Differentiation Question 181
Question: $ \underset{x\to 0}{\mathop{\lim }}( \frac{10\sin 9x}{9\sin 10x} )( \frac{8\sin 7x}{7\sin 8x} )( \frac{6\sin 5x}{5\sin 6x} )( \frac{4\sin 3x}{3\sin 4x} ) $
$ ( \frac{\sin x}{\sin 2x} )= $
Options:
A) $ \frac{63}{256} $
B) $ \frac{1}{6} $
C) $ \frac{6}{5} $
D) $ \frac{1}{2} $
Show Answer
Answer:
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} $
BETA
