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Let $f_n(x)$ converges to a function $f(x)$ and $x_n$ is a sequence converging to $x$. show that $ \lim\limits_{n\to\infty} f_n(x_n)=f(x). $

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    nour: What happened to the assumption that the functions $f_n$ are continuous?2012-06-10

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The current formulation of the question is not true. We need the convergence of $f_{n}$ to $f$ be uniform.

Consider $f_{n}:[0,1]\to\mathbb{R}$ given by $f_{n}(x)=x^{n}$. Now $f_{n}\to f$ pointwise (not uniformly!) where $f:[0,1]\to\mathbb{R}$ is given by \begin{align*} f(x)=\begin{cases}&0,\,\,\mathrm{if}\,\,x\in[0,1[\\ &1,\,\,\mathrm{if}\,\,x=1. \end{cases} \end{align*} Choose $x_{n}=1-\frac{1}{n}$ for all $n\in\mathbb{N}$, whence $x_{n}\to 1$ and $f(1)=1$. Yet \begin{align*} f_{n}(x_{n})=\left(1-\frac{1}{n}\right)^{n}\to\frac{1}{e}\neq 1 \end{align*} So we have that $f_{n}(x_{n})$ does not converge to $f(x)$.


Answer before question was edited: Let $\varepsilon>0$. Since $f_{n}\to g$ uniformly we find $n_{1}\in\mathbb{N}$ so that $|f_{n}(x)-g(x)| <\frac{\varepsilon}{2}$ for all $n\geq n_{1}$ and $x\in\mathbb{R}$. Similarly, since $x_{n}\to a$ and $g$ is continuous (you need to prove this: uniform limit of continuous functions is continuous) then $g(x_{n})\to g(a)$, so we find $n_{2}\in\mathbb{N}$ with $|g(x_{n})-g(a)|<\frac{\varepsilon}{2}$ for all $n\geq n_{2}$.

Since for all $n\in\mathbb{N}$ we have \begin{align*} |f_{n}(x_{n})-g(a)| &\leq |f_{n}(x_{n})-g(x_{n})|+|g(x_{n})-g(a)|, \end{align*} then what can you conclude from here?