A) 1/4
B) 2
C) 1/2
D) zero
Correct Answer: A
Solution :
\[\underset{x\to \pi }{\mathop{\lim }}\,\frac{\sqrt{2+\cos x-1}}{{{(\pi -x)}^{2}}}\] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{2+\cos (\pi +h)-1}}{(\pi -{{(\pi +h)}^{2}})}\] \[\Rightarrow \] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{2-\cosh }-1}{{{h}^{2}}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{2-\cosh -1}{{{h}^{2}}}\times \frac{1}{\sqrt{2-\cos x}+1}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{1-\cosh }{{{h}^{2}}}\times \frac{1}{\sqrt{2-\cos x}+1}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{1-(1-2{{\sin }^{2}}h/2)}{{{h}^{2}}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{\sin }^{2}}h/2}{4{{(1/2)}^{2}}}\times \frac{1}{\sqrt{2-\cos x}+1}\] \[=\frac{1}{2}\times \frac{1}{2}=1/4\]
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