Question

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

• A. $\frac{1}{9}$
• B. $\frac{1}{6}$
• C. $\frac{1}{3}$
• D. None of these

Answer 1

Answer: (A)

$\underset{x\to 1}{\lim }\frac{\sqrt[3]{x^2}-2\sqrt[3]{x}+1}{(x-1{)}^2}$
$=\underset{y\to 1}{\lim }\frac{y^2-2y+1}{{\left(y^3-1\right)}^2}$
$[$ Putting $\sqrt[3]{x}=y;$ as $x\to 1,y\to 1]$
$=\underset{y\to 1}{\lim }\frac{(y-1{)}^2}{(y-1{)}^2{\left(y^2+y+1\right)}^2}$
$=\underset{y\to 1}{\lim }\frac{1}{{\left(y^2+y+1\right)}^2}=\frac{1}{9}$