Using conjugates to solve indetermination on limit with cubic root

Question

I have the following problem:

$$\lim\limits_{x \to -8} \frac{\sqrt{1-x} -3}{2 + \sqrt[3]{x}}$$

which seems at first like a simple limit with square and cubic roots that I can solve using conjugates to eliminate them. However, as we have a ${a + b}$ in the denominator, I can only end up with ${a^3 + b^3}$, which would be

$$\lim\limits_{x \to -8} \frac{(\sqrt{1-x} -3)(4 - 2\sqrt[3]{x} + \sqrt[3]{x^2})}{8 + x}$$

and it is still a division by zero.

I've tried other solutions too, like multiplying everything by ${\sqrt{1-x}+3}$ to eliminate the square root in the numerator, but both parts of the fraction still evaluate to zero.

I know the answer is -2, but I can't even evaluate the limit to get to an answer. Am I missing something dumb?

Answer 1

Hint.

$$\lim\limits_{x \to -8} \frac{\sqrt{1-x} -3}{2 + \sqrt[3]{x}} = \lim_{y\to -2}\frac{\sqrt{1-y^3}-3}{2+y}$$ and now

$$ \frac{\sqrt{1-y^3}-3}{y+2}=\frac{1-y^3-3^2}{(y+2)\left(\sqrt{1-y^3}+3\right)} $$

and the problem now is to determine

$$ -\frac{1}{6}\lim_{y\to -2}\frac{y^3+2^3}{y+2} $$

or making $z = -y$

$$ -\frac{1}{6}\lim_{z\to 2}\frac{2^3-z^3}{2-z} $$

Answer 2

If Cesareo is right and the question should be $$\lim_{x\to -8} \frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}}$$, the solution is to let $y=x^{1/3}$ so the limit becomes $$\lim_{y\to -2} \frac{\sqrt{1-y^3}-3}{2+y}$$ $$=\lim_{y\to -2} \frac{1-y^3-9}{(2+y)(\sqrt{1-y^3}+3)}$$ $$=\lim_{y\to -2} \frac{-(y^3+8)}{(2+y)(\sqrt{1-y^3}+3)}$$ $$=\lim_{y\to -2} \frac{-(y+2)(y^2-2y+4)}{(2+y)(\sqrt{1-y^3}+3)}$$ $$=\lim_{y\to -2} \frac{-(y^2-2y+4)}{(\sqrt{1-y^3}+3)}=-2.$$