Question

求极限 lim(x->0+) (int[0, X] cos (t^2)dt)^2/(int[0, X] sin (t^2

Answer 1

用洛必达法则上下求导

原式=lim(x→0+) 2∫[0,x] cost^2dt*cos(x^2)/sinx^2

=lim(x→0+) 2∫[0,x] cost^2dt/x^2 (用洛必达法则上下求导)

=lim(x→0+) 2cosx^2dt/(2x)

=+∞