A) \[\sin 2\]
B) \[2\sin 2\]
C) \[2\cos 2\]
D) 2
Correct Answer: C
Solution :
Apply formula of \[\sin C-\sin D\], i.e., \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin (2+x)-\sin (2-x)}{x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{2\cos 2.\sin x}{x}\] \[L\,{g}'(0)=0\] You may also apply L-Hospital rule.
You need to login to perform this action.
You will be redirected in
3 sec
