A) \[\frac{{{a}^{2}}-{{b}^{2}}}{2}\]
B) \[\frac{{{b}^{2}}-{{a}^{2}}}{2}\]
C) \[{{a}^{2}}-{{b}^{2}}\]
D) \[{{b}^{2}}-{{a}^{2}}\]
Correct Answer: B
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\cos ax-\cos bx}{{{x}^{2}}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{2\,\sin \,\left( \frac{a+b}{2} \right)x\,.\,\sin \,\left( \frac{b-a}{2} \right)\,x}{\left( \frac{a+b}{2} \right)x\,.\frac{2}{a+b}.\frac{2}{b-a}.\left( \frac{b-a}{2} \right)x}=\frac{{{b}^{2}}-{{a}^{2}}}{2}\] Aliter : Apply L-Hospital?s rule, \[\underset{x\to 0}{\mathop{\lim }}\,\,\frac{\cos ax-\cos bx}{{{x}^{2}}}=\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{-\,a\sin ax+b\sin bx}{2x}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{-\,{{a}^{2}}\cos ax+{{b}^{2}}\cos bx}{2}=\frac{{{b}^{2}}-{{a}^{2}}}{2}.\]
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