<s>[INST] $\lim_{x\to 3} \frac{x^3-27}{\sqrt{x-3}}$ 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?
 [/INST] 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?
</s>