Calculus of a Single Variable, Chapter 3, 3.9, Section 3.9, Problem 19

lim_(x->0)3x-sin^2x
plug in the value of x
= 3*0-sin^2(0)
=0
lim_(x->oo) 3x-sin^2x
lim_(x->oo) x(3-(sin^2x)/x)
Apply the squeeze theorem to evaluate the limit of sin^2x/x
-1<=sinx<=1
0<=sin^2x<=1
lim_(x->oo)(0/x)<=lim_(x->oo)(sin^2x)/x<=lim_(x->oo)(1/x)
lim_(x->oo)(0/x)=0
lim_(x->oo)(1/x)=0
So, by the squeeze theorem,
lim_(x->oo)(sin^2x)/x=0
Therefore,
lim_(x->oo)3(x-(sin^2x)/x)=oo(3-0)=oo