A) 0
B) \[\pi \]
C) \[2\pi \]
D) none of these
Correct Answer: C
Solution :
[c] Given \[f(x)={{x}^{2}}-{{\pi }^{2}}\] \[\underset{x\to -\pi }{\mathop{\lim }}\,\frac{{{x}^{2}}-{{\pi }^{2}}}{\sin (sinx)}=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{(-\pi +h)}^{2}}-{{\pi }^{2}}}{\sin (sin(-\pi +h))}\] \[\underset{h\to 0}{\mathop{\lim }}\,\frac{-2h\pi +{{h}^{2}}}{-\sin (sin+h)}\] \[\underset{h\to 0}{\mathop{\lim }}\,\frac{h-2\pi }{\frac{-\sin (sinh)}{\sinh }\times \frac{\sinh }{h}}=2\pi \]
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