Question

Solve:
$\underset{x\to 27}{lim}\frac{(x^{1/3}+3)(x^{1/3}-3)}{x-27}$

Answer 1

We have,

$\underset{x\to 27}{lim}\frac{\left(x^{1/3}+3\right)\left(x^{1/3}-3\right)}{x-27}$

$=\underset{x\to 27}{lim}\frac{\left(x^{1/3}+3\right)\left(x^{1/3}-3\right)}{\left(x^{1/3}-3\right)\left(x^{2/3}+9+3x^{1/3}\right)}$ [cancle both $\left(x^{1/3}-3\right)$]

$=\frac{3+3}{9+9+3\times 3}$

$=\frac{6}{27}$

$=\frac{2}{9}$

Hence the required answer is $\frac{2}{9}$.