It is easy to see that $f_n(x) = \dfrac{x^2}{\sqrt{x^2+1/n}}$ converge pointwise to $|x|$
Question: is the convergence uniform?
Attempt:
Want to show $\forall \epsilon > 0$, ... , $|f_n(x) - f| < \epsilon$
$|f_n(x) - f| = |\dfrac{x^2}{\sqrt{x^2+1/n}} - \dfrac{x^2}{\sqrt{x^2}}|$
Take $x = \frac{1}{\sqrt{n}}$
$|f_n(x) - f| = |\dfrac{x^2}{\sqrt{x^2+1/n}} - \dfrac{x^2}{\sqrt{x^2}}| = (1-1/\sqrt{2})n$
So $(f_n)$ fails to converge uniformly, can someone confirm?
Also, can someone spot the place where uniform convergence fails for the above sequence in the accompanying figure below?
$$\left|\frac{x^2}{\sqrt{x^2+\frac{1}{n}}}-\frac{x^2}{\sqrt{x^2}}\right|=\left|\frac{x^2\left(\sqrt{x^2}-\sqrt{x^2+\frac{1}{n}}\right)}{\sqrt{x^2+\frac{1}{n}}\sqrt{x^2}}\right|\leq \left|\frac{x^2\left(\sqrt{x^2}-\sqrt{x^2+\frac{1}{n}}\right)}{\sqrt{x^2}\sqrt{x^2}}\right|$$ $$=\left|\sqrt{x^2}-\sqrt{x^2+\frac{1}{n}}\right|=\sqrt{x^2+\frac{1}{n}}-\sqrt{x^2}\leq \sqrt{x^2}+\sqrt{\frac{1}{n}}-\sqrt{x^2}=\sqrt{\frac{1}{n}},\,\forall x\in\mathbb R$$ In the last line we used the inequality $\sqrt{a+b}\leq \sqrt{a}+\sqrt{b},\,\forall a,b>0$ for $a=x^2,\, b=\frac{1}{n}$