Consider the MacLaurin series for cos theta:
cos theta = sum_(n=0)^oo (-1)^n theta^(2n)/((2n)!)
and let theta = 3x^2:
cos (3x^2) = sum_(n=0)^oo (-1)^n(3x^2)^(2n)/((2n)!)
cos (3x^2) = sum_(n=0)^oo (-1)^n3^(2n)x^(4n)/((2n)!)
Using the ratio test:
lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo) abs ( ( 3^(2(n+1))x^(4(n+1))/((2(n+1))!))/(3^(2n)x^(4n)/((2n)!)))
lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo) abs (3^(2(n+1))/3^(2n) * ((2n)!) /((2(n+1))!)* x^(4(n+1))/x^(4n))
lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo)3^(2n+2)/3^(2n) * ((2n)!)/((2n+2)!) * abs(x^(4n+4)/x^(4n))
lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo)(9x^4)/((2n+1)(2n+2))
so for any value of x in RR:
lim_(n->oo) abs (a_(n+1)/a_n) = 0
and the series is convergent, which means that the radius of convergence is R=oo.