Prove uniform convergence of $f_{n} (x) = \sqrt{\frac{1}{n^2} +x^2}$ to $f(x) = |x|$

Question

For each $n ∈ \Bbb N$, let $f_{n} (x) = \sqrt{\frac{1}{n^2} +x^2}. $ Show $(f_{n})^{\infty}_{n=0}$ converges uniformly on $\Bbb R$ to $f(x) = |x|$

I'm trying to find the $N$ such that for all $x \in \Bbb R$ and $n > N$, $$\Bigg|\sqrt{\frac{1}{n^2} +x^2} - |x|\Bigg| < \epsilon$$ but I'm struggling with proving convergence for all real $x$. I'v seen that for $x∈ [0,1]$, for $n$ greater than or equal to $1$, $$\Bigg|\sqrt{\frac{1}{n^2} +x^2} - |x|\Bigg|\leq \frac{1}{\sqrt{n}} $$ but I'm unsure of how to generalize. Any help is appreciated!

Answer 1

Note that$$\sqrt{x^2+\frac1{n^2}}-\lvert x\rvert=\frac{\frac1{n^2}}{\sqrt{x^2+\frac1{n^2}}+\lvert x\rvert}.$$Since the numerator is constant and the denominator is positive and increases to $\infty$ when $\lvert x\rvert\to\infty$, the maximum of $\sqrt{x^2+\frac1{n^2}}-\lvert x\rvert$ is attained when $x=0$. And that maximum is $\frac1n$. So$$(\forall x\in\mathbb{R}):\left\lvert\sqrt{x^2+\frac1{n^2}}-\lvert x\rvert\right\rvert\leqslant\frac1n,$$from which the uniform convergence follows.

Answer 2

If you use $\sqrt a -\sqrt b=\frac {a-b} {\sqrt a +\sqrt b}$ you will get the result easily.

Answer 3

Hint: We have $$|f_n(x) - |x|| = \frac{1}{n^2} \frac{1}{|x|+\sqrt{1/n^2 +x^2}} \le \frac{1}{n}.$$

Answer 4

Hint: $$\left(\vert x \vert +\frac{1}{n}\right)^2=x^2+\frac{1}{n^2}+2\vert x \vert \frac{1}{n}=f_n^2+2\vert x \vert \frac{1}{n} \geq f_n^2 \geq 0$$ so $$0 \leq f_n \le \vert x \vert+\frac{1}{n}$$ and so $$0 \leq f_n - \vert x \vert \leq \frac{1}{n} \rightarrow 0$$

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Moral:

Uniform limit of sequence of continuously differentiable function need not be continuously differentiable!