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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.

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3 Answers 3

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

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

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  • $\begingroup$ Thank you sir. I really liked the trick you did. $\endgroup$ Jul 4 at 6:00
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HINT: Let $a:=\sqrt[3]{x-6}$. Then \begin{align} x+2&=a^3+8\\ &=(a+2)(a^2-2a+4). \end{align}

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