Question

The value of $\underset{x\to 3}{lim}\frac{\left(x^{\frac{1}{3}}+3\sqrt{3}\right)\left(x^{\frac{1}{3}}-3\sqrt{3}\right)}{x-3}$ is $\frac{1}{9}$.

• A. True
• B. False

Answer 1

The correct option is A True

We have,

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

On apply L-hospital rule, we get

$\underset{x\to 3}{lim}\frac{\left(\frac{1}{3}{x}^{\frac{-2}{3}}+0\right)\left(\frac{1}{3}{x}^{\frac{-2}{3}}-0\right)}{1-0}$

$\underset{x\to 3}{lim}\frac{\left(\frac{1}{3}{x}^{\frac{-2}{3}}\right)\left(\frac{1}{3}{x}^{\frac{-2}{3}}\right)}{1}$

Therefore,

$=\frac{1}{3}(1{)}^{-2/3}\times \frac{1}{3}(1{)}^{-2/3}$

$=\frac{1}{9}$

Hnece, this is the answer.