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

Solution 1:

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 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.

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

Solution 2:

Many introductory calculus books define $\lim_{x\to a}f(x)$ under a premise that $f$ must be defined on an open interval containing $a$. Since any open interval containing $0$ contains negative numbers, it would seem that $\lim_{x\to 0}\sqrt{x}$ is undefined.

However, the definition of a limit in the introductory calculus books is a narrow version of a definition normally used in mathematical analysis. For example, according to Walter Rudin's "Principles of Mathematical Analysis" the limit can be defined at any limit point of the domain of $f$. Since the domain of $f(x)=\sqrt{x}$ is the interval $[0,\infty)$, and $0$ is a limit point for this interval, then by such definition, $\lim_{x\to 0}\sqrt{x}$ is defined and is equal to $0$.

Of course, the concept of a "limit point" would require a discussion that could make an introductory calculus books not so introductory. Hence the narrow definition of a limit. My personal impression is that the introductory calculus books could easily be improved to include a definition of a limit that leaves no doubt about the fact that $\lim_{x\to 0}\sqrt{x}=0$.