AMATH 唔識做！

Let x = (pi/2) - n

2012-11-25 08:47:10 補充：
so, when n->pi/2, x->0, and n = (pi/2) - x, therefore
lim(n->pi/2) (cos n) / [(pi/2) - n]
= lim(x->0) {cos[(pi/2) - x]} / x
= lim(x->0) (sin x) / x
= 1 ... [by definition]

lim(n→π/2)⁡(cos n)/(π/2 - n)

= lim(n→π/2)⁡(-sin n)/(-1)

= sin π/2

= 1