Question

第四题！求过程

Answer 1

(1)
lim(x->π/2) ln(2cos3x)
不存在
(2)
lim(x->0) ln( sinx/x)
=ln1
=0
(3)
lim(x->1) tan(x-1)/(x^2+x-2)
=lim(x->1) (x-1)/[(x-1)(x+2)]
=lim(x->1) 1/(x+2)
=1/3
(4)
lim(x->0) x(sin(1/x^2) - 1/sinx )
=lim(x->0) (xsin(1/x^2) - x/sinx )
=0-1
=1
(5)
lim(x->0) ln(1+x)/x (0/0)
=lim(x->0)[ 1/(1+x)]
=1

追问

第一题不是应该等于1吗

追答

lim(x->π/2) ln(2cos3x)
=ln(2cos(3π/2) )
=ln0
不存在

(4)
lim(x->0) x(sin(1/x^2) - 1/sinx )
=lim(x->0) (xsin(1/x^2) - x/sinx )
=0-1
=-1

追问

是1

追答

|sin(1/x^2) |≤1
=>lim(x->0) (xsin(1/x^2) =0

lim(x->0) x/sinx =1

(4)
lim(x->0) x(sin(1/x^2) - 1/sinx )
=lim(x->0) (xsin(1/x^2) - x/sinx )
=0-1
=-1

追问

第二题

追答

1.
(1)

lim(x->0-) f(x)
=lim(x->0-) (x^2-2)
=-2

lim(x->0+) f(x)
=lim(x->0+) ln(1+bx)/x (0/0)
=lim(x->0+) b/(1+bx)
=b
=lim(x->0-) f(x)
=>
b=-2

(2)

f(0) = f(0+)=f(0-)
a-1= -2 and b=-2
a=-1 and b=-2

2.
f(x)= (x^2-1)/(x^2-3x+2)
=(x-1)(x+1)/[ (x-2)(x-1) ]
断点 x=1 or x=2

x=1 : 可去间断点
x=2 : 无穷间断点