Uniform limit of continuously differentiable function

Question

Consider $$f_n(x)=\sqrt{x^2+\frac{1}{n}}$$ on $[-1,1]$

I know $f_n(x) \rightarrow \vert x \vert$ pointwise on $[-1,1]$

How to prove this convergence is also uniform? That is, how to make $$\Big\vert \;\sqrt{x^2+\frac{1}{n}} -\vert x \vert \;\; \Big\vert $$ as small as possible? Any hint?

Answer 1

For $a,b \ge 0$, we see $$(a+b)^2 \ge a^2 + b^2 \,\,\, \implies \,\,\, a + b \ge \sqrt{a^2 + b^2}.$$ This shows that $$\sqrt{x^2 + \frac 1 n} \le \lvert x \rvert + \frac 1 {\sqrt n}.$$ It's also clear that $\lvert x \rvert \le \sqrt{x^2 + \frac 1 n}$ and thus $$0 \le \sqrt{x^2 + \frac 1 n} - \lvert x \rvert\le \frac{1}{\sqrt n}$$ shows that $\sqrt{x^2 + \frac 1 n} \to \lvert x \rvert$ uniformly for $x \in \mathbb R$.

Answer 2

Note that $$ \lvert x\rvert \leq f_n(x) \leq \lvert x\rvert + \frac{1}{\sqrt{n}}\qquad \forall x\in[-1,1]\tag{$\dagger$} $$ the second inequality e.g. by observing that, for any $x$, $$ \left(\lvert x\rvert + \frac{1}{\sqrt{n}} \right)^2 = f_n(x)^2 + 2\frac{\lvert x\rvert}{\sqrt{n}} \geq f_n(x)^2\,. $$

Answer 3

Note that

$$\sqrt{x^2+\frac{1}{n}} -\vert x \vert =\left(\sqrt{x^2+\frac{1}{n}} -\vert x \vert\right) \frac{\sqrt{x^2+\frac{1}{n}} +\vert x \vert}{\sqrt{x^2+\frac{1}{n}} +\vert x \vert} = \frac{x^2 + \frac{1}{n}- x^2}{\sqrt{x^2+\frac{1}{n}} +\vert x \vert} \leqslant \frac{1}{\sqrt{n}} $$