Calculus of a Single Variable, Chapter 8, 8.7, Section 8.7, Problem 32

Given to solve,
lim_(x->oo) sinx/(x-pi)
This can be solved by applying the squeeze theorem and is as follows
as we know the limits or boundaries of sin(x) is
-1<=sin(x)<=1
Dividing the above expression with x-pi we get
-1/(x-pi)<=sin(x)/(x-pi)<=1/(x-pi)
now, let us apply the limits that x-> oo we get
lim_(x->oo)(-1/(x-pi))<=lim_(x->oo) sin(x)/(x-pi)<=lim_(x->oo) 1/(x-pi)
but
lim_(x->oo)(-1/(x-pi)) = -1/(oo -pi) = 0
lim_(x->oo)(1/(x-pi))= 1/(oo -pi) = 0
so,
0<=lim_(x->oo) sin(x)/(x-pi)<=0
so ,
lim_(x->oo) sin(x)/(x-pi) = 0