Question

跪求第三题解题过程，非常感谢

Answer 1

(1)
lim(x->a) (sinx -sina)/(x-a) (0/0)
=lim(x->a) cosx
=cosa

追问

能详细点吗

追答

(1)
lim(x->a) (sinx -sina)/(x-a) (0/0)
=lim(x->a) d/dx (sinx -sina)/d/dx (x-a)
=lim(x->a) cosx /1
=lim(x->a) cosx
=cosa

追问

还没学积分呢

追答

consider
sin(A+B)=sinA.cosB +cosA.sinB (1)
sin(A-B)=sinA.cosB -cosA.sinB (2)

(1)-(2)
sin(A+B)-sin(A-B) = 2cosA.sinB

=>
sinx - sina = 2cos [(x+a)/2] . sin [(x-a)/2]

lim(x->a) (sinx -sina)/(x-a) (0/0)
=lim(x->a) 2cos [(x+a)/2] . sin [(x-a)/2] /(x-a)
=lim(x->a) cos [(x+a)/2] . lim (x->a) sin [(x-a)/2] /[(x-a)/2]
=lim(x->a) cos [(x+a)/2]
=cosa

追问

大神，求问11题怎么做