How to calculate this limit? Is it possible to use L'Hopital's rule?

Question

$$\lim_{x \to -2}\frac{(x-6)^{\frac{1}{3}}+2}{x+2}$$

Update

What is wrong with WolframAlpha then? As it states that this limit does not exist here.

Answer 1

Yes, of course: $$\lim_{x\rightarrow-2}\frac{\sqrt[3]{x-6}+2}{x+2}=\lim_{x\rightarrow-2}\frac{1}{3\sqrt[3]{(x-6)^2}}=\frac{1}{12}.$$ Without: $$\lim_{x\rightarrow-2}\frac{\sqrt[3]{x-6}+2}{x+2}=\lim_{x\rightarrow-2}\frac{x-6+8}{(x+2)(\sqrt[3]{(x-6)^2}-2\sqrt[3]{x-6}+4)}=\frac{1}{12}.$$

Answer 2

Deriv of top is $(1/3)(x-6)^{(-2/3)}$ and of bottom is $1$. So no problem to use Lopital.

Answer 3

I'd say no, you can't use l'Hopital. This limit is the derivative of $(x-6)^{1/3}$ at $x=-2$, by definition of the derivative. If you don't already know this limit, how would you differentiate the numerator?

More concretely, if you try to apply l'Hopital, and use the definition of the derivative to do the calculations, the denominator becomes $1$, and the numerator turns into $$\lim_{x\to-2}\frac{(x-6)^{1/3}+2}{x-2}$$ So technically, you can use l'Hopital (the requirements of the theorem are fulfilled), but it won't help you progress at all.

Instead, since this limit is a derivative, I suggest you use more general derivative rules, like the derivative of $x^{1/3}$ and the chain rule, to find this derivative, and thus your answer.

Answer 4

If you want a "fancy", non-Hospital way, to evaluate the limit: let us only basic, though somewhat cumbersome, factoring.

From $\;a^3+b^3=(a+b)(a^2-ab+b^2)\iff a+b=\cfrac{a^3+b^3}{a^2-ab+b^2}\;$ , when we assume nothing is divided by zero, we get:

$$x+2=\left(x^{1/3}\right)^3-\left(2^{1/3}\right)^3=\left(x^{1/3}+2^{1/3}\right)\left(x^{2/3}-x^{1/3}2^{1/3}+2^{2/3}\right)$$

and likewise:

$$(x-6)^{1/3}+2=(x-6)^{1/3}+8^{1/3}=\frac{(x-6)+8}{(x-6)^{2/3}-8^{1/3}(x-6)^{1/3}+8^{2/3}}=$$

$$=\frac{x+2}{(x-6)^{2/3}-8^{1/3}(x-6)^{1/3}+8^{2/3}}$$

So we finally get that

$$\frac{(x-6)^{1/3}+2}{x+2}=\frac1{(x-6)^{2/3}-8^{1/3}(x-6)^{1/3}+8^{2/3}}\xrightarrow[x\to-2]{}\frac1{(-8)^{2/3}-8^{1/3}\cdot(-8)^{/13}+8^{2/3}}=$$

$$=\frac1{4-2\cdot(-2)+4}=\frac1{12}$$