# 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, $$\lim_{x \to -2} \frac{\sqrt[3]{x-6} + 2}{x^3 + 8}=\lim_{x\to-2}\frac{\sqrt[3]{x-6} + 2}{x+2}\cdot\lim_{x\to-2}\frac1{x^2-2x+4}=\frac1{12}\cdot\frac1{12}=\frac1{144}.$$

• The first part was the one i was looking for before applying the limit arithmetic's, this helps a lot. Thank you for answer. Jul 4 at 6:03
• @Blurred_Vision I'm glad that I can help. You are welcome.
– Feng
Jul 4 at 6:03

Hint:

First calculate $$\lim_{x\to-2}(x^2-2x+4)$$

Then set $$\sqrt[3]{x-6}+2=y\implies x=6+(y-2)^3=-2+12y-6y^2+y^3$$

$$\implies x+2=?$$

$$y\to0$$ as $$x\to-2$$

Can you take it from here?

• Thank you sir. I really liked the trick you did. Jul 4 at 6:00

HINT: Let $$a:=\sqrt[3]{x-6}$$. Then \begin{align} x+2&=a^3+8\\ &=(a+2)(a^2-2a+4). \end{align}