 Solution

Yes; in fact the limit is uniformly continuous. Given $\epsilon >0$, uniform convergence provides one $N$ with ${\sup }_x|f_N(x)-f(x)|<\epsilon /3$. Uniform continuity of this particular $f_N$ provides $\delta >0$ such that $|x-y|<\delta$ implies $|f_N(x)-f_N(y)|<\epsilon /3$. The triangle inequality then givesThis same $\delta$ works at every point of $\mathbb{R}$, proving uniform continuity and therefore continuity. The use of one global $N$ is exactly what fails for mere pointwise convergence.