Calculus of a Single Variable, Chapter 8, 8.7, Section 8.7, Problem 31

Given to solve ,
lim_(x->oo) cosx/x
by applying the squeeze theorem we can solve the limits
and it is as follows,
as we know the boundaries of the cos(x) as
-1<=cos(x)<=1
now dividing the above expression with x , we get
-1/x<=cos(x)/x<=1/x
now applying the limits of x-> oo for the above expression, we get
lim_(x->oo) (-1/x)<=lim_(x->oo) cos(x)/x<=lim_(x->oo) (1/x)
now upon x-> oo we get
lim_(x->oo) (-1/x) =(-1/oo) =0
and
lim_(x->oo) (1/x) = (1/oo)=0
so,
lim_(x->oo) (-1/x)<=lim_(x->oo) cos(x)/x<=lim_(x->oo) (1/x)
=>0<=lim_(x->oo) cos(x)/x<=0
=>lim_(x->oo) cos(x)/x =0