Or simply apply L'Hospital
show that -------
Lt (2+x)sin(2+x) - 2sin2 x = 2cos2 + 2sin2
x→ 0
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3 Answers
Anurag Ghosh
·2014-02-14 18:27:53
I think the RHS should be 2cos2 + sin2
Himanshu Giria yes... sorry ....
Upvote·0· Reply ·2014-02-14 20:05:36
Aditya Agarwal
·2014-02-14 18:44:35
the function given is f(x)= x.sin(x).
So we can find the limit of this function by simple differentiation.
(It will be continuous at x=2 since it is a composition of 2 continuous functions)
f'(x) = xcosx + sinx
f'(2) = 2cos2 + sin2