We have indeterminateness of type
0/0,
i.e. limit for the numerator is
x→0+lim(−x+tan(x))=0and limit for the denominator is
x→0+lim(x+2sin(x))=0Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
x→0+lim(x+2sin(x)−x+tan(x))=
x→0+lim(dxd(x+2sin(x))dxd(−x+tan(x)))=
x→0+lim(2cos(x)+1tan2(x))=
x→0+lim(dxd(2cos(x)+1)dxdtan2(x))=
x→0+lim(−2sin(x)(2tan2(x)+2)tan(x))=
x→0+lim(−sin(x)tan(x))=
x→0+lim(dxdsin(x)dxd(−tan(x)))=
x→0+lim(cos(x)−tan2(x)−1)=
x→0+lim(cos(x)−tan2(x)−1)=
0It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 3 time(s)