Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\text{x}-\sin2\text{x}}{\text{x}^3}$

Answer

$\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\text{x}-\sin2\text{x}}{\text{x}^3}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\text{x}-2\sin\text{x}\cos\text{x}}{\text{x}^3}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\text{x}(1-\cos\text{x})}{\text{x}^3}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\text{x}(1-\cos\text{x})}{\text{x}^3}\times\frac{1+\cos\text{x}}{1+\cos\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\big(1-\cos^2\text{x}\big)}{\text{x}^3(1+\cos\text{x})}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\big(\sin^2\text{x}\big)}{\text{x}^3(1+\cos\text{x})}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin^3\text{x}}{\text{x}^3(1+\cos\text{x})}$
$=2\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\sin\text{x}}{\text{x}}\Big)^3\times\lim\limits_{\text{x}\rightarrow0}\frac{1}{(1+\cos\text{x})}$
$=2\times1\times\frac{1}{(1+1)}$
$=1$

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