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  4. If L= displaystyle lim x arrow 2 (√[3]60+x2-4/ sin (x-2)), then the value of 3 L is

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

Limits and Derivatives

Solution:

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