第四题!求过程
1个回答
展开全部
(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
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
更多追问追答
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
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