<s>[INST] $\sqrt{x^{2}+\frac{1}{n}}$ converges uniformly to $\left|x\right|$ Im trying to prove that $f_{n}\left(x\right)=\sqrt{x^{2}+\frac{1}{n}}$  converges uniformly to $f(x) = \left|x\right|$ in $[-1,1]$. 
So for evary $\varepsilon$ exists $N \in \mathbb{N}$ s.t for all $n>N$ and for all $x \in [-1,1]$   $\left|f_{n}\left(x\right)-f\left(x\right)\right| \le \varepsilon$
And this how i tried to continue : 
$\left|f_{n}\left(x\right)-f\left(x\right)\right|=\left|\sqrt{x^{2}+\frac{1}{n}}-\left|x\right|\right|=\left|\sqrt{x^{2}+\frac{1}{n}}-\sqrt{x^{2}}\right|=\left|\frac{\left(\sqrt{x^{2}+\frac{1}{n}}-\sqrt{x^{2}}\right)\left(\sqrt{x^{2}+\frac{1}{n}}+\sqrt{x^{2}}\right)}{\sqrt{x^{2}+\frac{1}{n}}+\sqrt{x^{2}}}\right|=\left|\frac{x^{2}+\frac{1}{n}-x^{2}}{\sqrt{x^{2}+\frac{1}{n}}+\sqrt{x^{2}}}\right|=\left|\frac{\frac{1}{n}}{\sqrt{x^{2}+\frac{1}{n}}+\sqrt{x^{2}}}\right|$
Which lead me to the same problem of $\left|\sqrt{x^{2}+\frac{1}{n}}-\sqrt{x^{2}}\right|$
any advice ?
 [/INST] You can notice that
$$
\sqrt{x^2+\frac1n}\le\sqrt{x^2+2\frac{|x|}{\sqrt n}+\frac1n}=\sqrt{\left(|x|+\frac1{\sqrt n}\right)^2}=|x|+\frac1{\sqrt n}.
$$
So $0<f_n(x)-|x|\le\dfrac1{\sqrt n}$. Hence, the uniform convergence (on all $\mathbb R$, not only on $[-1,1]$).
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