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

Question

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?

Answer 1

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 $$0 \leq f_n -\vert x \vert\leq \frac{1}{\sqrt{n}} \longrightarrow 0$$