曲线y=1?e1xx+e1x+ln(1+ex)的渐近线条数为(  )A.0B.1C.2D.


曲线y=1?e1xx+e1x+ln(1+ex)的渐近线条数为(  )A.0B.1C.2D.3
函数y=y(x)=
1?e
1
x
x+e
1
x
+ln(1+ex)的所有间断点为:厅大x=0.
因为
lim
x→0+
1?e
1
x
x+e
1
x
=-1,
lim
x→0?
1?e
1
x
x+e
1
x
=-∞,
lim
x→0
ln(1+ex)
=ln2,
所以
lim
x→0+
y(x)
=ln2-1,
lim
x→0?
y(x)
=-∞,
故x=0为y(x)的无穷间断点,
从而x=0为y(x)的一个垂直渐近线.
因为
lim
x→+∞
ln(1+ex)
x
=
lim
x→+∞
(ln(1+ex))′
x′

=
lim
x→+∞
ex
1+ex

=1,
所以,
lim
x→+∞
y(x)
x
=
lim
x→+∞
(
1?e
1
x
x(x+e
1
x
)
+
ln(1+ex)
x
)
 
=0+1
=1;
又因为
 
lim
x→+∞
(ln(1+ex)?x)

=
lim
x→+∞
x(ln(1+e?x)?1)

=
lim
x→+∞
ln(1+e?x)?1
1
x

=
lim
x→+∞
e?x
1
x
(∵ln(1+x)~x(当x→0时))
=
lim
x→+∞
x
ex

=
lim
x→+∞
1
ex

=0,
所以,
lim
x→+∞
(y(x)?x)
=
lim
x→+∞
(
1?e
1
x
x+e
1
x
+ln(1+ex)?x)

=0+0
=0,
从而当x→+∞时,y(x)有渐近线局改y=x.
因为
lim
x→?∞
1?e
1
x
x+e
1
x
=0,
lim
x→?∞
ln(1+ex)=ln1=0,
所以,
lim
x→?∞
y(x)
=
lim
x→?∞
1?e
1
x
x+e
1
x
+
lim
x→?∞
ln(桐伏判1+ex) 
=0+0
=0,
故当x→-∞时,y(x)有水平渐近线y=0.
综上,y(x)有三条渐近线.
故选:D.