I want to show $\underset{n \to \infty}{\lim} (1+\frac{x}{n^2} )^n=1 \; \forall x \in \mathbb{R}$. But my solution only works for $x\leq 0$: $$\underset{n \to \infty}{\lim} (1+\frac{x}{n^2} )^n= \underset{n \to \infty}{\lim} (1-\frac{(-x)}{n^2} )^n=\underset{n \to \infty}{\lim} (1-\frac{\sqrt{-x}}{n} )^n(1+\frac{\sqrt{-x}}{n} )^n=e^{-\sqrt{-x}}+e^\sqrt{-x}=e^0 =1$$
The problem is, I cant use the $\exp$ identity with complex numbers because we only showed it for Reals. Is there a way to avoid the problem?
