I'm having some trouble with the logic of delta-epsilon arguments. Consider the following:

Prove that $\lim \limits_{x \to a} \sqrt{|x|}$ exists

$$\begin{equation} \begin{split} |x - a| < \delta &\Rightarrow \left | \sqrt{|x|} - \sqrt{|a|} \right | \\[10pt] &\Rightarrow \left | \frac{|x| - |a|}{\sqrt{|x|} + \sqrt{|a|}} \right | \end{split} \end{equation} $$

Suppose $\delta = \epsilon \sqrt{|a|} $

$$\begin{equation} \begin{split} &|x - a| < \epsilon \sqrt{|a|} \\[10pt] &\Rightarrow |x| - |a| < \epsilon \sqrt{a} \\[10pt] &\Rightarrow \frac{|x| - |a|}{\sqrt{|a|}} < \epsilon \\[10pt] &\Rightarrow \frac{|x| - |a|}{\sqrt{x} + \sqrt{|a|}} < \frac{|x| - |a|}{\sqrt{|a|}} < \epsilon \\[10pt] &\Rightarrow \frac{|x| - |a|}{\sqrt{x} + \sqrt{|a|}} < \epsilon \end{split} \end{equation}$$

The problem is

$$ \frac{|x| - |a|}{\sqrt{x} + \sqrt{|a|}} < \epsilon \not \Rightarrow \left | \frac{|x| - |a|}{\sqrt{|x|} + \sqrt{|a|}} \right | < \epsilon $$

Edit: I have a separate proof for when a = 0

  • $\begingroup$ Do you already know that composition of continuous functions is continuous? $\endgroup$
    – user251257
    Jan 8, 2019 at 15:56
  • $\begingroup$ Yes, I assumed a is not equal to zero $\endgroup$ Jan 8, 2019 at 16:01
  • $\begingroup$ @user251257 I didn't consider that. Thanks. But I still would like to know if it is possible to answer the question without resorting to that theorem. $\endgroup$ Jan 8, 2019 at 16:04
  • $\begingroup$ Your question is confusing. First you mention the rationale behind this sort of proof, but then you seem to be asking about how to prove this specifically, (and it's not just an example). What exactly do you want? $\endgroup$
    – Git Gud
    Jan 8, 2019 at 16:08
  • 2
    $\begingroup$ As the previous comment implies, all you missed was the opportunity to take the absolute value of $|x| - |a|$ when you could. You wrote $|x| - |a|$ instead of $\left\vert|x| - |a|\right\vert,$ which gave you a statement that was true but not as strong as you needed. $\endgroup$
    – David K
    Jan 8, 2019 at 16:51

1 Answer 1


In this particular case, you just need to do the lower inequality as well. However, I'm going to give an overview of how to answer these sorts of questions in a way that makes sense, always works, and doesn't give people a headache when they're trying to mark it. I'll write out the proof as I'd write it in Then


and write my commentary outside of them.

So, first, we notice that it's pretty obvious that the limit is going to be $\sqrt{|a|}$. Now, we'll write down some boilerplate that's identical for any such proof:

Given any $\varepsilon > 0$, let $\delta := $

We don't know what $\delta$ needs to be, yet, but we'll figure it out as we go, and just leave a space. Now, on with the boilerplate:

Then for any $x \in \mathbb{R}$ such that $|x - a| < \delta$, we have

And now we'll start with the actual work, and try to fiddle with $|\sqrt{|x|} - \sqrt{|a|}|$ to get something that looks like $|x - a|$, so we can get a $\delta$ out, then use whatever conditions we're going to put on $\delta$ to turn that into an $\varepsilon$. First off, we'll try to use a difference of two squares to neaten things up a bit.

\begin{align*}\left|\sqrt{|x|} - \sqrt{|a|}\right| &= \frac{\left|\sqrt{|x|}-\sqrt{|a|}\right|\left|\sqrt{|x|}+\sqrt{|a|}\right|}{\left|\sqrt{|x|}+\sqrt{|a|}\right|}\\&=\frac{||x|-|a||}{\left|\sqrt{|x|}+\sqrt{|a|}\right|}\end{align*}

Now, that thing on the top looks almost like $|x-a|$, but not quite. The reverse triangle inequality can fix that, though, and then we can get our $\delta$ in there:

\begin{align*}&\leq\frac{|x-a|}{\left|\sqrt{|x|}+\sqrt{|a|}\right|}\mbox{ by the reverse triangle inequality}\\&<\frac{\delta}{\left|\sqrt{|x|}+\sqrt{|a|}\right|}\end{align*}

Now, we're nearly done: just need to get rid of the $x$ at the bottom, but that's easy: $\sqrt{|x|} \geq 0$, so $\left|\sqrt{|x|} + \sqrt{|a|}\right| \geq \sqrt{|a|}$. Since it's in the denominator, we'll flip that around:

$$< \frac{\delta}{\sqrt{|a|}}$$

So, if we just chose $\delta = \sqrt{|a|}\varepsilon$, we'll be done:

$$ = \varepsilon.$$

Now, we'll go back and fill in our definition of $\delta$ at the top:

Given any $\varepsilon > 0$, let $\delta := \sqrt{|a|}\varepsilon > 0. $

and add some more boilerplate at the end:

Thus, $\lim_{x\to a}\sqrt{|x|} = \sqrt{|a|}$.

Putting that all together, the proof ends up as:

Given any $\varepsilon > 0$, let $\delta := \sqrt{|a|}\varepsilon > 0.$
Then for any $x \in \mathbb{R}$ such that $|x - a| < \delta$, we have

\begin{align*}\left|\sqrt{|x|} - \sqrt{|a|}\right| &= \frac{\left|\sqrt{|x|}-\sqrt{|a|}\right|\left|\sqrt{|x|}+\sqrt{|a|}\right|}{\left|\sqrt{|x|}+\sqrt{|a|}\right|}\\&=\frac{||x|-|a||}{\left|\sqrt{|x|}+\sqrt{|a|}\right|}\\&\leq\frac{|x-a|}{\left|\sqrt{|x|}+\sqrt{|a|}\right|}\mbox{ by the reverse triangle inequality}\\&<\frac{\delta}{\left|\sqrt{|x|}+\sqrt{|a|}\right|}\\&\leq\frac{\delta}{\sqrt{|a|}}\\&=\varepsilon.\end{align*} Thus, $\lim_{x\to a}\sqrt{|x|} = \sqrt{|a|}$.

  • $\begingroup$ The reverse triangle equality gives $\leq$ rather than $=,$ because you haven't ensured that $x$ and $a$ have the same sign. You could produce a more complicated formula for $\delta$ to fix that, or you could just write $\leq,$ which works fine. $\endgroup$
    – David K
    Jan 8, 2019 at 16:47

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