What is (1/0)^0?

>see order of operation
gb2/ 8th grade

(a/b)^n = (a^n)/(b^n) fucking retard

Let's fall for the le transcendental continuation meme.
[eqn]\left( \frac{1}{x} \right)^x \,=\, \mathrm e^{x\,\ln\,\frac{1}{x}} \,=\, \mathrm e^{-x\,\ln\,x}[/eqn]
Since [math]x\,\ln\,x \,\xrightarrow[x \,\rightarrow\, 0]{}\, 0[/math], we have [math]\lim_{x \,\rightarrow\, 0} \left( \frac{1}{x} \right)^x \,=\, \mathrm e^0 \,=\, 1[/math].

>not taking L'Hospitals rule

Sorry, you're absolutely right in that we cannot assume that they aren't since we have yet to establish So as it stands, this question is missing context or other details.

Lame.
I'll invent a number! It will be indivisible by every number! It won't even be divisible by 1 .! And it will have all signs at once! It'll be negative, positive, imaginary... everything! And its value won't even make sense!

Would that have any practical purpose?

In the case where the derivative is not defined, and thereby L'Hospitals rule won't yield any usable forms, does Squeeze theorem always apply?

>won't work because 1^0 will make it not indeterminate

What about sinx/x , x=0 ?
L'Hospitals rule yields cosx/1
literally 1/1 = 1

No... Not until some-one finds a use for it anyway.