Math 1100: Quantitative Analysis
Limits of Rational Functions
Limits of Polynomial Functions
Let f (x) be a polynomial (that is, a function of the type f (x) = an xn + an−1 xn−1 + · · · +
a1 x + a0 ). Then
lim f (x) = f (c)
x→c
Example 1.
lim x3 − 2x + 1 = (−3)3 − 2(−3) + 1
x→−3
= −27 + 6 + 1
= −20
Example 2.
lim 2x2 − x = 2(1)2 − 1
x→1
=2−1
=1
Limits of Polynomial Functions Suppose that we instead want to find the limit of a
(x)
, where f (x) and g(x) are polynomials. There are a few possibilities in
rational function fg(x)
this case.
• If g(x) 6= 0, then
f (x)
f (c)
=
x→c g(x)
g(c)
This case corresponds to a point where the function is continuous.
lim
f (x)
does not exist. This corresponds to a vertical
x→c g(x)
• If g(x) = 0 and f (x) 6= 0, then lim
asymptote of the graph.
• If g(x) = 0 and f (x) = 0, then we can factor (x − c) out of the numerator and
denominator and proceed with a new function. This case may correspond to a hole in
the graph (in which case the limit exists) or to a vertical asymptote (in which case it
does not).
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The next few examples will use the function h(x) =
shown below.
x2 −5x+6
.
x2 −x−2
The graph of this function is
5
2.5
-7.5
-5
-2.5
0
2.5
5
7.5
10
-2.5
-5
We define f (x) = x2 − 5x + 6 and g(x) = x2 − x − 2.
x2 − 5x + 6
x→4 x2 − x − 2
Example 3. Compute lim
f (x)
f (c)
=
. Thus
x→c g(x)
g(c)
Since g(4) = 42 − 4 − 2 = 10, we have lim
x2 − 5x + 6
42 − 5(4) + 6
=
x→4 x2 − x − 2
42 − 4 − 2
2
=
10
1
=
5
lim
x2 − 5x + 6
x→−1 x2 − x − 2
Since g(−1) = (−1)2 − (−1) − 2 = 0 and f (x) = (−1)2 − 5(−1) + 6 = 12, the limit does not
exist. Graphically, we can see that there is an asymptote at x = −1.
Example 4. Compute lim
2
x2 − 5x + 6
x→2 x2 − x − 2
2
Since g(2) = 2 − 2 − 2 = 0 and f (x) = 22 − 5(2) + 6 = 0, we are in the third indeterminate
case.
Example 5. Compute lim
x2 − 5x + 6
(x − 3)
= lim
2
x→2 x − x − 2
x→2 (x + 1)
x−3
= lim
x→2 x + 1
−1
=
3
lim
Graphically, we can see that there is a hole in the graph at the point (2, − 13 )
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