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

$0 \le {\sin}^{2} x \le 1 \implies \frac{0}{x} ^ 2 \le {\sin}^{2} \frac{x}{x} ^ 2 \le \frac{1}{x} ^ 2$
Now using the squeeze thereom and taking limits $x \to \infty$ 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