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Special Integrals - 3

lim x → 5 x- ...

Question

$\lim_{x \to 0} \frac{\sec 5 x - \sec 3 x}{\sec 3 x - \sec x}$

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Solution

$\lim_{x \to 0} [\frac{\sec 5 x - \sec 3 x}{\sec 3 x - \sec x}]$

$= \lim_{x \to 0} [\frac{\frac{1}{\cos 5 x} \frac{1}{\cos 3 x}}{\frac{1}{\cos 3 x} - \frac{1}{\cos x}}] = \lim_{x \to 0} [\frac{\cos 3 x - \cos 5 x}{\cos 5 x \cos 3 x \{\frac{\cos x - \cos 3 x}{\cos x \cos 3 x}\}}] = \lim_{x \to 0} [\frac{(\cos 3 x - \cos 5 x) \cos x}{\cos 5 x \{\cos x - \cos 3 x\}}] = \lim_{x \to 0} [\frac{- 2 \sin (\frac{3 x + 5 x}{2}) \sin (\frac{3 x - 5 x}{2}) \times \cos x}{\cos (5 x) [- 2 \sin (\frac{x + 3 x}{2}) \sin (\frac{x - 3 x}{2})]}] [∵ cosC - cosD = - 2 \sin (\frac{C + D}{2}) \sin (\frac{C - D}{2})] = \lim_{x \to 0} [\frac{\sin (4 x) \times \sin (- x) \times \cos x}{\cos (5 x) \times \sin (2 x) \times \sin (- x)}] = \lim_{x \to 0} [\frac{\sin (4 x)}{4 x} \times \frac{4 x}{\frac{\sin (2 x)}{2 x} \times 2 x} \times \frac{\cos x}{\cos 5 x}] = \frac{4}{2} \frac{\cos 0}{\cos 0} = 2$

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