A) \[\frac{63}{256}\]
B) \[\frac{1}{6}\]
C) \[\frac{6}{5}\]
D) \[\frac{1}{2}\]
Correct Answer: D
Solution :
[d] By using the rule, \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x}=1\] We get, req. limit \[=\frac{10}{9}.\frac{9}{10}.\frac{8}{7}.\frac{7}{8}.\frac{6}{5}.\frac{5}{6}.\frac{4}{3}.\frac{3}{4}.\frac{1}{2}\] \[=\frac{1}{2}\]
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