From $\phi'(0)=ia$, how can we get that $$\phi \left(\frac{t}{n}\right)^n\rightarrow e^{iat}$$, where $\phi$ is the characteristic function of a random variable $X$?
I know there is an approximation which is $\phi(t)=1+itEX+o({t^{2}})$. But I can't get farther to the answer. Thanks in advance!
For large $n$, $t/n$ is small, so we take a Taylor series of $\phi$ around zero: \begin{align} \lim_{n\rightarrow \infty}\,\phi\left(\frac{t}{n}\right)^n &=\lim_{n\rightarrow \infty}\,{\left[\phi\left(\frac{t}{n}\right)\right]}^n\\ &=\lim_{n\rightarrow \infty}\,{\left[\phi(0) \;+\; \frac{t}{n}\phi'(0) + O\left(\frac{1}{n^2}\right)\right]}^n\\ &=\lim_{n\rightarrow \infty}\,{\left[1 \;+\; \frac{t}{n}\, i a\right]}^n\\ &= e^{i a t} \end{align}
Here we have used the fact that $\phi(0) = 1$ for any characteristic function, and the fact that $\lim_{n\rightarrow \infty}\,{\left(1 \;+\; \frac{z}{n}\right)}^n = e^z$.