$\lim_{x\to 3} \frac{x^3-27}{\sqrt{x-3}}$ I was trying to calculate the following limit:
$$\lim_{x\to 3} \frac{x^3-27}{\sqrt{x-3}}$$
and, feeding it into WolframAlpha, I found the limit is $0$.
But;  the function isn't defined  from the left of $3$ 
My knowledge is that the function must be defined  in an  open interval  around $\{ x_0\}$ ; "neighbourhood"
So, is the answer of WolframAlpha correct,  or does the limit not exist?
 A: If a limit point is at the boundary of a domain, an open neighborhood may not contain points on all sides. For example, in the relative topology of $[3,\infty)$, $[3,4)$ is an open set, and the limit as $x\to3$ is a one-sided limit. Thus,
$$
\begin{align}
\lim_{x\to3}\frac{x^3-27}{\sqrt{x-3}}
&=\lim_{x\to3^+}\frac{x^3-27}{\sqrt{x-3}}\\
&=\lim_{x\to3^+}\sqrt{x-3}(x^2+3x+9)\\[3pt]
&=0
\end{align}
$$
A: The limit does exist (as a rule of thumb, Wolfram is more often right than humans are).
In the definition of a limit, you only consider the part of the open interval that belongs to the domain (here $\mathbb R^+$). As you can check on Wikipedia, $x\in D$ is required.
If that rule was not enforced, you would never have a limit at an endpoint nor in more complicated situations (think of $x\sqrt{\sin(1/x)}$), which would make it a poorer concept.

As an alternative way to compute this limit, you can shift the argument and write
$$
\lim_{x\to 3} \frac{x^3-27}{\sqrt{x-3}}=\lim_{y\to 0} \frac{(y+3)^3-27}{\sqrt{y}}=\lim_{y\to 0} \frac{y^3+9y^2+27y}{\sqrt{y}}=\lim_{y\to 0} (y^2+9y+27){\sqrt{y}}.
$$

Update:
By default Wolfram works in the complex. As the function is defined for all $z\ne0$, there is no domain limitation and Alpha does not answer the very question (though it does it implicitly by not raising a discussion on the sign).
A: First, note $x$ must be not less than $3$ for a root to be defined, and greater than $3$ for a denominator to be nonzero (hence a function defined). So any 'open interval' you may consider must be a subset of the domain, which is, at most, $(3,\infty)$.
Then:
$$\begin{align}
\frac{x^3-27}{\sqrt{x-3}} & = \frac{(x-3)(x^2+3x+9)}{\sqrt{x-3}} \\
& = \frac{\sqrt{x-3}\,\sqrt{x-3}\,(x^2+3x+9)}{\sqrt{x-3}} \\
& = \frac{\sqrt{x-3}}{\sqrt{x-3}}\cdot\sqrt{x-3}\,(x^2+3x+9) \\
& = \sqrt{x-3}\,(x^2+3x+9)
\end{align}$$
Now the existence of a limit is quite obvious, isn't it?
A: HINT: we have $$x^3-27=(x-3)(x^2+3x+9)$$
