Consider the sequence of functions defined by $$ f_n(x)= \left(1+ \frac{x} {n}\right )^{n^2} $$ for $x \in [-1,0]$. What can I say about the convergence?
My attempt
For every $n$, $f_n(0)=1$. Then using the Taylor expression, for every $x \in [-1,0)$ $$ \lim_{n \to +\infty} f_n (x) = \lim_{n \to +\infty} e^{\ln(\left(1+ \frac{x} {n}\right)^{n^2}) } = \lim_{n \to +\infty} e^{n^2 \ln(\left(1+ \frac{x} {n}\right)) } = \lim_{n \to +\infty} e^{n^2 \frac{x} {n}} = \lim_{n \to +\infty} e^{nx} = 0 $$ We have a limit function which is not continuous, whereas every $f_n$ is continuous. So we don't have uniform convergence.
I would like to be sure about it. Do you know other ways to solve the problem? Thanks for your comments and hints.
("extra" curiosity: I was tempted to write down something like (*edited) $ f_n(x)= e^{xn} $. Is such a thing possible or completely wrong? )