If a sequence of functions $f_n(x)$ converges uniformly to a function $f(x)$, and if each $f_n(x)$ is uniformly continuous, then does it follow that the limit function $f(x)$ is also uniformly continuous.
My try:
Since each $f_n(x)$ is uniformly continuous, we have $|f_n(x_n) - f_n(x_m)| < \epsilon/3$, for ....
and $f_n(x)$ converges uniformly to a function $f(x)$, we have $|f_n(x_m) - f(x_m)| < \epsilon/3$, for...
Thus, $$|f(x_m) - f(x_n)| \leq |f_n(x_m) - f(x_m)|+|f_n(x_n) - f_n(x_m)|+|f(x_n) - f_n(x_n)| \\< \epsilon/3 +\epsilon/3 +\epsilon/3 = \epsilon$$
Thus $f(x)$ is also uniformly continuous.
Is the proof correct?
Continuing from my two comments, my proof would look like:
Let $\epsilon > 0$. Then since $|f(x) - f(y)| \leq |f_{n}(x) - f(x)| + |f_{n}(x) - f_{n}(y)| + |f_{n}(y) - f(y)|$, choose $\delta>0$ (independent of $x$) such that the middle term is $< \epsilon/3$ -- we can do this since $f_{n}$ is uniformly continuous for each $n$. Then letting $N$ be large enough such that for every $x$, the first and last terms are each $< \epsilon/3$ (we can do this by uniform convergence), we get that if $|x - y| < \delta$, $|f(x) - f(x)| < 3( \epsilon/3) = \epsilon$, and we are done.