How to find this limit? $$\lim_{x \to -8} \frac{\sqrt{1 - x} - 3 }{ 2 + \sqrt[3] {x}}$$
Progress
I don't know Taylor series yet. I tried to change variable $x$ to $t^3$ but I couldn't remove $(x + 8)$ factor, which creates $0$ in the denominator.
2026-03-29 04:26:30.1774758390
Find $\lim_{x \to -8} \frac{\sqrt{1 - x} - 3 }{ 2 + \sqrt[3] {x}}$
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Hint
Why don't you multiply the top and bottom of the fraction by $\left(\sqrt{1-x}+3\right)$ and see what happens...
Hint 2
After you do simplify with hint 1, think of factoring $a^3 + b^3$...
Update
You get $$ \frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}} \times \frac{\sqrt{1-x}+3}{\sqrt{1-x}+3} = \frac{1-x - 3^2}{\left(2+\sqrt[3]{x}\right) \left(\sqrt{1-x}+3\right)} = \frac{-(x+8)}{\left(2+\sqrt[3]{x}\right) \left(\sqrt{1-x}+3\right)} $$ and the top factors as the sum of two cubes: $$ 8+x = 2^3+\left(\sqrt[3]{x}\right)^3 = \left(2+\sqrt[3]{x}\right)(\ldots) $$ which should cancel nicely with the denominator. Can you finish it?