I am taking a calculus course, and in one of the exercises in the book, I am asked to find the limits for both sides of $\sqrt{x}$ where $x \to 0$.
Graph for sqrt(x) from WolframAlpha:
This is how I solved the exercise:
For simplicity, I choose to disregard the negative result of $\pm\sqrt{x}$. Since we are looking at limits for $x \to 0$, both results will converge at the same point, and will thus have the same limits.
$\sqrt{x}$ = $0$ for $x = 0$.
$\sqrt{x}$ is a positive real number for all $x > 0$.
$\displaystyle \lim_{x \to 0^+} \sqrt{x} = \sqrt{+0} = 0$$\sqrt{x}$ is a complex number for all $x < 0$.
$\displaystyle \lim_{x \to 0^-} \sqrt{x} = \sqrt{-0} = 0 \times \sqrt{-1} = 0i = 0$
The solution in the book, however, does not agree that there exists a limit for $x \to 0-$.
I guess there are three questions in this post, although some of them probably overlaps:
- Does $\sqrt{x}$ have a limit for $x \to 0$?
- Are square root functions defined to have a range of only real numbers, unless specified otherwise?
- Is $\sqrt{x}$ continuous for $-\infty < x < \infty$?
WolframAlpha says the limit for x=0 is 0: limit (x to 0) sqrt(x)
And also that both the positive and negative limits are 0: limit (x to 0-) sqrt(x)
If my logic is flawed, please correct me.