Functions Question 43
Question: $ \underset{x\to 0}{\mathop{\lim }},\frac{\cos ax-\cos bx}{x^{2}}= $
[AI CBSE 1988]
Options:
A) $ \frac{a^{2}-b^{2}}{2} $
B) $ \frac{b^{2}-a^{2}}{2} $
C) $ a^{2}-b^{2} $
D) $ b^{2}-a^{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \underset{x\to 0}{\mathop{\lim }},\frac{\cos ax-\cos bx}{x^{2}} $ $ =\underset{x\to 0}{\mathop{\lim }}\frac{2,\sin ,( \frac{a+b}{2} )x,.,\sin ,( \frac{b-a}{2} ),x}{( \frac{a+b}{2} )x,.\frac{2}{a+b}.\frac{2}{b-a}.( \frac{b-a}{2} )x}=\frac{b^{2}-a^{2}}{2} $ Aliter : Apply L-Hospital?s rule, $ \underset{x\to 0}{\mathop{\lim }}\frac{\cos ax-\cos bx}{x^{2}}=\underset{x\to 0}{\mathop{\lim }},\frac{-,a\sin ax+b\sin bx}{2x} $ $ =\underset{x\to 0}{\mathop{\lim }}\frac{-,a^{2}\cos ax+b^{2}\cos bx}{2}=\frac{b^{2}-a^{2}}{2}. $
BETA
