I was checking the counter example for sequence of functions $f_n$ in $C^1[0,1]$ such that $\lim_{n\to \infty}f_n$ is not differentiable. I got the counter example from here: Sequence of differentiable functions. I am not able to prove $f_n:[0,1]\to \mathbb R$ defined as $f_n(x)=\sqrt{(x-1/2)^2+1/n}$, which converges uniformly to non-differentiable function $f(x)=|x-1/2|$.
My attempt:-
Consider $$ |\sqrt{(x-1/2)^2+1/n}-|x-1/2||=|\frac{(\sqrt{(x-1/2)^2+1/n}-|x-1/2|)(\sqrt{(x-1/2)^2+1/n}+|x-1/2|)}{\sqrt{(x-1/2)^2+1/n}+|x-1/2|}|$$
$$=|\frac{1/n}{\sqrt{(x-1/2)^2+1/n}+|x-1/2|}|(\because \sqrt{(x-1/2)^2}=|x-1/2|)$$
$$\leq\frac{1/n}{|x-1/2|}$$
Now I am not able to eliminate $x.$ Could you please suggest some technique to complete the proof?