Question

Find the following limit:
$\underset{x\to 9}{lim}\frac{\sqrt[3]{3x-1}-2}{x-9}.$

Answer 1

$\underset{x\to 9}{lim}\frac{\sqrt[3]{3x-1}-2}{x-9}$

$\Rightarrow$ Put $x-1=t^3$

$\Rightarrow x=t^3+1$ here $x\to 9$ so $t\to 2$

$\underset{t\to 2}{lim}\frac{\sqrt[3]{3t^3}-2}{t^3+1-9}=\underset{t\to 2}{lim}\frac{t-2}{t^3-8}$

$=\underset{t\to 2}{lim}\frac{t-2}{t^3-2^3}$

$=\underset{t\to 2}{lim}\frac{(t-2)}{(t-2)(t^2+2t+4)}$

$=\frac{1}{4+4+4}=\frac{1}{12}$.