The convergence of $\sqrt{x^2+\frac{1}{n}}$

Question

Let $f_n(x)=\sqrt{x^2+\frac{1}{n}}$.

i) Determine the limit-function $f$.

ii) Does $f_n(x)$ converge uniformly to $f$?

For the first: We have $\lim_{n\rightarrow \infty}\sqrt{x^2+\frac{1}{n}}=\sqrt{\lim_{n\rightarrow \infty}x^2+\frac{1}{n}}=\sqrt{x^2}=|x|$,

because $\sqrt{*}$ is continuous.

But I was not able to proof ii). I am pretty sure that the convergence is uniformly, but I dont know how to prove this..:/

I am thankful for any kind of help.

Answer 1

The definition for uniform converge is:

$\forall \epsilon >0, \exists N >0 , \forall n\geq N, \sup_{x\in \mathbb{R}}|f_n(x) -f(x)| < \epsilon $

In this case: $f_n = \sqrt{x^2+\frac{1}{n}}, f = |x|$

For $x \geq 0$, $$f_n(x) -f(x) = \sqrt{x^2+\frac{1}{n}} - x > x-x >0$$ $$f_n(x) -f(x) = \frac{1/n}{\sqrt{x^2+\frac{1}{n}} + x } \leq \frac{1/n}{\sqrt{0^2+\frac{1}{n}} + 0 } = \sqrt{1/n} \leq \sqrt{1/N}$$

In other words, $\sup_{x\in \mathbb{R^+\cup \{0\}}}f_n(x) -f(x) = \sqrt{1/N}$

For $x \leq 0$, in same manner, $$\sup_{x\in \mathbb{R^-\cup \{0\}}}f_n(x) -f(x) = \sqrt{1/N}$$

Let $\sqrt{1/N} = \epsilon$ So $|f_n(x) -f(x)|$ is well-bounded.

$\forall \epsilon >0, N = 1/\epsilon^2$ satisfying uniform convergence.

How can I come up with such idea:

Firstly, I do the x>0 case:

$$ \sqrt{x^2+\frac{1}{n}} - x < \epsilon \implies \sqrt{x^2+\frac{1}{n}} <\epsilon+x \implies x^2+\frac{1}{n} < x^2+2\epsilon x+ \epsilon^2$$.

On the left hand side the item only independent of x is $\frac{1}{n}$. One the RHS, it is $\epsilon^2$

So I let $1/n = \epsilon^2$ and find such method.

Answer 2

Using $\sqrt {a+b} \le \sqrt a + \sqrt b$ for $a,b\ge 0,$ we see that $\sqrt {x^2 + 1/n} \le |x| + \sqrt {1/n}.$ Thus, for any $n,$

$$0 \le \sqrt {x^2 + 1/n} - |x| \le \sqrt {1/n}$$

for all $x\in \mathbb R.$ This gives uniform convergence on $\mathbb R.$