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?
 A: 
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!
A: Use Taylor's polynomial at order $1$:
Set $x=-8+h$ $\;(h\to 0)$. Then


*

*$\sqrt{1-x}-3=\sqrt{9-h}-3=3\biggl(\sqrt{1-\dfrac h9}-1\biggr)=3\biggl(1 -\dfrac h{18}+o(h)-1\biggr)=-\dfrac h{6}+o(h)$,

*$2+\sqrt[3] x=2+\sqrt[3]{-8+h}=2\biggl(1-\sqrt[3]{1-\dfrac h8}\biggr)=2\biggl(1-\Bigl(1-\dfrac h{24}+o(h)\Bigr)\biggr)=\dfrac h{12}+o(h)$


So we obtain
$$\frac{\sqrt{1-x}-3}{2+\sqrt[3] x}=\frac{-\dfrac h{6}+o(h)}{\dfrac h{12}+o(h)}=\frac{-\dfrac 1{6}+o(1)}{\dfrac 1{12}+o(1)}=-2+o(1).$$
A: With substiuation $x+8=t$ we have
$$\lim_{x\to-8}\dfrac{\sqrt{1-x}-3}{2+\sqrt[3]{x}}=\lim_{t\to0}\frac{\sqrt{9-t}-3}{\sqrt[3]{t-8}+2}=\frac{\lim_{t\to0}\frac{\sqrt{9-t}-3}{t}}{\lim_{t\to0}\frac{\sqrt[3]{t-8}+2}{t}}=\frac{(\sqrt{9-t})'\Big|_{t=0}}{(\sqrt[3]{t-8})'\Big|_{t=0}}=\frac{-\frac16}{\frac{1}{12}}=\color{blue}{-2}$$
