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
show that -------
Lt (2+x)sin(2+x) - 2sin2 x = 2cos2 + 2sin2
x→ 0
I think the RHS should be 2cos2 + sin2
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