Question

If $L=\underset{x\to 2}{\lim }\frac{\sqrt[3]{60+x^2}-4}{\sin (x-2)}$, then the value of $3L$ is ___

Answer 1

Answer: 0.25

We have, $\underset{x\to 2}{\lim }\frac{\sqrt[3]{60+x^2}-\sqrt[3]{64}}{\sin (x-2)}$
$=\underset{x\to 2}{\lim }\frac{\left[{\left(60+x^2\right)}^{\frac{1}{3}}-64^{\frac{1}{3}}\right]\left[{\left(60+x^2\right)}^{\frac{2}{3}}+{\left(60+x^2\right)}^{\frac{1}{3}}64^{\frac{1}{3}}+64^{\frac{2}{3}}\right]}{(x-2)\frac{\sin (x-2)}{(x-2)}\left[{\left(60+x^2\right)}^{\frac{2}{3}}+{\left(60+x^2\right)}^{\frac{1}{3}}64^{\frac{1}{3}}+64^{\frac{2}{3}}\right]}$
$=\underset{x\to 2}{\lim }\frac{60+x^2-64}{(x-2)[16+4\times 4+16]}$
$=\underset{x\to 2}{\lim }\frac{(x-2)(x+2)}{48(x-2)}$
$=\underset{x\to 2}{\lim }\frac{x+2}{48}$
$=\frac{4}{48}=\frac{1}{12}$