The limit of periodic function sequence is periodic Suppose $f_n(x)$ is continuous and periodic, $f_n(x+T_n)=f_n(x)$. 
If we know that $f_n(x) \to f(x)$ uniformly and $T_n \to T$, 
can we conclude that $f(x+T)=f(x)$ ?
 A: Hint: 
$$|f(x+T) - f(x)| = |f(x+T) - f(x+T_n) + f(x+T_n) - f_n(x+T_n) + f_n(x+T_n) - f(x)|$$
Since the $f_n$ are continuous and converge uniformly, what do you know about $f$?
Hope this helps. 
A: There is a catch to the question as we can only assume that $f_n$ converges to $f$ uniformly on compact subsets
(for example $\sin (2\pi x/T_n)$ converges uniformly to $\sin(2\pi x/T)$ on compacts but not on ${\Bbb R}$, except if $T_n=T$)
Let $K_N=[-NT, NT]$  (compact interval). Then $f_{|K_{N+2}}=\lim (f_n)_{|K_{N+2}}$
 is continuous, being the uniform limit of continuous functions.
Let $\epsilon>0$ and find $0<\delta<T$ so that $x,y\in K_{N+2}, |x-y|< \delta \Rightarrow |f(x)-f(y)| < \epsilon/3$.
For $n$ large enough $\|f-f_n\|_{K_{N+2}}<\epsilon/3$ and $|T-T_n|<\delta<T$.
 Let $x\in K_N$. Then for any $x\in K_N$ we have  $x+T_n, x+T\in K_{N+2}$. So:
$$ |f(x)-f(x+T)|\leq \epsilon/3 + |f(x)-f(x+T_n)| \leq \epsilon + |f_n(x)-f(x+T_n)| = \epsilon$$ and $\epsilon>0$ was arbitrary. The above can surely be tidied up somewhat...
