= Solution
\b[Yes; in fact the limit is uniformly continuous.] Given $\varepsilon>0$, <uniform convergence> provides one $N$ with $\sup_x|f_N(x)-f(x)|<\varepsilon/3$. <Uniform continuity> of this particular $f_N$ provides $\delta>0$ such that $|x-y|<\delta$ implies $|f_N(x)-f_N(y)|<\varepsilon/3$. The <triangle inequality> then gives
$$
|f(x)-f(y)|\leq|f(x)-f_N(x)|+|f_N(x)-f_N(y)|+|f_N(y)-f(y)|<\varepsilon.
$$
This 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>.
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