展开全部
4. lim<x→0->f(x) = lim<x→0->(1-cosax)/x
= lim<x→0->(ax)^2/(2x) = 0 = f(0),
lim<x→0+>f(x) = lim<x→0+>[√(b+x^2)-1]/x
= lim<x→0+>[√b√(1+x^2/b)-1]/x
lim<x→0+>[√b+x^2/(2√b)-1]/x = 0
b = 1.
左导数 lim<x→0->[f(x)-f(0)]/(x-0)
= lim<x→0->(1-cosax)/x^2 = a^2/2
右导数 lim<x→0+>[f(x)-f(0)]/(x-0)
= lim<x→0+>[√(1+x^2)-1]/x^2
= lim<x→0+>(x^2/2)/x^2 = 1/2,
a^2/2 = 1/2, a = ±1.
1. (1) f(x) = (x-3)/[x((x-3)(x+2)]
间断点 x = -2, 0, 3,
x = -2, 0 为无穷间断点, x = 3 为可去间断点。
(2) f(x) = (x-1)(x+1)/[((x-1)(x-2)]
间断点 x = 1, 2,
x = 2 为无穷间断点, x = 1 为可去间断点。
= lim<x→0->(ax)^2/(2x) = 0 = f(0),
lim<x→0+>f(x) = lim<x→0+>[√(b+x^2)-1]/x
= lim<x→0+>[√b√(1+x^2/b)-1]/x
lim<x→0+>[√b+x^2/(2√b)-1]/x = 0
b = 1.
左导数 lim<x→0->[f(x)-f(0)]/(x-0)
= lim<x→0->(1-cosax)/x^2 = a^2/2
右导数 lim<x→0+>[f(x)-f(0)]/(x-0)
= lim<x→0+>[√(1+x^2)-1]/x^2
= lim<x→0+>(x^2/2)/x^2 = 1/2,
a^2/2 = 1/2, a = ±1.
1. (1) f(x) = (x-3)/[x((x-3)(x+2)]
间断点 x = -2, 0, 3,
x = -2, 0 为无穷间断点, x = 3 为可去间断点。
(2) f(x) = (x-1)(x+1)/[((x-1)(x-2)]
间断点 x = 1, 2,
x = 2 为无穷间断点, x = 1 为可去间断点。
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询