Evaluate the limit of $(\sqrt{(1-x)}-3)/(2+\sqrt[3]x)$ as $x\rightarrow-8$ Find the limit  $$\lim _{x\rightarrow-8}\frac {\sqrt{(1-x)}-3}{2+\sqrt[3]x}$$
I'm trying to solve this problem but strangely I'm getting different results. I think I have problems while factoring the roots,I used the basic formula $$a^3-b^3=(a+b)(a^2 - ab + b^2)$$ to factor out the bottom and the $(a^2-b^2)$ formula to factor the top, at the end I'm getting the result $-1$, while in online equation solver the result is $0$. I just wanna be sure if I'm doing it correct. Thanks
 A: HINT:
Write
$$\sqrt{1-x}-3=\frac{-(x+8)}{\sqrt{1-x}+3}$$
And
$$2+x^{1/3}=\frac{x+8}{x^{2/3}-2x^{1/3}+4}$$
The limit of interest is $-2$.
A: The basic formula that you are using is wrong, the formula actually is
$$ a^3-b^3 = (a-b)(a^2+ab+b^2). $$
I hope this helps you.
A: $$L=\lim _{x\rightarrow-8}\frac {\sqrt{(1-x)}-3}{2+\sqrt[3]x}$$
let $y=\sqrt[3]x$ then
$$L=\lim _{y\rightarrow-2}\frac {\sqrt{(1-y^3)}-3}{2+y} \tag 1$$
which is in the form of a derivative
$$L = f'(-2) \text{ where } f(x) =  \sqrt{1-y^3}$$
If you are not allowed to use derivatives just multiply numerator and denominator by  $\sqrt{(1-y^3)}+3$
p.s. the answer should be $L=-2$
A: Using the formulas above correct and with the help you provided me I was able to reach the result which in this case is: -2, which it seems to be correct, thanks for all you help.
A: A straightforward application of L' Hopital's rule gives
$\lim _{x\rightarrow-8}\frac {\sqrt{(1-x)}-3}{2+\sqrt[3]x} = \lim _{x\rightarrow-8}\frac{-\frac{1}{2 \sqrt{1-x}}}{\frac{1}{3 x^{2/3}}} = \lim _{x\rightarrow-8}-\frac{3 x^{2/3}}{2\sqrt{1-x}} = -2$
