<s>[INST] 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?
 [/INST] With substiuation $x+8=t$ we have
$$\lim_{x\to-8}\dfrac{\sqrt{1-x}-3}{2+\sqrt[3]{x}}=\lim_{t\to0}\frac{\sqrt{9-t}-3}{\sqrt[3]{t-8}+2}=\frac{\lim_{t\to0}\frac{\sqrt{9-t}-3}{t}}{\lim_{t\to0}\frac{\sqrt[3]{t-8}+2}{t}}=\frac{(\sqrt{9-t})'\Big|_{t=0}}{(\sqrt[3]{t-8})'\Big|_{t=0}}=\frac{-\frac16}{\frac{1}{12}}=\color{blue}{-2}$$
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