# If $f_n\to f$ pointwise and $f_n(x_n)$ converges, does $\lim f_n(x_n)=\lim f(x_n)$?

I have a sequence of real-valued functions $$(f_n)_{n\in\mathbb{N}}$$ which converge to $$f$$ pointwise. I also have a sequence of points (which do not necessarily converge) $$(x_n)_{n\in\mathbb{N}}$$ in a real Hilbert space such that $$(f_n(x_n))_{n\in\mathbb{N}}$$ converges. Does this imply that $$\lim_{n\to\infty} f_n(x_n)=\lim_{n\to\infty}f(x_n)?$$ I feel like the answer is probably not, but I have not been able to come up with a counterexample yet.

I found this link, but this is a counterexample to a distinct problem (they assume convergence of $$x_n$$ to a point $$x$$ and compare $$\lim f_n(x_n)$$ with $$f(x)$$; I am asking about the comparison with $$\lim_{n\to\infty} f(x_n)$$.

• Isn't your condition weaker, thus even less likely to be true? Commented Sep 18, 2023 at 9:46
• Unless I am mistaken, math.stackexchange.com/a/570519/42969 in the referenced thread is a counterexample. Commented Sep 18, 2023 at 9:57

$$f_n(x) = \frac{1}{1+(x-n)^2}, \qquad f(x) = 0, \qquad x_n = n.$$
Then $$f_n(x) \to f(x)$$ for every $$x \in \mathbb{R}$$, but
$$\lim_{n\to\infty} f_n(x_n) = 1 \neq 0 = \lim_{n\to\infty} f(x_n).$$