Evaluate the following limit: _ x 0 1- 2x 2x- 8x - Vidyadip

Question

Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{1-\cos2\text{x}}{\cos2\text{x}-\cos8\text{x}}$

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Answer

$\lim\limits_{\text{x}\rightarrow0}\frac{1-\cos2\text{x}}{\cos2\text{x}-\cos8\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin^2\text{x}}{-2\sin\big(\frac{2\text{x}+8\text{x}}{2}\big)\sin\big(\frac{2\text{x}-8\text{x}}{2}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\sin^2\text{x}}{\sin5\text{x}\times\sin(-3\text{x})}$
$=\frac{\lim\limits_{\text{x}\rightarrow0}\sin^2\text{x}}{-\big(\lim\limits_{\text{x}\rightarrow0}\sin5\text{x}\big)\big(-\lim\limits_{\text{x}\rightarrow0}\sin3\text{x}\big)}$
$=\frac{\big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}\big)^2\times\text{x}^2}{\big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin5\text{x}}{5\text{x}}\times5\text{x}\big(\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}}{3\text{x}}\big)\times3\text{x}}$
$=\frac{1\times\text{x}^2}{1\times5\text{x}\times1\times3\text{x}}$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big]$
$=\frac{\text{x}^2}{15\text{x}^2}$
$=\frac{1}{15}$

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