A) \[-\frac{1}{4}\]
B) \[\frac{1}{2}\]
C) 1
D) 2
Correct Answer: D
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{(1-\cos 2x)(3+\cos x)}{x\tan 4x}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{(2si{{n}^{2}}x)(3+\cos x)}{x\left( \frac{\tan 4x}{4x} \right)\times 4x}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{\sin }^{2}}x(3+\cos x)}{4{{x}^{2}}}=\frac{2}{4}(3+1)=2\]
You need to login to perform this action.
You will be redirected in
3 sec
