What is the limit of (sin^2x/x^2)(sin2xx2) as x approaches infinity?

1 Answer
Mar 21, 2018

lim_(x->oo)sin^2x/x^2=0

Explanation:

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