Functions Question 232
Question: The value of $ \underset{x\to 0}{\mathop{\lim }}\frac{(1-\cos 2x)\sin 5x}{x^{2}\sin 3x} $ is
[MP PET 2000; UPSEAT 2000; Karnataka CET 2002]
Options:
A) 10/3
B) 3/10
C) 6/5
D) 5/6
Show Answer
Answer:
Correct Answer: A
Solution:
$ \underset{x\to 0}{\mathop{lim}},\frac{(1-\cos 2x),\sin 5x}{x^{2}\sin 3x} $ $ =\underset{x\to 0}{\mathop{lim}},\frac{2{{\sin }^{2}}x,\sin 5x}{x^{2}\sin 3x} $ $ =\underset{x\to 0}{\mathop{lim}}( \frac{2{{\sin }^{2}}x}{x^{2}} )\frac{( \frac{\sin 5x}{x} )}{( \frac{\sin 3x}{x} )} $ $ =\underset{x\to 0}{\mathop{lim}},2,{{( \frac{\sin x}{x} )}^{2}}\times \frac{5\underset{x\to 0}{\mathop{lim}},( \frac{\sin 5x}{5x} )}{3\underset{x\to 0}{\mathop{lim}},( \frac{\sin 3x}{3x} )} $ $ =\frac{2\times 5}{3}=\frac{10}{3} $ .
BETA
