Hi I was looking for $f_n : [0,1] \to \mathbb{R}$ such that every $f_n$ is continuous and $f_n$ converges pointwise to a function $f : [0,1] \to \mathbb{R}$ that is continuous as well. Then I wanted to find a series with $x_n \to a \in [0,1] $ such that : $\lim_{n \to \infty} f_n(x_n) \neq f(a)$.
I was wondering if $f_n : [0,1] \to \mathbb{R} : x \to 2nx \space if \space 0 \leq x < \frac{1}{2n} , 2-2nx \space if \space \frac{1}{2n} \leq x \leq \frac{1}{n}, \space else \space 0 $ ($f_n \to f =0$) and $x_n = \frac{1}{2n}$ ($x_n \to 0$) would work. Because (correct me if I am wrong) $f_n(x_n) = 1$ and $f(x) = 0$.
Are there any simpler examples is my example correct?
Your example is good. Here is another example with each $f_n$ also being smooth:
Consider $f_n : [0, 1] \to \mathbb{R}$ given by $f_n (x) = n x \exp(-n x)$. Note that $f_n (0) = 0$ for all $n \in \mathbb{Z}^+$, and for $x \in (0, 1]$, the sequence $(f_n (x))_{n \in \mathbb{Z}^+}$ is nonzero but converges to $0$. Therefore, $(f_n)_{n \in \mathbb{Z}^+} \to f = 0$. The sequence $(a_n)_{n \in \mathbb{Z}^+} = \frac{1}{n}$ obviously converges to $0$, and we have that $f_n (a_n) = \exp (-1)$ for all $n \in \mathbb{Z}^+$. Then $(f_n (a_n))_{n \in \mathbb{Z}^+} \to \exp (-1) \ne f (0) = 0$.