What are the asymptote(s) and hole(s), if any, of f(x)=1+1x1x?
Answers
Answer:
f
(
x
)
has a hole (removable discontinuity) at
x
=
0
.
It also has a horizontal asymptote
y
=
1
.
It has no vertical or slant asymptotes.
Explanation:
Given:
f
(
x
)
=
x
sin
(
1
x
)
I will use a few of properties of
sin
(
t
)
, namely:
|
sin
t
|
≤
1
for all real values of
t
.
lim
t
→
0
sin
(
t
)
t
=
1
sin
(
−
t
)
=
−
sin
(
t
)
for all values of
t
.
First note that
f
(
x
)
is an even function:
f
(
−
x
)
=
(
−
x
)
sin
(
1
−
x
)
=
(
−
x
)
(
−
sin
(
1
x
)
)
=
x
sin
(
1
x
)
=
f
(
x
)
We find:
∣
∣
∣
x
sin
(
1
x
)
∣
∣
∣
=
|
x
|
∣
∣
∣
sin
(
1
x
)
∣
∣
∣
≤
|
x
|
So:
0
≤
lim
x
→
0
+
∣
∣
∣
x
sin
(
1
x
)
∣
∣
∣
≤
lim
x
→
0
+
|
x
|
=
0
Since this is
0
, so is
lim
x
→
0
+
x
sin
(
1
x
)
Also, since
f
(
x
)
is even:
lim
x
→
0
−
x
sin
(
1
x
)
=
lim
x
→
0
+
x
sin
(
1
x
)
=
0
Note that
f
(
0
)
is undefined, since it involves division by
0
, but both left and right limits exist and agree at
x
=
0
, so it has a hole (removable discontinuity) there.
We also find:
lim
x
→
∞
x
sin
(
1
x
)
=
lim
t
→
0
+
sin
(
t
)
t
=
1
Similarly:
lim
x
→
−
∞
x
sin
(
1
x
)
=
lim
t
→
0
−
sin
(
t
)
t
=
1
So
f
(
x
)
has a horizontal asymptote
y
=
1
graph{x sin(1/x) [-2.5, 2.5, -1.25, 1.25]}