Question

请教一道题目，感谢感谢！

Answer 1

f(x)
= ln|x|/(x^2-3x+2)
= ln|x|/[(x-1)(x-2)]
x=0, x=2 无穷间断点
lim(x->1 ) f(x)
=lim(x->1) ln|x|/[(x-1)(x-2)]
=lim(x->1) lnx/[(x-1)(x-2)]
=lim(x->1) ln(1-(1-x))/[(x-1)(x-2)]
=lim(x->1) -(1-x)/[(x-1)(x-2)]
=lim(x->1) 1/(x-2)
=-1
x=1 , 可去间断点

追问

谢谢！请问x=2时如何判断啊？

追答

x=2 无穷间断点
lim(x->2) ln|x|/[(x-1)(x-2)] ->∞

追问

懂了，感谢！

可以再请教您一道题目吗？

麻烦您了

追答

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

=[(n-1)!/n!]. { (n+1)!/[2(n!)] }
=(1/2)[(n-1)!.(n+1)!/(n!)^2]
=(1/2){ (n+1)!/[n.n!] }
=(1/2) [ (n+1)/n]
lim(n->∞)( 1-1/2^2)(1-1/3^2)...(1-1/n^2)
=lim(n->∞) (1/2) [ (n+1)/n]
=1/2

追问

如果您有时间的话，我还想咨询您一些数学问题

实在是麻烦了

我没有思路

追答

lim(x->4) [√(2x+1) -3 ]/[√(x-2) -√2 ]

有理化分母
=lim(x->4) [√(2x+1) -3 ].[√(x-2) +√2 ]/[(x-2) -2 ]

=2√2.lim(x->4) [√(2x+1) -3 ]/(x-4)
有理化分子
=2√2.lim(x->4) [(2x+1) -9 ]/{(x-4). [√(2x+1) +3 ] }
=(√2/3).lim(x->4) 2(x-4)/(x-4)
=2√2/3
(6)
泰勒公式
y->0
(1-y)^(2/3) =1 -(2/3)y -(1/9)y^2 +o(y^2)
(1-y)^(1/3) = 1-(1/3)y -(1/9)y^2 +o(y^2)
-2(1-y)^(1/3) = -2+(2/3)y +(2/9)y^2 +o(y^2)
(1-y)^(2/3)-2(1-y)^(1/3) +1 =(1/9)y^2+o(y^2)
x=1-y

lim(x->1) [ [x^(2/3) -2x^(1/3) +1]/(x-1)^2
=lim(y->0) [ [ (1-y)^(2/3) -2(1-y)^(1/3) +1]/y^2
=lim(y->0) (1/9)y^2/y^2
=1/9

追问

太感谢您了！！

有时间还请您看下这两道题目！！谢谢🙏