Convergence of functions in a certain point

Question

Assume that $(f_n)$ is a sequence of continuous functions with $f_n:[0,1] \to \mathbb{R}$. Assume now that $(x_n)$ is a sequence of elements from the interval $[0,1]$ such that $x_n \to 0$ and $f_n(x_n)=0$ for each $n$. Does it follow that $f_n(0) \to 0$ for $n \to \infty$?

(Background is a bit difficult to explain. I am for days trying to understand a particular definition of the osculating circle as it occurs in the differential geometry of curves and now I reached a point where the above problem occurs. However, it is possible that the above assertion is utterly wrong and I am still on the completely wrong way...)

Answer 1

No. Take, for instance,$$f_n(x)=\begin{cases}1-nx&\text{ if }x\leqslant\frac1n\\0&\text{ otherwise.}\end{cases}$$Then each $f_n$ is continuous, $(\forall n\in\Bbb N):f_n\left(\frac1n\right)=0$, but $(\forall n\in\Bbb N):f_n(0)=1$.

Answer 2

Let $$f_n(x) = \begin{cases}0 & x \notin [-1/n, 1/n] \\ nx+1 & x \in [-1/n, 0] \\ -nx + 1 & x \in [0, 1/n]\end{cases}.$$

This is a tent function that is supported on $[-1/n, 1/n]$.

Then $x_n = 2/n$ satisfies $f_n(x_n)=0$ and $x_n \to 0$, but $f_n(0) = 1$ for all $n$.