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Mathematics Question on Derivatives

lim⁡x→0(109sin 9xsin 10x)(87sin 7xsin 8x)(65sin 5xsin 6x)(43sin 3xsin 4x)(sin xsin 2x)\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

63256\frac{63}{256}

B

16\frac{1}{6}

C

65\frac{6}{5}

D

12\frac{1}{2}

Answer

12\frac{1}{2}

Explanation

Solution

lim⁡x→0(109sin⁡9xsin⁡10x)(87sin⁡7xsin⁡8x)\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)
(65sin⁡5xsin⁡6x)(43sin⁡3xsin⁡4x)(sin⁡xsin⁡2x)\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)
=lim⁡x→0(sin⁡9x9x⋅1sin⁡10x10x)(sin⁡7x7x⋅1sin⁡8x8x)=\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)
=(sin⁡5x5x⋅1sin⁡6x6x)(sin⁡3x3x⋅1sin⁡4x4x)=\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)
(sin⁡xx1sin⁡2x2x⋅12)\left(\frac{\sin x}{x} \frac{1}{\frac{\sin 2 x}{2 x}} \cdot \frac{1}{2}\right)
=12[∵lim⁡f(x)→0sin⁡f(x)f(x)=1]=\frac{1}{2}\left[\because \lim _{f(x) \rightarrow 0} \frac{\sin f(x)}{f(x)}=1\right]