A) 1/120
B) ?1/120
C) 1/20
D) None of these
Correct Answer: A
Solution :
Expand \[\sin x\] and then solve. Aliter : Apply L-Hospital?s rule \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x-x+\frac{{{x}^{3}}}{6}}{{{x}^{5}}}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\cos x-1+\frac{3{{x}^{2}}}{6}}{5{{x}^{4}}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{-\sin x+\frac{6x}{6}}{20{{x}^{3}}}=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{-\cos x+1}{60{{x}^{2}}}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{120\,x}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{\cos x}{120}=\frac{1}{120}.\]
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