$f_n \colon \mathbb{R} \to \mathbb{R}$ converges continuously to $f$ if for $x_n \to x_0$, $f_n(x_n) \to f(x)$.
Assuming above, say $a_n \to \infty$. Then does $f_{a_n}(x_n) \to f(x)$ too?
I want to prove this because I want to prove that if $n_k$ is a subsequence of $n$, then $f_{n_k}(x_k) \to f(x)$ as $k \to \infty$.
We can consider $\left|f_{n_k}(x_k) - f_0(x_0) \right| \leq \left|f_{n_k}(x_k) - f_0(x_k) \right| + \left| f_0(x_k) - f_0(x_0) \right|$, but this seems like it would be the continuity of $f_0$ and $\left|f_{n_k}(x_k) - f_0(x_k) \right|$ seems hard to manipulate.
I'm using this to prove that $f_n \to f$ continuously if and only if $f_n \to f$ uniformly.
Thank you.
Here is the context:
We assume that $a_k$ is a strictly increasing sequence of natural numbers. Let $x_k\to x.$ Define the sequence $\tilde{x}_n$ by $\tilde{x}_n=x_k$ if $a_k\le n<a_{k+1}.$ For $n<a_1$ we may set $\tilde{x}_n=x_1.$ Then $\tilde{x}_n\to x_0.$ By assumptions $f_n(\tilde{x}_n)\to f(x_0).$ In particular $f_{a_k}(\tilde{x}_{a_k})\to f(x_0).$ But $\tilde{x}_{a_k}=x_k,$ hence $f_{a_k}(x_k)\to f(x_0).$
If $n_k$ is a subsequence of natural numbers then $a_k=n_k$ is strictly increasing, hence $x_k\to x_0$ implies $f_{a_k}(x_k)\to f(x_0).$