Prove the following function is pointwise convergent, and following this prove further whether or not it is uniformly convergent: $(e^{-nx^2})_{n\in \mathbb{N}}$ $ (x\in\mathbb{R})$
I understand the definition of pointwise convergent, but Im just unsure how to use the definition to prove the convergence.
Set $\epsilon > 0$ then $f_n(x)=(e^{-nx^2})\rightarrow0$ $(n\rightarrow\infty$) to is pointwise convergent to $f(x)=0$
For $n\neq0$ $|f_n(n)-f(n)|=|e^{-nn^2}|=|e^{-n^3}|\rightarrow 0$
For $n=0 |f_n(n)-f(n)|=|e^{-nn^2}|=|e^0|\rightarrow 1$
Therefore as the limit function is not continuous it is not uniformly convergent Is this proof correct?
If $x_n \to \infty$ then $e^{-x_n} \to 0$. Now if $x\neq 0$ then $nx^2 \to \infty$ and $e^{-nx^2} \to 0$. If $x=0$ then $nx=0 \forall n$ and $e^{-nx^2}=1$ does not change. Out of this it follows that $\forall x \in \mathbb{R} e^{-nx^2}$ converges as $n\to \infty$ to some number. This is the statement that the sequence of functions $e^{-nx^2}$ converges pointwise to another function, in this case to the function $f(x)=\begin{cases}0 \quad x\neq 0\\1 \quad x=0\end{cases}$.
Since $e^{-nx^2}$ is continuous for all $n$ but the limit is not, the sequence does not converge uniformly. (As the uniform limit of continuous functions is again continuous).