Let $g \in C^1[0,\infty)$ and assume $g'$ is bounded and $g(0)=0$. Find $\lim_{n \to \infty}\int_{0}^{\infty}\frac{g(x/n)}{x}e^{-x}$

Question

Let $g \in C^1[0,\infty)$ and assume $g'$ is bounded and $g(0)=0$. Find $\lim_{n \to \infty}\int_{0}^{\infty}\frac{g(x/n)}{x}e^{-x}$

My solution:

Since the derivative is bounded we get that $|g(x)|\leq M|x|$ for some $M$. Now let $u=x/n$ then the integral is $\int_{0}^{\infty}n\frac{g(u)}{u}e^{-nu}du$ now bvy previous observation $n\frac{|g(u)|}{u}e^{-nu}\leq Mne^{-nu}$ where we know that the sequence $\int_{0}^{\infty} Mne^{-nu}$ converges and thus by generalized dominated convergence $\lim_{n \to \infty}\int_{0}^{\infty}\frac{g(x/n)}{x}e^{-x}=0$ Is this correct? I feel like the problem wants you to use integration by parts, but I am not sure how. I would appreciate input and perhaps other proofs.