微积分求解极限问题

如图

(1)

lim(x->0) arcsinx/x
=lim(x->0) sinx/x
=1
(2)
x->0
1-cosx = 2[sin(x/2)]^2 ~ (1/2)x^2
lim(x->0) ( 1- cosx) /(x^2/2)
=lim(x->0) (x^2/2) /(x^2/2)
=1
(3)
lim(x->0) [f(x) -2]/x - sinx/x^2 ] =1, find lim(x->0) f(x)
solution :
lim(x->0) [f(x) -2]/x - sinx/x^2 ] =1
lim(x->0) [ x[f(x) -2] - sinx ]/x^2 =1
=>
x->0
x[f(x) -2] - sinx ~ x^2
x[f(x) -2] - x ~ x^2
xf(x) - 3x ~ x^2
=>
f(x) ~ 3 +x
lim(x->0) f(x) = 3

1 (1) 令 u = arcsinx， 则 x = sinu，x→0 时 u→0，则
limarcsinx/x = limu/sinu = 1
(2) lim(1-cosx)/(x^2/2) = lim2[sin(x/2)]^2/(x^2/2)
= lim[sin(x/2)/(x/2)]^2 = 1^2 = 1
2. 原式 = lim [xf(x)-2x-sinx]/x^2 (0/0)
= lim [f(x)+xf'(x)-2-cosx]/(2x) = 1
则 lim[f(x) + xf'(x) - 2 - cosx] = 0
limf(x) = 2+cos0 = 3