$\begingroup$ $0^0=999$ would be a contradiction to the power laws, because then $(0^0)^2 = 999^2 \ne 0^{0\cdot2} = 999$. The only two values for $0^0$ consistent with the power laws are $0$ and $1$. $\endgroup$ –

Nov 30, 2020 · Or as @user854214 puts it well, $(x,y)\mapsto x^y$ has a point of discontinuiry at $(x,y)=(0,0)$, which is utterly fine. Even if one person chose to define $0^0=0$ and another as $0^0=1$ that'd be fine as that just means $0^x,x^0$ aren't both continuous. $\endgroup$ –

Every number (including complex numbers) make this equation true. So we can assume that 0/0 equal all numbers at the same time. This kind of values are called "indeterminate", meaning that there cannot be a specific value assigned for it. Call it 0^0=0/0 by the example shown above, 0^0 is just as indeterminate as 0/0. $\endgroup$ –

In the context of limits, $0^0$ is an indeterminate form because if the "limitand" (don't know what the correct name is) evaluates to $0^0$, then the limit might or might not exist, and you need to do further investigation. Outside of limits, it's best to define $0^0$ as $1$ because the empty product - the product of no numbers - is defined as one.

In short, an exponent 'transforms' the number $1$, so $3$ (or $3^1$). The exponent $1$ 'gives the number $1$ the power to transform into $3$. The exponent $0$ provides $0$ power (i.e. gives no power of transformation), so $3^0$ gives no power …

There is no universal value for $\infty^0$. It is indeterminate , and the value depends on how you are getting the $\infty$ and the $0$. Some other indeterminate forms are $0^0, 1^\infty, \infty\times0,\frac00, 1$.

Possible Duplicate: Zero to zero power I'm wondering why $0^0$ is considered undefined. Why isn't 1 considered a valid solution? Considering $0^0 = 1$ seems reasonable to me for two reasons: $\\

An annoying thing about this discontinuity, is that if we approach it by any line, we get an answer. In particular, the line with slope $\beta$ through the origin always has the limit $$\lim_{t\rightarrow 0}t^{\beta t}$$ and this always equals $1$.

Jan 22, 2017 · 1) x^a × x^b = x^a+b; for x = 0 and a = 0, you would get 0^0 × 0^b = 0^b = 0, so we can't tell anything -- except confirm that 0^0 = 1 still works here! 2) x^{-a}=1/{x^a} -- so when a = 0 , x^{-0} = 1/x^0 = x^0 , which again does work for 0^0 = 1 ; 3) {x^a}^b = x^{a×b} , thus x^(1/n) is the n-th root -- and 1/n = 0 for no value of n , so ...

What is $0^0$? What is the present agreement to this? I feel that it should be 1 as anything to the power zero is one.... I am currently a school student so i would like a more of a school based answer.. So incase it comes in my exam i should know what to write:)