Understanding why a limit proof using another limit works

Question

Sorry for the title, hopefully I can explain it better. I think the title is about as good as I could get in terms of description.

I have a problem:

Let $x_n \ge 0$ for all $ N \in \mathbb{N}$

If $(x_n) \to x$, show that $(\sqrt{x_n}) \to \sqrt(x)$

Assume that we have already proved the limit going to zero.

My proof was as follows:

Our goal is to find an $N$ that satisfies the inequality:

$|\sqrt{x_n|} - \sqrt{x}| \lt \epsilon$

with epsilon being arbitrary.

So:

$|\sqrt{x_n|} - \sqrt{x}| \lt \epsilon$

$|\sqrt{x_n|}| \lt \epsilon + \sqrt{x}$

$|\sqrt{x_n|}|^2 \lt (\epsilon + \sqrt{x})^2$

$|x_n| \lt (\epsilon + \sqrt{x})^2$

$|x_n| \lt \epsilon^2 + 2 \epsilon \sqrt{x} + x$

$|x_n - x| \lt \epsilon^2 + 2 \epsilon \sqrt{x}$

Since we already know $|x_n - x|$ can be made arbitrarily small we are ready to proceed.

Then allow $\epsilon > 0$ to be arbitrary and choose an $N \in \mathbb{N}$ satisfying:

$|x_n - x| \lt \epsilon^2 + 2 \epsilon \sqrt{x}$

For $n \ge N$ we find after some algebra (to save typing the above backwards)

$|\sqrt{x_n|} - \sqrt{x}| \lt \epsilon$

Which shows that given the limit we can choose an $N$ for any given $\epsilon$ and find that all $n \ge N$ will be inside the $\epsilon$-neighborhood of $\sqrt{x}$.

Where I am confused is how I am using the given limit $(x_n) \to x$. I am sort of following a template here from the author. Adding this extra limit has confused me.

How does reducing the inequality $|\sqrt{x_n|} - \sqrt{x}| \lt \epsilon$ to $|x_n - x| \lt \epsilon^2 + 2 \epsilon \sqrt{x}$ and then knowing "we can make it arbitrarily small" help us prove the given limit? What is the intuition?

Answer 1

I want to stress a point made by Ovi in his comments that can lead you to big problems in lots of other cases, when working with inequalities.

Your argument starts with (I'm removing a few superflous "|"):

$|\sqrt{x_n} - \sqrt{x}| \lt \epsilon$

$|\sqrt{x_n}| \lt \epsilon + \sqrt{x}$

This correct in the sense of

$$|\sqrt{x_n} - \sqrt{x}| \lt \epsilon \Rightarrow |\sqrt{x_n}| \lt \epsilon + \sqrt{x}$$

However, that is not the implication that you want, because you are working backwards: The part $|\sqrt{x_n} - \sqrt{x}| \lt \epsilon$ is what you want to prove in the end, and you are looking for ways to get there.

That means what you want is an implication of the sort

$$ ???\Rightarrow |\sqrt{x_n} - \sqrt{x}| \lt \epsilon.$$

The important thing to note is that if you substitute $|\sqrt{x_n}| \lt \epsilon + \sqrt{x}$ for the "???" part, you get something that is patently untrue:

$$|\sqrt{x_n}| \lt \epsilon + \sqrt{x} \Rightarrow |\sqrt{x_n} - \sqrt{x}| \lt \epsilon $$

With "patently untrue" I mean that you can find values for $x_n$ and $x$ that fulfill the assumption of the implication, but the difference $|\sqrt{x_n} - \sqrt{x}|$ is not just a little bit bigger than $\epsilon$, but vastly bigger:

Set $x_n=0$, $x=1,000,000$ and we of course have that the assumption is fullfilled, but $|\sqrt{x_n} - \sqrt{x}|$ is not smaller than $\epsilon$, but $1,000$.

When working backwards with inequalities, always use extreme care to either use only transformations that are equivalent for inequalities (like multiplication with a positive value on both sides), or make steps of the form "If I could prove this, what I want to prove in the end would follow".

Answer 2

If $x=0$, the proof is trivial.

If $x\neq 0$, $|\sqrt{x_n}-\sqrt{x}|=\frac{|x_n-x|}{|\sqrt{x_n}+\sqrt{x}|}$.

Then we can find a $N$, s.t. $|\sqrt{x_n}+\sqrt{x}|>a>0$ when $n>N$, where $a$ is a constant.

$|\sqrt{x_n}-\sqrt{x}|<\frac{|x_n-x|}{a}\rightarrow0$.

Answer 3

You're using the fact that $x_n \rightarrow x$ in the very first step: that's how you know that you can make $|x_n - x|$ arbitrarily small. And you need to be able to make that difference arbitrarily small for the rest of the proof to work.