How to show that this limit goes to $e^{-t}$

Question

Assume that $t \in \mathbb{R}$ and $h(t): \mathbb{R}\rightarrow\mathbb{C}$. Also assume that $h$ is so that $|h(t)|/|t|$ goes to zero as $t$ goes to zero.

How do we show that $$(1-t/n+h(t/n))^n\rightarrow e^{-t}?$$ We have from calculus that $(1-t/n)^n$ goes to $e^{-t}$, but now that $h(t)$ part complicates things.

I was thinking about using L'Hôpital's rule but we do now know if $h$ is differentiable, and we do technically we must split in in real and imaginary parts to use it.

Answer 1

Hint: $\operatorname{Log}$ denoting the principal branch of logarithm we have $\dfrac {\operatorname{Log} (1-z)} z \to -1$ as $ z \to 0$ (by L'Hopital's Rule). So $$n \frac{\operatorname{Log} (1-t/n+h(t/n))}{n(t/n-h(t/n))} \to -1$$ and this gives $n \operatorname{Log} (1-t/n+h(t/n)) \to -t$. Take exponential to finish.

Remark: For real valued $h$, we can give a proof using Squeeze Theorem.