Let $p\in[1,\infty)$. We know that $C_c^\infty(\mathbb{R}^n)$ is a dense subspace of $W^{k,p}(\mathbb{R}^n)$ and if $\Omega$ open in $\mathbb{R}^n$ then $C^\infty(\Omega) \cap W^{k,p}(\Omega)$ is dense in $W^{k,p}(\Omega)$.
In my notes there is:
if $p=\infty$ these results don't hold, in fact, the limit in $W^{1,\infty}$ of a sequence in $C^\infty \cap W^{1,\infty}$ is a $C^1$ function. As counterexample take $u(x)=|x|$ is in $W^{1,\infty}(-1,1) \setminus C^1(-1,1)$.
The question is: is true that the limit in in $W^{1,\infty}$ of a sequence in $C^\infty \cap W^{1,\infty}$ is $C^1$? Does the sequence $\sqrt{x^2+1/n}$ converges at $|x|$ in $W^{1,\infty}(-1,1)$?
The sequence $$ f_n(x)=\sqrt{x^2+1/n} $$ does not converge to $f(x)=|x|$ in the $W^{1,\infty}-$norm. In fact, $\{f_n\}$ does not converge to any function in the $W^{1,\infty}-$norm, since for every $m\in\mathbb N$, $$ \lim_{n\to\infty}\|f_m-f_n\|_{W^{1,\infty}} \ge\lim_{n\to\infty}\|f_m'-f_n'\|_{L^\infty} =\lim_{n\to\infty}\sup_{|x|\le 1}|f_m'(x)-f_n'(x)|=1. $$