Section 2.6
1
Limits at Infinity
We have talked about looking at limx→c f (x) where c is a real number. Now
we want to consider what happens when x approaches infinity.
As a precaution we should look at the definition of this type of limit.
Definition: [Limit at Infinity] limx→∞ f (x) = L if for all ε > 0 there exists
an N in the natural numbers such that when ever x > N , then
|f (x) − L| < ε.
I probably will not have you guys prove anything using this definition
since the tecniques are the same as with the ε − δ definition of a limit. But
be mindful of it. We can adapt our definition to also take into account when
x approaches −∞. Here are some examples of how we can deal with rational
functions and limits as we go to infinity.
First we establish a fact limx→∞ x1r = 0, same with −∞.
Example 1: Evaluate limx→∞ 3x−2
. See if you can follow my reasoning, we
2x+1
went over these in depth in class.
Solution:
3x − 2 1/x
3x − 2
= lim
·
x→∞ 2x + 1 1/x
x→∞ 2x + 1
3 − 2/x
= lim
x→∞ 2 + 1/x
3
=
2
lim
(1)
(2)
(3)
From line 1 we divide through by the the greatest power of the denominator. Line 2 is just simplification. Line 3 we notice that out
function is continuous when we are away from 0 and we use the fact
that limx→∞ x1r = 0.
Example 2: Evaluate limx→∞
1−x2
.
x3 −x+1
Solution:
1
1 − x2
1 − x2
1/x3
=
lim
·
x→∞ x3 − x + 1
x→∞ x3 − x + 1 1/x3
1/x3 − 1/x
= lim
x→∞ 1 − 1/x2 + 1/x3
=0
lim
(1)
(2)
(3)
It is up to you to fill in the details.
Example 3: Evaluate limx→∞
x4 −3x2 +x
.
x3 −x+2
Solution:
x4 − 3x2 + x
x4 − 3x2 + x 1/x3
=
lim
·
x→∞ x3 − x + 2
x→∞ x3 − x + 2
1/x3
x − 3/x + 1/x2
=
1 − 1/x2 + 2/x3
=∞
lim
(1)
(2)
(3)
The above examples actutally show us something interesting about limits
at infinity and horizontal asympototes. The following result shows us the
connection.
2
Horizontal Asymptotes as Limits
Theorem 1 Suppose that P(x) and Q(x) are polynomial functions. Then
P (x)
has :
the rational function Q(x)
i) a horizontal asymptote 0 if deg P (x) < deg Q(x)
ii) no horizontal asymptote if deg P (x)) > deg Q(x)
iii) a horizontal asymptote p/q where p is the leading coefficient of P(x) and
q is the leading coefficient of Q(x), when deg P (x) = deg Q(x).
Proof:
2
The proof is actually strait forward and uses the method we used to solve
the above examples. Just consider general polynomials. But the short of it
is we essentially do what we did in lines 1, 2 and 3 of the examples. Divide
through by the largest power in the denominator, simplify, and use that fact
that limx→∞ x1r = 0. We’re done!
Q.E.D.
3