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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?

daw
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  • $\sup\left|x-\dfrac12\right|\left(\sqrt{1-\dfrac{1}{n\left(x-\dfrac12\right)^2}}-1\right)\rightarrow 0$ – Piquito Dec 01 '22 at 00:43
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    $1/\sqrt{n}$ is a better bound to replace your last line. It is independent of $x$ and tends to zero, giving the result. – Joshua Tilley Dec 01 '22 at 00:54
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    More generally, one can prove that $$\sqrt{a+b} \leq \sqrt{a} + \sqrt{b}$$ for any $a,b \geq 0$ ("square both sides"). Then setting $a = (x-1/2)^2$ and $b=1/n$ yields $$\sqrt{(x-1/2)^2+1/n} - |x-1/2| \leq \sqrt{1/n}.$$ – Brian Moehring Dec 01 '22 at 07:04

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