Question

$\underset{x\to 1}{lim}\frac{\sqrt[3]{3x^2}-2.\sqrt[3]{3x}+1}{(x-1{)}^2}$ is equal to

• A. $\frac{1}{9}$
• B. $\frac{1}{6}$
• C. $\frac{1}{3}$
• D. $1$

Answer 1

Answer: (A)

$\frac{1}{9}$

$\underset{x\to 1}{lim}\frac{\sqrt[3]{3x^2}-2.\sqrt[3]{3x}+1}{(x-1{)}^2}$
$=\underset{x\to 1}{lim}{\left(\frac{x^{\frac{1}{3}}-1}{x-1}\right)}^2$
$=(\frac{1}{3}{)}^2$ ($\because {lim}_{x\to 1}\frac{x^n-a^n}{x-a}=n{a}^{n-1}$)
$=\frac{1}{9}$