Question

高数题如图？

Answer 1

解答如下：

Answer 2

ans: D

f(x) =ln|x|/ { (x^2-x).[1+2^(1/(x-2))]}

lim(x->1) f(x)

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

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

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

=1/(1+1/2)

=2/3

x=1, 可去间断点

lim(x->0+) f(x)

=lim(x->0+) ln|x|/ { (x^2-x).[1+2^(1/(x-2))]}

=(-2/3 ) .lim(x->0+) ln|x|/ x 

->+无穷

x=0,  无穷间断点

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

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

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

=0

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

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

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

=(1/2)ln2. [ 1/ (1+0)] 

=(1/2)ln2

x=2, 跳跃间断点