Question

hopital rule answer with explanation please ​

Answer 1

lim x->0 (2sinx-sin2x)/(x-sinx)

since it is 0/0 form we can diff. num. and den. wrt. x and the answer of limit doest get affected.

lim x->0 (2cox-2cos2x)/(1-cosx) again 0/0, so we diff.

Iim x->0 (-2sinx+4sin2x)/(sinx) again 0/0, so we diff.

lim x->0 (-2cosx+8cos2x)/cosx this expression is not in 0/0 form, so we put

limiting value of x

{-2(1)+8(1)}/(1) = 6