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.


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

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

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

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

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

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

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?

  • $\begingroup$ Hmm unfortunately I don’t think your proof works, because going backwards, I don’t think you can justify every step. For example, I don’t think you can deduce the first step from the second (counting from after the “so”). The place where you use that x_n goes to x is where you fix epsilon and you say that there exists an N s.t. after N, we have |x_n-x|< epsilon. If you didn’t know that x_n goes to x, you couldn’t say that. I am going to sleep now, but if no one answers this just ping me and I can give more details tomorrow. $\endgroup$
    – Ovi
    Feb 15, 2019 at 3:49
  • $\begingroup$ @Ovi That would be wonderful, to be honest it has been a while since I had to remember all the ways to manipulate absolute values. $\endgroup$
    – CL40
    Feb 15, 2019 at 4:47

3 Answers 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.


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".

  • $\begingroup$ Very insightful. Im sorry if this mistake seemed trivial. I am self teaching from a real analysis book after having been out of school for 3 years (I'm planning on starting my MSc in fall). Thank you for taking the time to write this. This chapter has been particular discouraging because its really testing me, haha. The author had made it seem you always just work to find something you can transform backwards into the required epsilon inequality. It seems thats the goal in general, I hadn't realized you can't just "add across inequalities" like I did. Thanks again. $\endgroup$
    – CL40
    Feb 15, 2019 at 15:33
  • $\begingroup$ I compare this to cooking recipes: In order to learn cooking, you follow recipes, so you get a feel for 'how things work'. But when you need to come up with your own recipes, you need to know more: Why do you even need (or want) to "cook" food, why do certain ingrefient always seem to go together, etc. With math it can be similar: Following a proof is one thing, but if you need to come up with one on your own, you need to know more, or at least apply things a bit differently. Just like cooking, not all ideas work out ;-) $\endgroup$
    – Ingix
    Feb 15, 2019 at 22:58

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.


  • $\begingroup$ So I take it my squaring solution was too complicated? What was the idea that gave you the simpler fraction solution? $\endgroup$
    – CL40
    Feb 15, 2019 at 4:19

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