Question

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

• A. 1/6
• B. 1/9
• C. -1/6
• D. -1/9

Answer 1

Answer: (B)

1/9

Given,

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

apply L-Hospital's rule

$=\underset{x\to 1}{lim}\left(\frac{\frac{2x^{\frac{5}{3}}-2{\left(x^2\right)}^{\frac{2}{3}}}{3x^{\frac{2}{3}}{\left(x^2\right)}^{\frac{2}{3}}}}{2(x-1)}\right)$

$=\underset{x\to 1}{lim}\left(\frac{x^{\frac{5}{3}}-{\left(x^2\right)}^{\frac{2}{3}}}{3x^{\frac{2}{3}}{\left(x^2\right)}^{\frac{2}{3}}(x-1)}\right)$

apply L-Hospital's rule

$=\underset{x\to 1}{lim}\left(\frac{\frac{5x^{\frac{2}{3}}}{3}-\frac{4x}{3\sqrt[3]{3x^2}}}{\frac{2{\left(x^2\right)}^{\frac{2}{3}}(x-1)}{x^{\frac{1}{3}}}+\frac{4x^{\frac{5}{3}}(x-1)}{\sqrt[3]{3x^2}}+3x^{\frac{2}{3}}{\left(x^2\right)}^{\frac{2}{3}}}\right)$

$=\underset{x\to 1}{lim}\left(\frac{\sqrt[3]{3x}(5x^{\frac{2}{3}}\sqrt[3]{3x^2}-4x)}{3(2{\left(x^2\right)}^{\frac{2}{3}}\sqrt[3]{3x^2}(x-1)+4x^{\frac{5}{3}}\sqrt[3]{3x}(x-1)+3x^{\frac{2}{3}}\sqrt[3]{3x}{\left(x^2\right)}^{\frac{2}{3}}\sqrt[3]{3x^2})}\right)$

$=\frac{\sqrt[3]{31}(5\cdot 1^{\frac{2}{3}}\sqrt[3]{31^2}-4\cdot 1)}{3(2{\left(1^2\right)}^{\frac{2}{3}}\sqrt[3]{31^2}(1-1)+4\cdot 1^{\frac{5}{3}}\sqrt[3]{31}(1-1)+3\cdot 1^{\frac{2}{3}}\sqrt[3]{31}{\left(1^2\right)}^{\frac{2}{3}}\sqrt[3]{31^2})}$

upon solving the above equation, we get,

$=\frac{1}{9}$