Function Sequence on Sequence Another analysis question:
$f_n\to F$ uniformly on $D$; $f_n$ is continuous on $D$. $x_n$ is a sequence with $x_n\to x$.
Prove that:
$$\lim_{n\to\infty}f_n(x_n)=F(x)$$
I've done some basic fudging, but really have no idea where to go from here once again.
Thanks!
 A: Hint: Assume that $f_n\xrightarrow{\text{uniformly}} F$ on $D$ and $f_n$'s are continuous. Then $F$ is continuous on $D$. Assume that $x_n\to x$ on $D$. Hence prove the following steps to obtain $f_n(x_n)\to F(x)$.


*

*First show that $f_n\to F$ uniformly on $D\Longleftrightarrow M_n:=\sup\limits_{t\in D}|f_n(t)-F(t)|\to 0$ as $n\to \infty.$

*Show that for each $n\in\Bbb N$, $|f_n(x_n)-F(x_n)|\le M_n\to 0$   (use step-1). 

*Use the continuity of $F$ to show that $|F(x_n)-F(x)|\to 0$ as $n\to \infty$.

*Hence conclude that $|f_n(x_n)-F(x)|\le |f_n(x_n)-F(x_n)|+|F(x_n)-F(x)|\to 0$ as $n\to\infty$.
A: Let $\varepsilon > 0$. Since $f_n \to F$ uniformly on $D$, there exists a positive integer $N$ such that if $n \ge N$, then $|f_n(t) - F(t)| < \varepsilon/2$ for all $t\in D$. Since $F$ is continuous (being the uniform limit of a sequence of continuous functions), there exists $\delta > 0$ such that for all $t\in D$, $|t - x| < \delta$ implies $|F(t) - F(x)| < \varepsilon/2$. As $x_n \to x$, there exists a positive integer $k$ such that $|x_n - x| < \delta$ for all $n \ge k$. Let $j = \max\{k,N\}$. If $n \ge j$, then $|x_n - x| < \delta$, which implies $|F(x_n) - F(x)| < \varepsilon$, and consequently
$$|f_n(x_n) - F(x)| \le |f_n(x_n) - F(x_n)| + |F(x_n) - F(x)| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.$$
Since $\varepsilon$ was arbitrary, $\lim\limits_{n\to \infty} f_n(x_n) = F(x)$.
A: $f_n$ converges to $f$ uniformly and $f$ is the limit of continuous functions, so $f$ is continuous.
Let $\epsilon>0$. Then $\exists N_1\in\mathbb{N}:\forall n\geq N_1\implies |f(x_n)-f(x)|<\frac{\epsilon}{2}$. 
Because we have that each $f_n$ converges uniformly to $f$, there is $N_2\in\mathbb{N}:\forall n\geq N_2\implies |f_n(y)-f(y)|<\frac{\epsilon}{2}$ for $\forall y\in D$.
Then let $N=\max\{N_1,N_2\}$. Then for $\forall n\geq N$ we have that 
$$|f_n(x_n)-f(x)|\leq |f_n(x_n)-f(x_n)|+|f(x_n)-f(x)|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$
