Evaluating $\lim\limits_{x \to 8} \frac{x^{2/3}-4}{x^{1/3}-2}$ without L'Hospital

Question

The problem $$\lim\limits_{x \to 8} \frac{x^{2/3}-4}{x^{1/3}-2}$$ is on my problem set due tomorrow.

In class, we only addressed limits with square roots, and we would just multiply by a conjugate to solve the limit. However, here multiplying by $$\left(\frac{x^{1/3}+2}{x^{1/3}+2}\right)^2$$ only produces another fraction where the denominator is zero when x=8. We have not yet learned L'Hospital's or derivatives in general so that is not an option. Is there perhaps some way to use limit squeeze theorem on this problem, or am I forgetting about some really easy limit law that solves this problem

Answer 1

You can sub $u=x^{1/3}$ then you have $$ \frac{u^2-4}{u-2} $$

Answer 2

Hint:

$$\lim\limits_{x \to 8} \frac{x^{2/3}-4}{x^{1/3}-2}=\lim\limits_{x\to8}(x^{1/3}+2)$$

Answer 3

Hint:

Numerator:

$x^{2/3}-4=(x^{1/3})^2-2^2=$

$(x^{1/3}-2)(x^{1/3}+2)$;

Denominator:

$x^{1/3}-2$.

Now consider $\lim x^{1/3} \rightarrow 2$.

Used: $a^2-b^2=(a-b)(a+b)$.