Pointwise convergence of $f_n(x+1/n)$ Assume that $f_n \to f$ pointwise (that is for all $x$ we have that $f_n(x) \to f(x)$). Let $x\in \mathbb{R}$. Is the following statement true?
$$
\lim_{n \to \infty}f_n \left(x+\frac{1}{n}\right) = f(x)
$$
I am not sure why this would be true. I know that the limit of $x+1/n$ is $0$ but it doesn't seem that this is true.
 A: Consider the function $f_n(x) = x^n$ for $x \in [0,1]$. Note that $f_n(1) = 1^n = 1$ for any $n$. This function converges pointwise to the function $f(x) = 0$ for $x \in [0,1)$ and $f(1) = 1$ but not uniformly convergence. Consider 
$$
\lim_{n \to \infty} f_n\left(x - \frac1n \right) = \lim_{n \to \infty}\left(x - \frac1n \right)^n = 0
$$
for $x \in [0,1)$. However, for $x = 1$
$$
\lim_{n \to \infty} f_n\left(1 - \frac1n \right) = \lim_{n \to \infty}\left(1 - \frac1n \right)^n = e^{-1} \neq 1
$$
Your argument works if the functions converge uniformly and also continuous. Rigorously, 

Exercis 7.9 (Principles of Mathematical Analysis, Rudin):
If $\{f_n\}$ is a sequence of continuous functions which converges uniformly to a function $f(x)$ on a set $E$, then 
  $$
\lim_{n \to \infty} f_n(x_n) = f(x)
$$
  for every sequence of points $x_n \in E$ such that $x_n \to x$ and $x \in E$.

A: In context: 
Exercise 7.9 Baby Rudin quoted by Tab1e.
Show that $\lim_n f_n(x_n)=f(x)$, where 
$x_n \rightarrow x$, and $f_n$ continuos and uniformly convergent to $f$ on $E$.
1) $f_n$ uniformly convergent to $f$:
$\epsilon >0$ given, there is a $n_0(\epsilon)$ s.t.
$|f_n(y)-f(y)| \lt \epsilon$ for $n\ge n_0$, and $y
\in E$.
2) Since $f$ is continuos:
There is a $n_1(\epsilon,x)$ s.t. for $n\ge n_1$ 
$|f(x_n)-f(x)|\lt \epsilon$.
3) For $n\ge \max (n_0,n_1)$ we have 
$|f_n(x_n)-f(x)| =$
$|f_n(x_n)-f(x_n)| +|f(x_n)-f(x)| \lt 2\epsilon.$
NB: $f_n$ continuos and uniformly convergent to $f$ implies $f$ is continuos.
