We have indeterminateness of type
oo/oo,
i.e. limit for the numerator is
x→∞limx3=∞and limit for the denominator is
x→∞limex=∞Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
x→∞lim(e−xx3)=
Let's transform the function under the limit a few
x→∞lim(x3e−x)=
x→∞lim(dxdexdxdx3)=
x→∞lim(3x2e−x)=
x→∞lim(dxdexdxd3x2)=
x→∞lim(6xe−x)=
x→∞lim(dxdexdxd6x)=
x→∞lim(6e−x)=
x→∞lim(6e−x)=
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)