A) 3
B) \[-3\]
C) 6
D) 0
E) 9
Correct Answer: B
Solution :
\[\underset{x\to \frac{\pi }{6}}{\mathop{\lim }}\,\frac{2{{\sin }^{2}}x+\sin x-1}{2{{\sin }^{2}}x-3\sin x+1}\] \[=\underset{x\to \frac{\pi }{6}}{\mathop{\lim }}\,\frac{4\sin x\cos x+\cos x}{4\sin x\cos x-3\cos x}\] (by LHospitals rule) \[=\underset{x\to \frac{\pi }{6}}{\mathop{\lim }}\,\frac{\cos x(4\sin x+1)}{\cos x(4\sin x-3)}=\frac{4\sin \frac{\pi }{6}+1}{4\sin \frac{\pi }{6}-3}\] \[=-3\]
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