# Question of basic calculus

by Caesar_Rahil
Tags: basic, calculus
 P: 69 can someone please convince me that lim x->0 sqrt(x) = 0 Who of you say it doesn't exist???
 HW Helper P: 1,372 If you mean $$\lim_{x \to 0^+} \sqrt{x}$$ (limit as x approaches 0 through the positives), then the limit is zero. Notice that this is a one-sided limit. This $$\lim_{x \to 0^-} \sqrt x$$ means you are trying to approach 0 from the left - through negative numbers. I feel confident in saying this limit does not exist. Why might that be? Finally, remember that the ordinary limit exists as a real number if, and only if, the two one-sided limits exist as real numbers and are equal. The comment and observation I made earlier combine to say (fill in the blank yourself) about $$\lim_{x \to 0} \sqrt x$$
HW Helper
Thanks
P: 26,148
 Quote by Caesar_Rahil can someone please convince me that lim x->0 sqrt(x) = 0 Who of you say it doesn't exist???
HI Caesar_Rahil!

lim x -> 0+ sqrt(x) = 0, but lim x->0- sqrt(x) depends on you define sqrt(x) for negative x …

how are you defining it?

Do you define it as not existing, or do you define it as an imaginary number?

 P: 69 Question of basic calculus I'm confining myself to real valued functions since left hand limit does not exit, that means sqrt(x) does not have a limit there so it does not exist is this interpretation correct?