I'm stumped on this question:
Show that a if series of continuous functions $f_n$ in the interval $[a,b]$ uniformly converges on $(a,b]$, then the limit $lim_{n \to \infty}f(a)$ exists.
I tried using Cauchy's Criterion for convergence and the fact the series is continuous but my proof doesn't seem right.
Was it this proof? Fix $\epsilon>0$. By the Cauchy criterion for uniform convergence, there is $N$ such that for all $n,m>N$ we have $$\vert f_n(x)-f_m(x) \vert<\epsilon$$ for all $x\in (a,b]$. Fix any $n,m>N$. Since $f_n$ and $f_m$ are continuous on $[a,b]$, there is $\delta_n>0$ and $\delta_m>0$ such that $$\vert f_n(a)-f_n(x)\vert<\epsilon$$ and $$\vert f_m(a)-f_m(x)\vert<\epsilon$$ for $\vert x-a\vert<{\rm min}\{\delta_n,\delta_m\}$. Then for any such $x$ \begin{eqnarray} \vert f_n(a)-f_m(a)\vert &=& \vert f_n(a)-f_m(a)+f_n(x)-f_n(x)+f_m(x)-f_m(x) \vert \\ &\leq& \vert f_n(a)-f_n(x)\vert +\vert f_m(x)-f_m(a)\vert + \vert f_n(x)-f_m(x)\vert <3\epsilon \end{eqnarray} Since $n,m\geq N$ were arbitrary, it follows that $f_n(a)$ is a Cauchy sequence in $\mathbb{R}$ and thus converges.