I want to show if the following sequence of functions converges pointwise and converges uniformly on R? $ f_n(x) = e^{-nx^2}. $
Here's what I did: -Fix $\displaystyle x\in R, \lim_{n\rightarrow\infty } f_n(x) = \lim_{n\rightarrow\infty }e^{-nx^2}=0 $, so $f_n $ converges point wisely to $f $ s.t. $f(x) = 0.$ Now for uniform convergence: $\displaystyle \|f_n - f\|_{\sup} = \sup_{x\in R}\left|e^{-nx^2}-0\right| $. Now how do I show that $\displaystyle \sup_{x\in R}\left|e^{-nx^2}\right| $ goes to $0$ as $n\rightarrow \infty $ and is thus uniformly convergent?
As you've noted in your question and the comments, we have two cases:
If $x \ne 0$, then the sequence is of the form $(e^{-x^2})^n$ which is a decaying exponential.
If $x = 0$, the sequence of values is constant $1$.
Hence, we have convergence everywhere. The limit is discontinuous (since it is $1$ at $0$ and $0$ otherwise), so the convergence is non-uniform.