Question

高数求间断点个数，谢谢

Answer 1

lim(x-> 1) f(x)

=lim(x-> 1) [ln|x|/(x^2-1) ] .[e^[1/(x-2)]/( 1+ e^[2/(x-2)] ) ]

=[e^(-1)/( 1+ e^(-2) )] . lim(x-> 1) ln|x|/(x^2-1)

=[e^(-1)/( 1+ e^(-2) )] . lim(x-> 1) lnx/[ (x-1)(x+1)]

=[e^(-1)/( 1+ e^(-2) )] . lim(x-> 1) ln( 1+(x-1)) /[ (x-1)(x+1)]

=[e^(-1)/( 1+ e^(-2) )] . lim(x-> 1) (x-1) /[ (x-1)(x+1)]

=[e^(-1)/( 1+ e^(-2) )] . lim(x-> 1) 1 /(x+1)

=(1/2) [e^(-1)/( 1+ e^(-2) )]

x=1 , 可去间断点

lim(x-> -1) f(x)

=lim(x-> -1) [ln|x|/(x^2-1) ] .[e^[1/(x-2)]/( 1+ e^[2/(x-2)] ) ]

=[e^(-1/3)/( 1+ e^(-2/3) )] . lim(x-> -1) ln|x|/(x^2-1)

=[e^(-1/3)/( 1+ e^(-2/3) )] . lim(x-> -1) ln(-x)/[(x-1)(x+1)]

=[e^(-1/3)/( 1+ e^(-2/3) )] . lim(x-> -1) ln[1+(-x-1)]/[(x-1)(x+1)]

=[e^(-1/3)/( 1+ e^(-2/3) )] . lim(x-> -1) (-x-1)/[(x-1)(x+1)]

=[e^(-1/3)/( 1+ e^(-2/3) )] . lim(x-> -1) -1/(x-1)

=(1/2)[e^(-1/3)/( 1+ e^(-2/3) )]

x=-1 , 可去间断点

lim(x-> 2+) f(x)

=lim(x-> 2+) [ln|x|/(x^2-1) ] .[ e^[1/(x-2)]/( 1+ e^[2/(x-2)] ) ]

=ln2 .lim(x-> 2+) e^[1/(x-2)]/( 1+ e^[2/(x-2)] ) 

分子分母同时除e^[2/(x-2)]

=ln2 .lim(x-> 2+) { 1/e^[1/(x-2)] } /( 1/e^[2/(x-2)]+ 1  )  

=ln2 . [ 0/( 0+ 1 ) ]

=0

lim(x-> 2-) f(x)

=lim(x-> 2-) [ln|x|/(x^2-1) ] .[ e^[1/(x-2)]/( 1+ e^[2/(x-2)] ) ]

=ln2 .lim(x-> 2-) e^[1/(x-2)]/( 1+ e^[2/(x-2)] ) 

=ln2 . [e^0/( 1+ e^0 ) ]

=(1/2)ln2

≠lim(x-> 2+) f(x)

x=2, 跳跃间断点

ie

2 个可去间断点 ： x=1 , x=-1

 

追问

问一下那里为什么是ln2呢，谢谢

追答

我错了，该是但影响不了 x=2跳跃间断点 的结果

lim(x-> 2+) f(x)

=lim(x-> 2+) [ln|x|/(x^2-1) ] .[ e^[1/(x-2)]/( 1+ e^[2/(x-2)] ) ]

=(1/3)ln2 .lim(x-> 2+) e^[1/(x-2)]/( 1+ e^[2/(x-2)] ) 

分子分母同时除e^[2/(x-2)]

=(1/3ln2 .lim(x-> 2+) { 1/e^[1/(x-2)] } /( 1/e^[2/(x-2)]+ 1  )  

=(1/3)ln2 . [ 0/( 0+ 1 ) ]

=0

lim(x-> 2-) f(x)

=lim(x-> 2-) [ln|x|/(x^2-1) ] .[ e^[1/(x-2)]/( 1+ e^[2/(x-2)] ) ]

=(1/3)ln2 .lim(x-> 2-) e^[1/(x-2)]/( 1+ e^[2/(x-2)] ) 

=(1/3)ln2 . [e^0/( 1+ e^0 ) ]

=(1/6)ln2

≠lim(x-> 2+) f(x)

x=2, 跳跃间断点

Answer 2

该函数即分段函数：
f(x)=sinπx，、x、<1/2;
f(x)=0，、x、>1/2
f(x)=-1/2，x=-1/2
f(x)=1/2，x=1/2
函数有2个跳跃间断点x=-1/2，x=1/2。