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#