Question

Lt x=0 (cos7X-cos9X)/(cosX-cos5X)

Answer 1

$\textbf{To find:}$

$\text{The value of}\;\displaystyle\lim_{x{\to}0}\;\dfrac{cos7x-cos5x}{cosx-cos5x}$

$\textbf{Solution:}$

$\text{Consider,}$

$\displaystyle\lim_{x{\to}0}\;\dfrac{cos7x-cos5x}{cosx-cos5x}$

$\text{Using the identity}$

$\boxed{\bf\,cosC-cosD=-2\,sin(\frac{C+D}{2})\,sin(\frac{C-D}{2})}$

$=\displaystyle\lim_{x{\to}0}\;\dfrac{-2\,sin8x\;sin(-x)}{-2\,sin3x\;sin(-2x)}$

$=\displaystyle\lim_{x{\to}0}\;\dfrac{sin8x\;sinx}{sin3x\;sin2x}$

$=\displaystyle\lim_{x{\to}0}\;\dfrac{\frac{sin8x\;sinx}{x^2}}{\frac{sin3x\;sin2x}{x^2}}$

$=\displaystyle\lim_{x{\to}0}\;\dfrac{(\frac{sin8x}{x})(\frac{sinx}{x})}{(\frac{sin3x}{x})(\frac{sin2x}{x})}$

$=\displaystyle\dfrac{\lim_{x{\to}0}\;(\frac{sin8x}{x}){\times}\lim_{x{\to}0}\;(\frac{sinx}{x})}{\lim_{x{\to}0}\;(\frac{sin3x}{x}){\times}\lim_{x{\to}0}\;(\frac{sin2x}{x})}$

$=\displaystyle\dfrac{8\lim_{x{\to}0}\;\frac{sin8x}{8x}{\times}\lim_{x{\to}0}\;\frac{sinx}{x}}{3\lim_{x{\to}0}\;\frac{sin3x}{3x}{\times}2\lim_{x{\to}0}\;\frac{sin2x}{2x}}$

$\text{Using,}\boxed{\bf\,\lim_{x{\to}0}\;\dfrac{sin\theta}{\theta}=1}$

$=\dfrac{8{\times}1}{3{\times}2}$

$=\dfrac{4}{3}$

$\therefore\bf\displaystyle\lim_{x{\to}0}\;\dfrac{cos7x-cos5x}{cosx-cos5x}=\dfrac{4}{3}$

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