We have indeterminateness of type
0/0,
i.e. limit for the numerator is
x→0+lime−x1=0and limit for the denominator is
x→0+limcot(x)1=0Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
x→0+lim(e−x1cot(x))=
Let's transform the function under the limit a few
x→0+lim(e−x1cot(x))=
x→0+lim(dxdcot(x)1dxde−x1)=
x→0+lim(x2(cot2(x)+1)e−x1cot2(x))=
x→0+lim(x2(cot2(x)+1)e−x1cot2(x))=
0It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)