This question is in the context of the convergence of characteristic functions. If we have an i.i.d sequence of random variables $\{ X_n \}_{n = 1}^\infty$ each with mean $\mathbf{E} [X_n] = 0$ and finite variance $\mathbf{Var}[X_n] = \sigma^2 > 0$, and denote the partial sums by $S_n = \sum_{i = 1}^n X_i$ then the characteristic function of $$ Z_n = \frac{S_n}{\sigma \sqrt{n}} $$ will be $$ \mathbf{E} \left[ \exp{ \left( i t Z_n \right) } \right] = \mathbf{E} \left[ \exp{ \left( \frac{itX_1}{\sigma \sqrt{n}} \right) } \right]^n = \left(1 + \frac{i \mathbf{E}[X_1]}{\sigma \sqrt{n}} - \frac{t^2 \mathbf{E}[X_1^2]}{2 \sigma^2 n} + o \left( \frac{1}{n} \right) \right) = \left( 1 - \frac{t^2}{2n} + o \left( \frac{1}{n} \right) \right)^n $$ I'm not sure how to prove that the vanishing factor of $o(1/n)$ can be ignored so that the limit still goes to $e^{-t^2/2}$ as $n \to \infty$ as expected, could someone explain?
Edit: I think I've come up with a solution. We can follow along the same lines of the proof that the expression without the error term converges to the exponential. We have
$$ \lim_{n \to \infty} \log\left( 1 - \frac{t^2 + o(1)}{2n} \right)^n = \lim_{n \to \infty} n \log\left( 1 - \frac{t^2 + o(1)}{2n} \right) \\ = - \lim_{n \to \infty} \frac{\log\left( 1 - \frac{t^2 + o(1)}{2n} \right) - \log{1}}{ -\frac{t^2 + o(1)}{2n}} \lim_{n \to \infty} \frac{t^2 + o(1)}{2} \\ = - \frac{d}{dt} \log{t} \bigg\rvert_{t = 1} \lim_{n \to \infty} \frac{t^2 + o(1)}{2} = -\frac{t^2}{2} $$ And hence $$ \lim_{n \to \infty} \left( 1 - \frac{t^2 + o(1)}{2n} \right)^n = \exp{\left( \lim_{n \to \infty} \log\left( 1 - \frac{t^2 + o(1)}{2n} \right)^n \right)} = e^{-t^2/2} $$ Is all of this correct?