If $0^0 = 1$ what is $\lim_{x\rightarrow 0} 0^{x}?$ If $0^0 = 1$ what is $\lim_{x \rightarrow 0} 0^{x}?$ Intuitively, it would appear to be equal to 1 as well since $\lim_{x \rightarrow \infty} f(x) = x = 0$. At the same time, if I consider a member of the set of points approaching the left-sided limit, something like $0^{-1}$, then $0^{-1}$ would appear to be something like $\dfrac{1}{0}$ and that can't be right. S0 there would be a discontinuity at $x = -1$ for $0^{x}$. But then it appears there would be no discontinuity at $x = 0$ either then. In any case, I'm unsure how to approach answering this problem rigorously and would appreciate some advice.
 A: Defining $0^0=1$ is a good idea when doing discrete mathematics, or set theory, or other things related to exponentiation and natural numbers.
It is less fortuitous in the context of analysis, since one can argue that $0^x$ and $x^0$ will disagree on the limit as $x\to0$, not to mention that if $x_n\to 0$ and $y_n\to 0$, then $\lim {x_n}^{y_n}$ could be a great deal of things. So it is not a bad idea to leave $0^0$ as undefined.
But looking at $\lim_{x\to0} 0^x$, we see that for every $\varepsilon>0$, there is some $\delta>0$ such that for all $x\in(0,\delta)$, $0^x=0$, and therefore this is the limit over the constant function $0$. So the limit is $0$, witnessing the discontinuity under the definition $0^0=1$.
(It should, perhaps, be remarked that $\lim_{x\to 0}x^x=1$, which a good argument in favor of this definition, but generally this exercise shows that $\lim_{(x,y)\to(0,0)}x^y$ is undefined, which supports the idea of leaving this as an undefined term.)

As a footnote, we only consider right-sided limits. Exponentiation with negative numbers involved becomes a real hassle in the domain of the real numbers. I'll leave it to you to figure out why defining this using continuity and rational sequences will not work.
A: The function $f(x)=0^x$ is constant and zero where defined (and it is defined only for $x>0$). Any limit of the function is $0$.
