# Proving the sequence $f_{n} = \sqrt{x^{2} + 1/n}$ converges uniformly to $f(x) = |x|$ on $(-1, 1)$.

I have the following exercise from my book:

For each $$n \in \mathbb{N}$$ and each $$x\in (-1, 1),$$ define

$$f_{n}(x) = \sqrt{x^{2} + \frac{1}{n}}$$

and define $$f(x) = |x|$$. Prove that the sequence $$\{f_{n}\}$$ converges uniformly on the open interval $$(-1, 1)$$ to the function $$f$$. Check that each function $$f_{n}$$ is continuously differentiable, whereas the limit function $$f$$ is not differentiable at $$x = 0$$.

My attempt:

First, we need to show $$\forall \epsilon > 0$$, there exists an index $$N$$ such that

$$|f(x) - f_{n}(x)| < \epsilon$$

for all $$n \geq N$$ and all $$x \in D$$. So, we have

$$\left||x| - \sqrt{x^2 + 1/n}\right|.$$

But since $$x^{2} + 1/n > 0$$ always, we have

$$\left||x| - \sqrt{x^2 + 1/n}\right| = \left||x| - |\sqrt{x^{2} + 1/n}|\right| \leq \left|x - \sqrt{x^{2} + 1/n}\right|,$$

where the last equality follows from the reverse triangle inequality. Then,

$$\left|x - \sqrt{x^{2} + 1/n}\right| \leq |x| < \epsilon.$$

So, choose $$N = \lceil{\epsilon}\rceil + 1$$. Does this choice of $$N$$ work? I don't really know. Can someone help me with the rest of the problem?

• Why is $|x|<\epsilon$? Your choice of $\epsilon$ can't depend on $x$. Also, if $x<0$, then $|x-\sqrt{x^2+1/n}|>|x|.$ My professor taught me a cool method to solve this problem using the squeeze theorem for uniform convergence. Can you show $|x|\leq\sqrt{x^2+1/n}\leq(|x|+1/n)?$ – Melody Dec 10 '18 at 1:50
• I cannot prove both inequalities. Can you help me? – joseph Dec 10 '18 at 1:54
• Consider $|x|^2\leq|x|^2+1/n\leq|x|^2+2|x|/n+1/n^2$. Now taking square roots gives us the inequality I mentioned earlier. – Melody Dec 10 '18 at 1:59

Observe that $$\left(\vert x\vert +\frac{1}{\sqrt{n}}\right)^2=x^2+\frac{1}{n}+2 \vert x \vert \frac{1}{\sqrt{n}}=f_n^2+2 \vert x \vert \frac{1}{\sqrt{n}} \geq f_n^2 \geq 0$$ so $$0 \leq f_n -\vert x \vert\leq \frac{1}{\sqrt{n}} \longrightarrow 0$$
• computing $f_n'$ is easy and observe each $f_n$ is differentiable and $f_n'$ is continuous, so $f_n$ is continuously differentiable. – Chinnapparaj R Dec 10 '18 at 2:24