More about slant asymptotes:
Consider the hyperbola
x
2
–
y
2
= 1:

2
 1
0
1
2
 2
 1
0
1
2
To see that the hyperbola
x
2
–
y
2
= 1 has the line
y
=
x
as a
slant asymptote, take the upper branch
y
= sqrt(
x
2
– 1) and
show that it gets arbitrarily close to the line
y
=
x
as
x
→
+
∞
: “rationalizing the numerator”, we get
lim
x
→
+
∞
x
– sqrt(
x
2
– 1) = lim
x
→
+
∞
1/(
x
+ sqrt(
x
2
– 1)) = 0.
You can likewise show that the line
y
= –
x
is also a slant
asymptote.
Indeed, both lines are twosided asymptotes.
Problem 4.4.49:
xy
=
x
2
+ 4
⇒
y
=
x
+ 4/
x
⇒
y
–
x
= 4/
x
,
which goes to 0 as
x
→
+
∞
and
as
x
→
–
∞
.
So
y
=
x
is a
twosided slant asymptote.
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10
 5
5
10
 10
 5
5
10
Problem 4.4.50:
y
=
e
x
–
x
⇒
y
– (–
x
) =
y
+
x
=
e
x
, which
goes to 0 as
x
→
–
∞
.
So
y
= –
x
is a onesided slant
asymptote.

3
 2
 1
1
2
3
 3
 2
 1
1
2
3
[Digression: How do we show that
e
x
–
x
goes to infinity as
x
→
+
∞
?
I gave a proof in class, but here’s a better way
that doesn’t require integration: Write
e
x
–
x
=
e
x
(1 –
x
/
e
x
).
We know
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 Fall '11
 Staff
 Calculus, Asymptotes, Slant Asymptote

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