Problem 2.3.1: Limits and square roots

Suppose that $x_n\rightarrow x$, where $x>0$. Show that $\sqrt{x_n}\rightarrow\sqrt{x}$.

Are we only assuming the limit $x$ is positive? Can we assume every element of the sequence is also positive?

@violincounter The text actually doesn't assume either, but I think we do need $x>0$. If $x>0$ and $x_n\rightarrow x$, then $x_n>0$ for sufficiently large $n$. So I don't think we need that additional assumption.

Here is my proof-in-progress:

First, we need to establish that $x_n$ is positive for large enough $n$. Choose $N^* \in \mathbb{N}$ so that $n\geq N^*$ implies $|x_n - x| < x$, which we may do since $x>0$. This is the same as saying (for $n \geq N^*$)

$$
-x < x_n - x < x,
$$

which if we add $x$ to the inequality gives $ 0< x_n $.

Now, let $\epsilon > 0$. Choose $N$ so that $N \geq N^*$ and $n \geq N$ implies that $|x_n-x| < \epsilon^2$. Then, for $n \geq N$,

$$
|\sqrt{x_n} - \sqrt{x}| \leq \sqrt{|x_n-x|} < \sqrt{\epsilon^2} = \epsilon.
$$

Here is what is troubling me. The inequality above, $|\sqrt{a} - \sqrt{b}| \leq \sqrt{|a-b|}$, is valid. We can show this by letting $\sqrt{a}$ and $\sqrt{b}$ be the hypotenuse and one side of a right triangle, and then using the geometrical triangle inequality to solve for $\sqrt{a} - \sqrt{b}$. Say $\sqrt{a}$ was the hypotenuse (i.e. $\sqrt{a}$ is larger than $\sqrt{b}$), then,

$$\sqrt{a} \leq \sqrt{a-b} + \sqrt{b} \quad \Rightarrow \quad \sqrt{a} - \sqrt{b} \leq \sqrt{a-b}.$$

The absolute values amount to nothing more than allowing either $\sqrt{a}$ or $\sqrt{b}$ to be the hypotenuse. My question is, how can we show this is true in a more formal, analysis-y way? Or do we need to? Finally, if I have made a mistake anywhere, I would love to know that too!

I'm having a hard time understanding what your question is.

I think you're proof that $\sqrt{x_n} \rightarrow \sqrt{x}$ is perfectly formal and "analysis-y", and I don't think there's anything wrong with using a geometric proof to validate your claims about those inequalities either. It seems that most of the fancier machinery that's being built in these first two chapters is being built on fundamental mathematics + the axiom of completeness. So there shouldn't be anything wrong with doing exactly what you did, I think all this is great!

After talking to Mark, I have revised my proof - I will leave the old post intact to preserve the context of Spiff's reply.

Here is the updated version:

First, we need to establish that $x_n$ is positive for large enough $n$. Choose $N^* \in \mathbb{N}$ so that $n\geq N^*$ implies $|x_n - x| < x$, which we may do since $x>0$. This is the same as saying (for $n \geq N^*$)

$$
-x < x_n - x < x,
$$

which if we add $x$ to the inequality gives $ 0< x_n $.

Let $\epsilon>0$, and choose $N \in \mathbb{N}$ so that $n \geq N$ implies $x_n$ is positive (as per above), and so that $|x_n - x|< \epsilon \sqrt{x}$.

Then, $n \geq N$ implies

$$ |\sqrt{x_n} - \sqrt{x}| = \left| (\sqrt{x_n} - \sqrt{x})\frac{\sqrt{x_n} + \sqrt{x}}{\sqrt{x_n} + \sqrt{x}}\right| = \left| \frac{x_n - x}{\sqrt{x_n}+\sqrt{x}} \right|;$$

the denominator is manifestly positive, so

$$\left| \frac{x_n - x}{\sqrt{x_n}+\sqrt{x}} \right| = \frac{|x_n-x|}{\sqrt{x_n}+\sqrt{x}} < \frac{|x_n-x|}{\sqrt{x}} < \frac{\epsilon \sqrt{x}}{\sqrt{x}} = \epsilon.$$