#lim_(x->0^-) (1/sqrt(x)) = ?# How do we calculate this?

I'm going to answer the question myself, but I want you people to validate my arguments, and please share if you have any critique or different approaches! :)

2 Answers
Jun 27, 2017

If we are working in #RR# the limit does not exist. (I'd say there is no definition for the limit.)

Explanation:

I am used to definitions of one sided limits that presuppose that the domain includes values on the appropriate side.

In #RR#, the domain of #1/sqrtx# includes no value less that #0#, so there is no definition for the limit as #x# approached #0# from the left.

Jun 27, 2017

#lim_(x->0^-) (1/sqrt(x)) = (-i)oo#

Explanation:

Here is my argument:
#lim_(x->0^-) (1/sqrt(x)) iff lim_(x->0^+) (1/sqrt((-x))) = lim_(x->0^+) (1/(isqrt(x))) = lim_(x->0^+) (i^-1/(sqrt(x))) = lim_(x->0^+) ((-i) * 1/(sqrt(x))) = -i * lim_(x->0^+) (1/sqrtx) = (-i)oo#

Do you approve?

My biggest concern is actually with the very first step. I didn't use mathematical laws, I purely used logic, so I'm not sure if it's legal or not.