Is this limit $\lim\limits_{x \to\, -8}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}} = 0$? Is this limit $\lim\limits_{x \to\, -8}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}} = 0$?
It's in indeterminant form, 0/0 when $x$ approaches $-8$. So I used LHopital's rule and got $$-\frac{3x^{\frac{2}{3}}}{2\sqrt{1-x}}$$ plug in $-8$ it is $-2(-1)^{2/3}$ which is imaginary. I used wolframalpha, the answer is $0$. So, which is correct?


Wolfram Alpha is not correct. 

You're almost correct, except that we have
$$\left(-8\right)^{2/3}=(64)^{1/3}=4$$ or alternatively $$\left(-8\right)^{2/3}=(-2)^{2}=4$$
Hence, applying L'Hospital's Rule 
$$\begin{align}
\lim_{x\to -8}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}}&=\lim_{x\to -8}\left(-\frac{3x^{2/3}}{2\sqrt{1-x}}\right)\\\\
&=-\frac{12}{6}\\\\
&=-2
\end{align}$$


An alternative approach would be to rationalize the numerator and denominator such that

$$\begin{align}
\lim_{x\to -8}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}}&=\lim_{x\to -8}\left(\left(\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}}\right)\left(\frac{\sqrt{1-x}+3}{\sqrt{1-x}+3}\right)\left(\frac{x^{2/3}-2x^{1/3}+4}{x^{2/3}-2x^{1/3}+4}\right)\right)\\\\
&=\lim_{x\to -8}\left(-\frac{x^{2/3}-2x^{1/3}+4}{\sqrt{1-x}+3}\right)\\\\
&=-2
\end{align}$$
as expected!
