Evaluate: * lim _ x 0 tan x - sin x ^ 3 x . - Vidyadip

Question

Evaluate:$\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{x}\rightarrow0}\frac{\text{tan x - sin x}}{\sin^{3}\text{x}}$.

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Answer

$\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{x}\rightarrow0}\frac{\text{tan x - sin x}}{\sin^{3}\text{x}}=\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{x}\rightarrow0}\frac{\text{tan x}\cdot\text{(1 - cos x)}}{\sin^{3}\text{x}}$
=$\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{x}\rightarrow0}\frac{\text{tan x}\text{(1 - cos x)}}{\text{x}^{3}\frac{\sin^{3}\text{x}}{\text{x}^{3}}}=\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}\cdot\frac{1 - \cos\text{x}}{\text{x}^{2}}\cdot\frac{1}{\frac{\sin^{3}\text{x}}{\text{x}^{3}}}$
$=\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}\cdot\frac{\sin^{2}\text{x}}{\text{x}^{2}}\cdot\frac{1}{1+\text{cos x}}\cdot\frac{1}{\frac{\sin^{3}\text{x}}{\text{x}^{3}}}$
= $1.1\cdot\frac{1}{2}\cdot\frac{1}{1}=\frac{1}{2}$.

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