Let $f_n(x)=e^{-nx}$ on the interval [0,1]. Explain why the sequence of functions {$f_n$} converges pointwise on [0,1]. What is the limiting function and is it continuous? Is the convergence uniform?
I think I have an intuitive idea of how to do this, but formalizing it and the portion on uniform convergence trip me up.
On
First, We need to observe that $$ \lim_{n\to \infty}e^{-nx}=\begin{cases} 1,x=0\\ 0,x\in (0,1] \end{cases} $$ Thus the limiting function is not continuous at $x=0$. In addition, $$ \lim_{n\to \infty}e^{-n \times (1/n)}=e^{-1}, $$ which means that it doesn't converges uniformly. In fact, $$ \exists \epsilon=e^{-1}, \forall \delta>0, |e^{-n\times (1/n)}-0|=e^{-1}\ge \epsilon $$
To prove that $f(x)=e^{-nx}$ converges pointwise, pick a fixed x first. If $x = 0$, $f(x)=1$ as $n \rightarrow \infty$. If $x>0$, $f(x)=0$ as $n \rightarrow \infty$. So $f(x)$ does converges pointwise. Now the limiting function is not continuous at $0$ as you can see.
The idea of uniformly convergent is that your convergence does not depend on $x$. As far as the intuition goes, think about they gave you any large number $N$, as you pick $x$ close enough to zero, $f_n(x)$ will approach $1$ eventually as $x$ gets close enough to $0$, which makes it impossible to make $|f_n(x)-f(x)|<\epsilon$