during exam preparation I found the following question which I could not solve: $$\lim_{n \rightarrow \infty}\left(m_X\left(\frac{t}{n}\right)\right)^n = \exp\left(t\cdot\mathbb{E}(X)\right)$$
where $m_X(t)$ is the moment generating function.
As hint was given to use L'Hopital and to interchange limit and expected value without argumentation. However, I have no idea how to apply L'Hospital since there is no fraction.
I tried to verify the formula for an exponentially distributed random variable but it turned out that either I have a calculation error or the formula is wrong.
Would really appreciate if anyone can help me through the exercise or give me a hint how to start!
First notice that, as $n\to \infty$, you have $m_X(t/n) \to m_X(0) = 1$. Hence the limit you want to evaluate is of the form $\lim_{n\to\infty} f(n)^{g(n)}$ where $$\begin{cases}\lim_{n\to\infty} f(n) = 1\\ \lim_{n\to\infty} g(n) = +\infty\end{cases} $$
By a known identity, we have in this setting that $$\lim_{n\to\infty} f(n)^{g(n)} = \exp\left(\lim_{n\to\infty}g(n)\cdot(f(n) - 1)\right) $$
Plugging this back in the limit we're interested in yields : $$\begin{align*}\lim_{n \rightarrow \infty}\left(m_X\left(\frac{t}{n}\right)\right)^n &= \exp\left(\lim_{n\to\infty}n\cdot\left(m_X\left(\frac{t}{n}\right)-1\right)\right) \\ &=\exp\left(\lim_{n\to\infty}\frac{m_X\left(\frac{t}{n}\right)-1}{1/n}\right)\\ &=\exp\left(\lim_{h\to0}\frac{m_X\left(t\cdot h\right)-1}{h}\right)\end{align*} $$
Can you finish ?