n\leqslant g$ where $$g:t\mapsto\mathrm e^{-2t}\mathbf 1_{t\gt 0}.$$ The function $g$ is integrable and $g_n\to g$ pointwisely hence $$\lim_{n\to\infty}\int g_n=\int g=\frac12.$$