Suppose that $\{f_n\}$ is a sequence of real valued functions on $\mathbb{R}$ that converges uniformly to $f$ and let $\{x_n\}$ be a real sequence converging to $x$.
Show that $\lim_{n \rightarrow \infty} f_n(x_n) = f(x)$.
It seems like the origin Uniform Convergence with a catch , I will be thankful is somecan help me with that :).
First of all, it should be noted that the $f_n$ must be continuous for the result to hold. For example, if we take $f_n \equiv f$ for all $n$ and $$ f(x) := \left\{\begin{array}{lll} 1 & \text{if} & x > 0, \\ 0 & \text{if} & x \leq 0, \end{array} \right. $$ then taking the sequence $x_n = 1/n$ we have for any $n$ that $f_n(x_n) = 1$, but $x_n \to 0$ and $f(0) = 0$.
Adding to the hypothesis that the $f_n$ are continuous, since $f_n \to f$ uniformly we know that $f$ is also continuous. By the triangle inequality, $$ |f_n(x_n) - f(x)| = |f_n(x_n) - f(x_n) + f(x_n) - f(x)| \leq |f_n(x_n) - f(x_n)| + |f(x_n) - f(x)|. $$ Let $\varepsilon > 0$. Since $f_n \to f$ pointwise, then $f_n(x_n) \to f(x_n)$, and since $f$ is continuous and $x_n \to x$, then $f(x_n) \to f(x)$. It follows that there exists $N$ such that for all $n \geq N$, $$ |f_n(x_n) - f(x_n)| < \frac{\varepsilon}{2} \qquad\text{and}\qquad |f(x_n) - f(x)| < \frac{\varepsilon}{2}. $$ Hence, by the inequality previously obtained, for all $n \geq N$, $$ |f_n(x_n) - f(x)| < \varepsilon, $$ which proves that $\lim_{n\to\infty} f_n(x_n) = f(x)$.