Clarification on $\frac{0^n}{0}$ when $n>0$ If this is a duplicate, I will gladly delete this if there is a duplicate but I've had difficulty finding one.  I had, until recently, believed that we don't define $\frac 0 0$ as the limits coming from different directions vary widely and so no value works for all cases. I thought that $0^0$ wasn't defined as $0^1*0^{-1}=0^0=\frac 0 0$.  I have been corrected on this but am now confused about such numbers as $\frac {0^2} 0$.
I had believed that as $\frac{0^2}{0} = 0^2*0^{-1} = 0^1 = 0$.  Is $\frac {0^n} 0$ for $n>1$ equal to zero as I had previously believed?  To be clear, I am mostly curious if I can state that $y=\frac{x^2}{x}$ is zero when $x=0$ or whether I have to specify that the limit as $x \rightarrow 0$ is $0$.
 A: For $f(x)=\frac{x^2}{x}$, $f(0)$ is undefined, but $\lim_{x\to 0}f(x)=0$.  The reason is that $f(x)=x$ for all nonzero $x$.
Please don't write things like $\frac{0}{0}$ or $0^{-1}$ unless you also say that this is a form rather than a number.
A: There is quite a big distinction between numbers and limits. In the case of numbers $0^2=0$ and so, no matter how much algebra you do, $0^2 \div 0 = 0\div 0$. You already know that $0\div 0$ is undefined and then so tell is $0^n$ for all $n \ge 1$ because $0^n = 0$ for all $n \ge 1$.
Limits are really very different. Even though $x^2 \to 0$ as $x\to 0$, the limit
$$\lim_{x \to 0} \frac{x^2}{x}=0$$ is well-defined because $x^2$ tends to zero "faster" than $x$ does. It's all to do with how things behave on their way to the limit. It's not the actual destination that matter. Similarly
$$\lim_{x \to 0} \frac{x}{x^2} = \infty$$
because $x^2$ gets smaller faster than $x$ and so you have something much smaller on the bottom and something much bigger on the top.
