Question

高数极限习题求lim(x->1)(1-X)tan(πx/2)的极限

Answer 1

令1-x=u,原式化为：

lim{u->0}utan[π(1-u)/2]

=lim{u->0}ucot(πu/2)

=lim{u->0}ucos(πu/2)/sin(πu/2)

=lim{u->0}cos(πu/2)*(2/π)lim{u->0}(πu/2)/sin(πu/2)

=1*(2/π)

=2/π.