# Another simple/conceptual limit question

Further to my previous post Very simple limits question to clarify my understanding , here's a related question. Let $f(x)=\sqrt x,x\geq 0$. What is the limit of $f$ as $x$ tends to $0$?

I think the answer is $0$, but my textbook claims that the limit doesn't exist because $f$ is not defined on any punctured neighbourhood of $0$, ie. $f$ doesn't have a left-sided limit. But since f is undefined on the negative reals, would you say that $f$'s limit at $0$ is $0$ (and not require a left limit in this case), or would you say that $f$'s limit at $0$ is non-existent (because it has no left-sided limit)?

ps. It's at Example 2c if you type in page 184 in the box on http://www.scribd.com/doc/74564079/Mathematical-Analysis.

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I'd say the book is kind of strange to insist that the limit doesn't exist (of course, the author is free to define his terms any way he pleases, so if he wants the limit to means two-sided limit, there's nothing we can do about it).

But I think the more natural way here is to declare that the limit exists. This is because the domain of $f$ is $[0, \infty)$ and so $f$ doesn't really live on $\mathbb R$ but rather on its subspace $[0, \infty)$ with the induced topology. But the neighborhoods of $0$ in this subspace are $[0, a)$ for any $a > 0$ and $f$ is obviously well-defined on all of these neighborhoods and the limit of $f$ at $0$ exists.

So the answer depends, on what space we are working with, really. But again, it's strange to insist here on $\mathbb R$ because $f$ is not a function on this space.

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The function $f:\mathbb{[0,\infty)}\rightarrow\mathbb{R}$ given by $f(x)=\sqrt{x}$ is what you are talking about. $0$ is a limit point of the domain. As such, you can ask yourself if the limit exists at $0$. And it does, and it is zero, and it is relatively easy to prove.

The fact that it is not "defined in every neighborhood" taking in account the real line is quite meaningless. Everything you need is in the "domain space", the "rest" of the real line does not exist for your purposes.

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Thank you (and to Marek) for your complete answers. – Ryan Jan 22 '13 at 15:42