Question

求解极限问题😭😭😭😭㈷

求解极限问题😭😭😭😭😭😭😭😭😭明天考试了，还有两道题不会





Answer 1

(3) lim(x→π)lncos(2x)/(x-π)^2
=lim(x→π)[2/cos(2x)]/[2(x-π)]——洛必达法则
=lim(x→π)1/[cos(2x)(x-π)]
=∞
(5) lim(x→0+)sin(ax)/√(1-cosx)
=lim(x→0+)sin(ax)/sin(x/2)——sin(ax)~ax，sin(x/2)~x/2
=lim(x→0+)(ax)/(x/2)
=2a

追问

那个√(1-cosx)怎么得了sin(x/2？？？)

追答

√(1-cosx)=√[2sin^2(x/2)]
=√2sin(x/2)
请改一下：
lim(x→0+)sin(ax)/√(1-cosx)
=lim(x→0+)sin(ax)/[√2sin(x/2)]——sin(ax)~ax，sin(x/2)~x/2
=lim(x→0+)(ax)/(√2x/2)
=√2a

Answer 2

会不会 l'Hôpital's Rule?
这两题都是 0/0 的题型