i.e. limit for the numerator is x→0+limx=0 and limit for the denominator is x→0+limcot(2x)1=0 Let's take derivatives of the numerator and denominator until we eliminate indeterninateness. x→0+lim(xcot(2x)) = x→0+limdxdcot(2x)1dxdx = x→0+lim2cot2(2x)+21cot2(2x) = x→0+lim2cot2(2x)+21cot2(2x) = 2 It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)