Question

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

• A. $\frac{1}{9}$
• B. $\frac{1}{6}$
• C. $\frac{1}{3}$
• D. $\frac{1}{2}$

Answer 1

Answer: (A)

$\frac{1}{9}$

$\underset{x\to 1}{lim}\frac{\sqrt[3]{3x^2}-2\sqrt[3]{3x}+1}{{(x-1)}^2}=\underset{x\to 0}{lim}\frac{{(\sqrt[3]{3x}-1)}^2}{{(\sqrt[3]{3x}-1)}^2{(1+\sqrt[3]{3x}+\sqrt[3]{3x^2})}^2}$
$=\underset{x\to 0}{lim}\frac{1}{{(1+\sqrt[3]{3x}+\sqrt[3]{3x^2})}^2}=\frac{1}{9}$
Hence, option 'A' is correct.