Expert: Ahmed Salami - 7/3/2010

Question
Hi.
I have this very basic  limit question which I thought I knew the answer to but somehow I am wrong.
It is the limit x ---> 0 (x)^(1/2)
I first thought that since the limit as x --> 0+ f(x)  =0 while the square root function is not defined for x < 0 in real number therefore limit x -->0- f(x) should be undefined so the limit x--->0+ f(x) =!  limit x--->0- f(x) so the limit doesn't exist ,however, how can this be explained?   http://www.wolframalpha.com/input/?i=limit+x---%3E0+%28x%29^%281%2F2%29

Answer
Hi Hamad,
First, i'd like to say that a function is never properly defined until its domain is specified. For instance, the function f(x) = x² defined for all real numbers x > 0 is different from the function g(x) = x² defined for all real x because they have different domains, even though they have the same values at every point where both of them are defined.
A real function is a function whose input and output values are real numbers. When the domain of such a function is not specified explicitly, we will assume that the domain is the largest set of real numbers to which the function assigns real values. Thus, if we talk about the function x² without specifying a domain, we mean the function g(x) above.
Now, the domain of the function f(x) = √ x is the interval [0, ∞), since negative numbers do not have real square roots and so the limit as x approaches 0 wouldnt exist. However, if the function is explicitly defined for all x (i.e including complex values) then the limit would be 0 as shown in the example.

Regards