Limit of $\sqrt x$ as $x$ approaches $0$

Question

What is the limit of $\sqrt x$ as $x$ approaches $0$? I asked a few people and they all gave me two different answers. Some said that the limit is $0$, and other say that the limit is not defined because the right-hand limit is $0$ while the left-hand limit is undefined. I'm confused. please help!!

Answer 1

It may depend on definitions: for some, to even talk about

$$\;\lim_{x\to 0}f(x)\;$$

the function must be defined on a complete neighborhood of $\;x_0\;$ , i.e. in an open interval of the form $\;(x_0-\epsilon\,,\,x_0+\epsilon)\;,\;\;\epsilon >0\;$ .

For others, which is what Daniel seems to point at, it is enough to talk of the function within its definition domain.

Thus, under the first point of view, we can not talk about

$$\lim_{x\to 0}\sqrt x\;\;\text{since this function's defined only for $\;x\ge 0\;$}$$

meaning: there is no "left" neighborhood of $\;0\;$ in the definition domain and thus we can only talk of the right-sided limit. It depends on the context.

Answer 2

From what I understand, the limit as $x$ approaches $0$ of $\sqrt{x}$ is in fact $0$. I think I understand where your confusion comes from. If I understand correctly then you will have been taught that

$$\lim_{x\rightarrow 0^-} \sqrt{x} = \lim_{x\rightarrow 0^+} \sqrt{x}$$

then

$$\lim_{x\rightarrow 0} \sqrt{x} $$ will exist or in other words, both the left and right side limit must exist for you to find both limits. However, the subtle "thing" one must understand is that the function has to be defined in the negative spectrum for $x$, i.e. it must exist in the negative domain to begin with. The square root function isn't defined in the negative domain and therefore, by definition, you cannot take the limit in that domain. This doesn't mean that the limit does not exist there, it just means that you cannot take a limit there to begin with. Thus, the limit at $0$ of the $f(x) = \sqrt{x}$ is in fact $0$ i.e.

$$\lim_{x\rightarrow 0} \sqrt{x} = 0$$