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. none of these

Answer 1

The correct option is A $\frac{1}{9}$

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

$=\underset{x\to 1}{lim}\frac{(\sqrt[3]{3x}-1{)}^2}{(\sqrt[3]{3x}-1{)}^2(\sqrt[3]{3x^2}+\sqrt[3]{3x}+1{)}^2}$ .................. Using $a^3-b^3=(a-b)(a^2+ab+b^2)$

$=\underset{x\to 1}{lim}\frac{1}{(\sqrt[3]{3x^2}+\sqrt[3]{3x}+1{)}^2}=\frac{1}{(1+1+1{)}^2}=\frac{1}{9}$