Question

高数题，如图，在线的，急

如图

Answer 1

掌握一个准则：可导必连续

Answer 2

可导必连续，在分界点 x = 0 处
左极限是 lim<x→0->f(x) = lim<x→0->[1-√(1-x)]/x = lim<x→0->1/[1+√(1-x)] = 1/2
右极限是 lim<x→0+>f(x) = lim<x→0+>a+bx = a = f(0), 故得 a = 1/2.
左导数是 lim<x→0->[f(x)-f(0)]/x = lim<x→0->{[1-√(1-x)]/x-1/2}/x
= (1/2)lim<x→0->[2-2√(1-x)-x]/x^2 (0/0)
= (1/2)lim<x→0->[1/√(1-x)-1]/(2x) = (1/4)lim<x→0->[1-√(1-x)]/[x√(1-x)]
= (1/4)lim<x→0->1/{√(1-x)[1+√(1-x)]} = 1/8
右导数是 lim<x→0+>[f(x)-f(0)]/x = lim<x→0+>bx/x = b, 故得 b = 1/8.

Answer 3

设g(x)=(1-√1-x)/x，x∈R
则lim(x--＞0)g(x)=1/2
g'(0)=1/4所以f(0)=1/2 即a=1/2
f'(0)=g'(0) 即b=1/4

Answer 4

证明0左右极限存在并且相等，应该是用定义

Answer 5