limit of function near a given point Find the limit of the given function as $x\to -2$ without using L’Hôpital’s Rule:
$$\displaystyle \lim_{x \to -2} \frac{\sqrt[3]{x-6} + 2}{x^3 + 8}$$
I used the identity: $x^3 + 2^3 = (x+2)(x^2 -2x +4)$ at the denominator. Then i tried to use the same identity at the numerator and didnt succeed. So i tried to expand the numerator, but i got different answer by checking the limits from both of the sides near $x = -2$.
The answer is $\frac{1}{144}$ which is an approximation of the limit at the given point.
Any suggestions and support would be kindly appreciated.
Edit: Just watching it from the side and I can rewrite the 2 at the numerator as $\sqrt[3]2$. Yet, its not helping a lot.

As you have noticed that $x^3 + 2^3 = (x+2)(x^2 -2x +4)$, we first calculate the limit
$$\lim_{x\to-2}\frac{\sqrt[3]{x-6} + 2}{x+2}.$$
Let $t=\sqrt[3]{x-6}$, then $t\to-2$ as $x\to-2$ and $x+2=t^3+8=(t+2)(t^2-2t+4)$, so
$$\lim_{x\to-2}\frac{\sqrt[3]{x-6} + 2}{x+2}=\lim_{t\to-2}\frac{t + 2}{(t+2)(t^2-2t+4)}=\lim_{t\to-2}\frac1{t^2-2t+4}=\frac1{12}.$$
Therefore,
$$\lim_{x \to -2} \frac{\sqrt[3]{x-6} + 2}{x^3 + 8}=\lim_{x\to-2}\frac{\sqrt[3]{x-6} + 2}{x+2}\cdot\lim_{x\to-2}\frac1{x^2-2x+4}=\frac1{12}\cdot\frac1{12}=\frac1{144}.$$
