Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}-\sin\text{x}}{\sin3\text{x}-3\sin\text{x}}$
$\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}-\sin\text{x}}{\sin3\text{x}-3\sin\text{x}}$
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Answer
$\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}-\sin\text{x}}{\sin3\text{x}-3\sin\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\frac{\sin\text{x}}{\cos\text{x}}-\sin\text{x}}{3\sin-4\sin^3\text{x}-3\sin\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}\big(\frac{1}{\cos\text{x}}-1\big)}{-4\sin^3\text{x}}$
$=\frac{-1}{4}\lim\limits_{\text{x}\rightarrow0}\frac{\frac{1}{\cos\text{x}}-1}{\sin^2\text{x}}$
$=\frac{-1}{4}\lim\limits_{\text{x}\rightarrow0}\frac{\frac{1-\cos\text{x}}{\cos\text{x}}}{1-\cos^2\text{x}}$
$=\frac{-1}{4}\lim\limits_{\text{x}\rightarrow0}\frac{1\cos\text{x}}{(\cos\text{x})(1-\cos\text{x})(1+\cos\text{x})}$
$=\frac{-1}{4}\lim\limits_{\text{x}\rightarrow0}\frac{1}{(\cos\text{x})(1+\cos\text{x})}$
$=\frac{-1}{4}\times\frac{1}{1(1+1)}$
$=\frac{-1}{4}\times\frac12$
$=-\frac{1}{8}$
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