Question

高数求间断点

Answer 1

f(x) = ln|x|sin(πx/2)/|x^2-1|, 间断点是 x = -1， 0， 1.
x = -1 处左极限 lim(x→-1-)f(x) = lim(x→-1-)ln(-x)sin(πx/2)/(x^2-1)
= - lim(x→-1-)ln(-x)/(x^2-1) (0/0)
= - lim(x→-1-)(1/x)/(2x) = -1/2；
x = -1 处右极限 lim(x→-1+)f(x) = - lim(x→-1+)ln(-x)sin(πx/2)/(x^2-1)
= lim(x→-1-)ln(-x)/(x^2-1) (0/0)
= lim(x→-1-)(1/x)/(2x) = 1/2。
x = -1 是跳跃间断点。
x = 1 处左极限 lim(x→1-)f(x) = - lim(x→1-)lnxsin(πx/2)/(x^2-1)
= - lim(x→1-)lnx/(x^2-1) (0/0)
= - lim(x→1-)(1/x)/(2x) = -1/2；
x = 1 处右极限 lim(x→1+)f(x) = lim(x→1+)lnxsin(πx/2)/(x^2-1)
= lim(x→1+)lnx/(x^2-1) (0/0)
= lim(x→1+)(1/x)/(2x) = 1/2.
x = 1 是跳跃间断点。
x = 0 处左极限 lim(x→0-)f(x) = - lim(x→0-)ln(-x)sin(πx/2)/(x^2-1)
= lim(x→0-)ln(-x)/csc(πx/2) (∞/∞)
= lim(x→0-)(1/x)/[-(π/2)csc(πx/2)cot(πx/2)]
= (-2/π)lim(x→0-)sin(πx/2)tan(πx/2)/x = 0,
x = 0 处右极限 lim(x→0+)f(x) = - lim(x→0+)lnxsin(πx/2)/(x^2-1)
= lim(x→0+)lnx/csc(πx/2) (∞/∞)
= lim(x→0+)(1/x)/[-(π/2)csc(πx/2)cot(πx/2)]
= (-2/π)lim(x→0+)sin(πx/2)tan(πx/2)/x = 0。
x = 0 是可去间断点。