Question

$\displaystyle \lim_{x \to 0} \left(\frac{10}{9}\frac{sin \, 9x}{sin \, 10x}\right) \left(\frac{8}{7}\frac{sin \, 7x}{sin \, 8x}\right) \left(\frac{6}{5}\frac{sin \, 5x}{sin \, 6x}\right) \left(\frac{4}{3}\frac{sin \, 3x}{sin \, 4x}\right) \left(\frac{sin \, x}{sin \, 2x}\right) $ =

• A. $\frac{63}{256}$
• B. $\frac{1}{6}$
• C. $\frac{6}{5}$
• D. $\frac{1}{2}$
• E. $\frac{256}{63}$

Answer 1

Answer: (D)

$\underset{{x \rightarrow 0}}{\lim}\left(\frac{10}{9} \frac{\sin 9 x}{\sin 10 x}\right)\left(\frac{8}{7} \frac{\sin 7 x}{\sin 8 x}\right)$
$\left(\frac{6}{5} \frac{\sin 5 x}{\sin 6 x}\right)\left(\frac{4}{3} \frac{\sin 3 x}{\sin 4 x}\right)\left(\frac{\sin x}{\sin 2 x}\right)$
$=\underset{{x \rightarrow 0}}{\lim} \left(\frac{\sin 9 x}{9 x} \cdot \frac{1}{\frac{\sin 10 x}{10 x}}\right)\left(\frac{\sin 7 x}{7 x} \cdot \frac{1}{\frac{\sin 8 x}{8 x}}\right) $
$ =\left(\frac{\sin 5 x}{5 x} \cdot \frac{1}{\frac{\sin 6 x}{6 x}}\right)\left(\frac{\sin 3 x}{3 x} \cdot \frac{1}{\frac{\sin 4 x}{4 x}}\right) $
$ \left(\frac{\sin x}{x} \frac{1}{\frac{\sin 2 x}{2 x}} \cdot \frac{1}{2}\right) $
$=\frac{1}{2}\left[\because \lim _{f(x) \rightarrow 0} \frac{\sin f(x)}{f(x)}=1\right]$