Question

How do you use the Squeeze Theorem to find `lim (sin^2x/x^2)` as x approaches infinity?

Answer 1

As

`0<=sin^2x<=1=>0/x^2<=sin^2x/x^2<=1/x^2`

Now using the squeeze thereom and taking limits `x->oo` in the last inequality we have that

#lim_(x->oo)0/x^2<=lim_(x->oo)sin^2x/x^2<=lim_(x->oo) 1/x^2=> lim_(x->oo) sin^2x/x^2=0#