For limits involving trigonometric functions divided by something, we're best off using the Squeeze Theorem, which tells us that if we have a function f(x), we can definite h(x) and g(x) such that
h(x)<=f(x)<=g(x)
And that if
lim_(x->a)h(x)=lim_(x->a)g(x), then lim_(x->a)f(x)=lim_(x->a)g(x)=lim_(x->a)g(x).
In other words, if the two functions between which f(x) lies have the same limit when approaching the same value, f(x) must also have that limit.
So, recall that
-1<=sinx<=1
This means that
0<=sin^2x<=1 (Sine squared can only be positive/zero, and less than one).
We have sin^2x/x^2, so to accommodate this in our inequality, we can divide everything by x^2:
0/x^2<=sin^2x/x^2<=1/x^2
0<=sin^2x/x^2<=1/x^2
Now, let's take the limits as x->oo of h(x)=0, g(x)=1/x^2:
lim_(x->oo)0=0
lim_(x->oo)1/x^2=1/oo^2=0
So, since the limits of h(x), g(x) are equal, the limit of f(x)=sin^2x/x^2 must also be the limits of h(x), g(x).
Thus,
lim_(x->oo)sin^2x/x^2=0