Question

求函数 f(x)=actan1/(x-1/x) 的间断点,并判断其类型

Answer 1

 f(x)=actan[1/(x-1/x)]

lim(x->0+) actan[1/(x-1/x)]  = -π/2

lim(x->0-) actan[1/(x-1/x)]  = π/2

x=0, 跳跃间断点

lim(x->1+) actan[1/(x-1/x)]  = π/2

lim(x->1-) actan[1/(x-1/x)]  = -π/2

x=1, 跳跃间断点

Answer 2

f(x) = actan1/(x-1/x), 间断点 x = 0, x = 1.
lim<x→0->f(x) = lim<x→0->actan1/(x-1/x) = 0
lim<x→0+>f(x) = lim<x→0+>actan1/(x-1/x) = 0
x = 0 是可去间断点。
lim<x→1->f(x) = lim<x→1->actan1/(x-1/x)
= lim<x→1->actanx/(x^2-1) = -π/2,
lim<x→1+>f(x) = lim<x→1+>actan1/(x-1/x)
= lim<x→1+>actanx/(x^2-1) = π/2,
x = 1 是跳跃间断点。