Question

Solve $\underset{x\to 0}{lim}\frac{\cos 7x-\cos 9x}{\cos 3x-\cos 7x}=$

• A. $\frac{5}{2}$
• B. $\frac{2}{5}$
• C. $\frac{7}{5}$
• D. $\frac{4}{5}$

Answer 1

Answer: (D)

$\frac{4}{5}$

$\underset{x\to 0}{lim}\frac{\cos 7x-\cos 9x}{\cos 3x-\cos 7x}$

$\Rightarrow \underset{x\to 0}{lim}\frac{\left(2\sin \left(\frac{7x+9x}{2}\right)\sin \left(\frac{9x-7x}{1}\right)\right)}{\left(2\sin \left(\frac{3x+7x}{2}\right)\sin \left(\frac{7x-3x}{2}\right)\right)}$

$\Rightarrow \underset{x\to 0}{lim}\frac{\sin 8x.\sin x}{\sin 5x.\sin 2x}$

$\Rightarrow \underset{x\to 0}{lim}\frac{\frac{\sin 8x}{8x}\times \left(8x\right)\frac{\sin x}{x}\times x}{\frac{\sin 5x}{5x}\times \left(5x\right)\frac{\sin 2x}{2x}\left(2x\right)}$

$\underset{x\to 0}{lim}\frac{[\frac{sin8x}{8x}.\frac{sinx}{x}]}{[\frac{sin5x}{5x}.\frac{sin2x}{2x}]}\times \frac{\left(8x\right)\left(x\right)}{\left(5x\right)\left(2x\right)}$

$\Rightarrow \underset{x\to 0}{lim}\frac{8x^2}{10x^2}$

$\Rightarrow \frac{8}{10}=\frac{4}{5}$

$\therefore \underset{x\to 0}{lim}\frac{(\cos 7x-\cos 9x)}{(\cos 3x-\cos 7x)}=\frac{4}{5}$