A) \[\frac{1}{16}\]
B) \[\frac{1}{8}\]
C) \[\frac{1}{4}\]
D) \[\frac{\pi }{2}\]
Correct Answer: A
Solution :
| \[\underset{x\to \pi /2}{\mathop{\lim }}\,\frac{\cot x-\cos x}{{{\left( \pi -2x \right)}^{3}}}=\underset{x\to \pi /2}{\mathop{\lim }}\,\frac{\cot x(1-\sin x)}{{{\left( \pi -2x \right)}^{3}}}\] |
| \[=\,\,\underset{x\to \pi /2}{\mathop{\lim }}\,\frac{\tan (\pi /2-x)\,(1-\cos (\pi /2-x)}{{{(\pi -2x)}^{3}}}\] |
| \[=\,\,\underset{x\to \pi /2}{\mathop{\lim }}\,\frac{\tan (\pi /2-x).2{{\sin }^{2}}(\pi /4-(x/2))}{2((\pi /2)-x).16{{((\pi /4)-(x/2))}^{2}}}\] |
| \[=\frac{2}{32}\,\underset{x\to \pi /2}{\mathop{\lim }}\,\frac{\tan \,((\pi /2)-x)}{\pi /2-x}\times \underset{x\to \pi /2}{\mathop{\lim }}\,\,{{\left[ \frac{\sin (\pi /4-(x/2))}{(\pi /4)-(x/2)} \right]}^{2}}\] |
| \[=\frac{2}{32}=\frac{1}{16}\] |
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