Disclaimer: I am an adult learning Calculus. This is not a student posting his homework assignment. I think this is a great forum!
$$\lim_{x\to8}{\frac{\sqrt[3] x-2}{x-8}}$$
How do I factor the top to cancel the $x-8$ denominator?
$\endgroup$ 12 Answers
$\begingroup$The objective here is to factor the denominator, viewing it as the difference of cubes.
This might be easier to see if you put $t=\sqrt[\large 3]x.\; $ Then $\;x = t^3,\;$ and the corresponding (equivalent) limit becomes $$\lim_{t\to 2}\; \dfrac {t - 2}{t^3 - 8} = \lim_{t\to 2}\;\frac{t-2}{(t-2)(t^2 + 2t + 4)}$$
$\endgroup$ 3 $\begingroup$Actually you need to factor the denominator: $x-8 = (\sqrt[3]{x})^3-2^3 = (\sqrt[3]{x}-2)\left((\sqrt[3]{x})^2+2\sqrt[3]{x}+2^2\right)$.
$\endgroup$ 1
