# Does $\lim_{x\to 0^+} \sqrt{x}$ exist or not?

Does this right-handed limit exist or not? $$\lim_{x\to 0^+} \sqrt{x}$$

Schaum's Easy Outline of Calculus (Second Edition) says it does. And doesn't.

An example in the book states:

The function $f(x)=\sqrt{x}$; then $f$ is defined only to the right of zero.

Okay. So the right-handed limit does exist.

Hence, $\lim_{x\to 0} \sqrt{x}=\lim_{x\to 0^+} \sqrt{x}=0$.

Okay. I'm still with you. The limit is $0$.

Of course, $\lim_{x\to 0^+} \sqrt{x}$ does not exist,...

Qué, Mr. Fawlty?

... since $\sqrt{x}$ is not defined when $x<0$.

Okay, so are they messing with me? Is my coffee too weak? Too strong? Is there some subtle truth about limits that escapes me?

Or is that a typo? Did they mean in that last line to omit the '$+$' by the '$0$' and write "Of course, $\lim_{x\to 0} \sqrt{x}$ does not exist,..."?

EDIT:

• That's probably a typo. Are you absolutely sure that's what the textbook says? Can you provide a screenshot?
– 5xum
Sep 10, 2015 at 12:16
• I did question my aging eyes... I'll see what I can come up with. Sep 10, 2015 at 12:18
• It is a typo. Take a look at: math.stackexchange.com/a/637299/137035
– user137035
Sep 10, 2015 at 12:21
• Looking at the book (I found only 1st edition), there seem to be a misprint, yes. Sep 10, 2015 at 12:22
• @Quality - The typos keep you on your toes. :D Sep 10, 2015 at 12:41

Both $$\lim_{x\to 0} \sqrt{x} \quad \text{and}\quad \lim_{x\to 0^+}\sqrt{x}$$ exist. In general for a function $f$ with domain $D(f)$, recall the definition of the $$\lim_{x\to a} f(x) = L.$$ The definition says that this means that: For all $\epsilon >0$ there is a $\delta >0$ such that if $x\in D(f)$ and $0<\lvert x - a \rvert < \delta$ then $\lvert f(x) - L\rvert<\epsilon$. Often we don't write in the requirement that $x$ be in the domain of $f$, but this is a requirement.

Likewise the right hand limit exists.

See this Wikipedia article for more on this: https://en.wikipedia.org/wiki/%28%CE%B5,_%CE%B4%29-definition_of_limit#Precise_statement

• The key element here seems to be the stipulation that $x\in D(f)$. That makes sense to me. We don't have to worry about whether the left handed limit exists, because we're not expected to look there. Sep 10, 2015 at 13:02
• @AdamHrankowski: That's right. Sep 10, 2015 at 13:03

Yes it exists, as you can check by $\varepsilon$-analysis.

Taking any $\varepsilon > 0$, we have $\sqrt{x} < \varepsilon$ if $0 < x < \varepsilon^{2}$.

Put a comma after "defined" in the last sentence. Or equivalently, brackets between "since" and "defined".

Shows why textbook authors really need good English as well as good maths! This happens an awful lot in academia. Very often when there's a bump, you're spending time solving a grammar problem! Understandable for research papers, but who killed off all the editors when it came to textbooks?

• It seems to me that there is a typographical error in the last sentence, which should say that $\lim_{x \to 0^-} \sqrt{x}$ does not exist. Sep 9, 2018 at 11:34
• the problem is not resolved with putting the comma, because first it states the limit exists (=0) and then it says the same limit does not exist. So, the last limit must read $x\to 0-$ Sep 9, 2018 at 12:56