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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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Is this a homework? If yes, what have you tried so far? –  Thomas E. Jun 9 '12 at 23:38
    
I tried to solve this as follows: Since $f_n$ is uniformly continous then $| f_n(y)-f(y)|$<$\epsilon/2$ for all y. and since lim $X_n$=x then $X_n$ is cauchy. and by continuity of f, there exists a $\delta$>0 such that $|x_n-x|$<$\delta$ implies $|f(n_x)-f(x)|<\epsilon/2$ thus by the inequality $|f_n(x_n)-f(x)|=|f_n(x_n)-f(x_n)+f(x_n)-f(x)|<|f_n(x_n)-f(x_n)|+|f(x_n)-f(x)|$<‌​$\epsilon$ .... I used f(x) instead of g(x), but I feel something missing and unsufficent...any help? –  nour Jun 9 '12 at 23:55
    
I would suggest first proving that $g$ is continuous as a uniform limit of continuous functions. Use this to work around with the hint I gave below. In your comments, is $f(x)$ same as $g(x)$? –  Thomas E. Jun 10 '12 at 0:00
    
I edited my answer according to the newly edited question, and as far as I'm concerned, the current claim does not hold. –  Thomas E. Jun 10 '12 at 15:11
    
nour: What happened to the assumption that the functions $f_n$ are continuous? –  Did Jun 10 '12 at 16:57

1 Answer 1

up vote 1 down vote accepted

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?

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