Limit problem involving cube roots (without use of L'Hôpital's rule or Taylor series)

Question

How to find this limit without the help of L'Hôpital's rule nor expansion to Taylor series?

Limit:

$$\lim_{x\to -8}\frac{ (9+ x)^{1/3}+x+7}{(15+2 x)^{1/3}+1} $$

Answer 1

Let's change variable first: $u=x+8$.

Your limit becomes: $$ \lim_{u\rightarrow 0} \frac{(1+u)^{1/3}-1+u}{1-(1-2u)^{1/3}}. $$

Now use my comment above: $$ (1+u)^{1/3}-1=\frac{u}{(1+u)^{2/3}+(1+u)^{1/3}+1} $$ and $$ 1-(1-2u)^{1/3}=\frac{2u}{1+(1-2u)^{1/3}+(1-2u)^{2/3}} $$ Now your function becomes: $$ \frac{u(1+(1-2u)^{1/3}+(1-2u)^{2/3})}{2u((1+u)^{2/3}+(1+u)^{1/3}+1)} + \frac{u(1+(1-2u)^{1/3}+(1-2u)^{2/3})}{2u}. $$ Simplify the $u$'s and find that the limit is $1/2+3/2=2$.

Answer 2

Hint: As $$\sqrt[3]a+\sqrt[3]b=\frac{a-b}{\sqrt[3]a^2-\sqrt[3]ab+\sqrt[3]b^2}$$ we have $$\lim_{x\to -8}\frac{\sqrt[3]{9+x}+x+7}{\sqrt[3]{15+2x}+1}=\lim_{x\to -8}\frac{9+x+(x+7)^3}{\sqrt[3]{9+x}^2-\sqrt[3]{9+x}(x+7)+(x+7)^2}\frac{\sqrt[3]{15+2x}^2-\sqrt[3]{15+2x}+1}{15+2x+1^3} = \lim_{x\to -8}\frac{9+x+(x+7)^3}{3}\frac{3}{16+2x}=\lim_{x\to -8}\frac{9+x+(x+7)^3}{16+2x} $$ Things should be straighforward from here on.

EDIT: The question was changed to $x\to -\infty$. Again everything reduces to computing $$\lim_{x\to -\infty}\frac{9+x+(x+7)^3}{16+2x}=\lim_{x\to -\infty}x^2\frac{\frac9{x^3}+\frac1{x^2}+(1+\frac7{x})^3}{\frac{16}x+2} $$ which is also simple.