Dividing by 0 yields 0?

I was messing around with this function trying to find it’s critical points and stumbled upon the following conundrum:

for some reason, Desmos seems to believe that 0^(1/0) is 0. furthermore, it seems to believe the function x^(1/0) is defined in [-1,1]:

does anybody have an explenation for this? Thank you!

Hey there!
1/0 in the calculator behaves like +Infinity for many purposes, even though it displays as undefined . 0/0 and 1/0 have different calculation rules, even though they display the same way.

A fair amount of our behavior here is inherited from the IEEE standard for floating point arithmetic, which defines +Infinity , -Infinity , and the rules for what happens when you raise other numbers to those powers.