Limits at infinity
If you have a limit of a rational function (i.e. a quotient of polynomials), then there are
three cases.
(1) Degree of numerator < degree denominator: In this case, the denominator dominates
the quotient, so you do the trick we did in discussion, namely multiply the top and
bottom by 1 over the the highest degree term. As an example, if we have
x2 + 3x + 1
,
lim
x→∞
x3 − 2
then the degree of the bottom is 3 and the top has degree 2. In this case, we multiply
the top and bottom by 1/x3 to get
(x2 + 3x + 1) x13
= lim
x→∞
x→∞
(x3 − 2) x13
lim
1
x
+ x32 +
1 − x23
1
x3
.
At this point, the usual limit laws work, and since the top goes to 0 and the bottom
goes to 1, the limit is 0/1 = 0.
(2) Degree of numerator = degree of denominator: In this case, you do the same
thing as above. For example, if the problem is
−3x2 + x + 1
lim
,
x→∞
x2 − 2
then we can multiply the top and bottom by 1/x2 , and we find
−3x2 + x + 1
=
lim
x→∞
x2 − 2
(−3x2 + x + 1) x12
lim
x→∞
(x2 − 2) x12
−3 + x1 +
= lim
x→∞
1 − x22
1
x2
,
and now we use the usual limit laws to get −3/1 = −3.
(3) Degree of numerator > degree of the denominator: In this case, the limit will
be ∞ or −∞. While the method I showed you in discussion works, also be comfortable
and be able to apply the following logic, as this is probably what you should do on the
test. Instead of multiplying by the reciprocal of the highest degree term of both the
top and bottom, just consider the highest degree term on the bottom. For example,
in the limit
x3 − 1
lim
,
x→∞ x + 1
the degree of the top is greater than the degree of the bottom, and the bottom has
degree 1. So we multiply the top and bottom by 1/x (not 1/x3 ) to get
(x3 − 1) x1
x2 − x1
lim
= lim
.
x→∞ (x + 1) 1
x→∞ 1 + 1
x
x
Now in this case, the bottom is going to 1 and the top is going to ∞ (because of the
lingering x2 term), so use what the professor called “extended limit laws” to say that
this goes to ∞. To be safe, on the test you may want to do this.
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