Why the power rule cannot be applied to solve this limit I'm asked to find:
\begin{align*}
\lim_{x\to 0}\frac{1}{x^{2/3}}
\end{align*}
I tried to apply the power rule for limits, and reduce it to:
\begin{align*}
\left(\lim_{x\to 0}\frac{1}{x}\right)^{2/3}
\end{align*}
And since $\lim_{x\to 0}1/x$ does not exist, I wrongly concluded that the asked limit does not exist.
Where does my reasoning go wrong?
 A: You cannot apply the power rule because the limit $\displaystyle \lim_{x\to 0}\dfrac1x$ does not exist. Remember that in order to apply it, it must exist:
If $\displaystyle \lim_{x\to a}f(x)$ exists, then
$$\lim_{x\to a}\left( f(x)\right)^n =\left( \lim_{x\to a}f(x)\right)^n.$$
But, it's not your case. So, you have to obtain directly the limit:
$$\lim_{x\to 0}\dfrac{1}{x^{2/3}}=+\infty,$$
so the limit doesn't exist.
A: The reason your reasoning was wrong (even though the conclusion was correct) is that the function $x^{-2/3}$ is not continuous at $x=0$. That is, you can only conclude
$$\lim_{x\to a}f(g(x))=f(\lim_{x\to a}g(x))$$
if $f(x)$ is continuous at $\lim_{x\to a}g(x)$. In this case, $g(x)=\frac{1}{x}$ and $a=0$ so the limit does not exist.
Indeed, this question warrants a little clarification (although it is technically correct) as the function is not defined for $x<0$. This isn't actually a problem since normally a limit is taken over elements in the domain of the function in question (in this case $x>0$) but it would probably be clearer if written: What is
$$\lim_{x\to 0^{+}} x^{-2/3}$$
Regardless, a proper way of going about this limit is to note that for $x<1$ we have
$$\frac{1}{x^{2/3}}>\frac{1}{x}$$
and therefore
$$\lim_{x\to 0^{+}}\frac{1}{x^{2/3}}\geq \lim_{x\to 0^{+}}\frac{1}{x}=\infty$$
