# Rational Functions

I got the following question by email today. I think a lot
of you may have a similar question, so

I'm answering it here.

Brad- I didn't get the chance to ask you in class the other day, but could you
please

explain to me again how you find the limit of a complex rational ? And also how
one

can graph this limit. Thank you.

I'll answer the two questions I think you might be asking.

1. How do you find the limit, as x approaches c, of a rational function?

Say the rational function is p(x)/q(x), where p and q are polynomials.

Solution : Try substituting c for x .

•
If q(c) ≠ 0, you have your limit: p(c)/q(c). Problem solved .

•
On the other hand, if the denominator is zero , it gets a little more
interesting. The

rational function is not defined at x = c (because dividing by zero is never
stylish), but

the limit may or may not exist. This breaks down into two cases:

(a) The numerator, p(c), is also zero. This means we're looking at the
"indeterminate"

0/0 case. It's an algebraic fact (called the Fundamental Theorem of Algebra)
that

whenever c is a root of a polynomial (such as p or q), the quantity (x-c) is a
factor

of that polynomial. So, if q(c) = 0, you can rewrite q(x) as

q(x) = (x - c)(some other polynomial):

Using this idea: if plugging in c for x gives us 0/0, then c is a root of both
polynomials.

Therefore (x - c) is a factor of both polynomials. Factor it out on top and
bottom,

cancel, and start over. (There is a chance that by cancelling this factor, you
have

changed the behavior of the function at x = c-but changing the behavior at a

single point has absolutely no effect on the limit, as described at the top of
p.53.)

See example 5 on p.54, or 1.5 #37-44.

(b) p(c) ≠ 0. In this case, the limit doesn't exist. We want to be more specific
than

this, if at all possible. The limit could be: ∞, -∞, or neither. The examples in
the

book don't fully explore this, but Example 10 on p.57 is a start. Let me give
you

another example:

Find .

To find the limit, we start by trying the substitution x = 2. Our rational
function

becomes

At this point we know the limit doesn't exist, because 6/0 is not a real number .

Nevertheless, we should see if we can learn anything more precise. One of the
best

ways to get information about a rational function is to factor it. Now, the
numerator

is a quadratic . We can factor that without too much trouble. The hard part is
the

denominator, a cubic. However, we know that 2 is a root of it, since setting x =
2

makes it zero. Thus, x-2 is a factor. You may use long division, synthetic
division,

or something else to factor it: x^{3} -x^{2} -8x+12 = (x-2)(x^{2} +x-6). (Personally, I

use "something else"...but that's another blog.) It's not too hard to finish
factoring

the denominator: (x-2)(x^{2} +x-6) = (x-2)(x-2)(x+3) = (x-2)^{2}(x+3). Our

whole rational function is now

Let's recall the big picture: we know that x = 2 is a
vertical asymptote for this

graph, because when we set x = 2, the function took the form (nonzero number
over

zero). Now we want to determine how the graph behaves near the asymptote, on

each side of it. The options for each side are to dive to -∞ or to skyrocket to
∞.

If both sides go up, the limit is ∞, if both sides go down, the limit is -∞, if
one

goes up and the other down, then we simply say the limit does not exist. (Read
the

Study Tip on p.57.)

So how do we determine which infinity the function approaches? It's all a
question

of sign. As the function gets close to the vertical asymptote x = 2, is the
curve

above the x-axis, or below it? The only ways for a function to get across the
x-axis

are (1) to actually cross it (creating an x-intercept in the process), or (2) to
jump

over it (at a discontinuity). The "actual crossings" happen when y = 0, and this

happens only when the numerator is zero. A discontinuity happens only when the

denominator is zero. Thus, if we find the roots of the numerator and denominator,

then we have found the only values of x at which the sign of y could change. In
the

current example, those roots are -4, 1, 2, and -3.

These four numbers break the real line into five pieces: (-∞,-4), (-4,-3), (-3,
1),

(1, 2), and (2,∞). The only intervals here that interest me are the ones very
close to

the x = 2 asymptote, that is, (1, 2) and (2,∞). We know that the function cannot

change sign inside either of these intervals (because we've identified the only
four

points where a sign change could happen, and they aren't inside those
intervals).

Now we imagine that x is somewhere inside the interval (1, 2), and look back at
our

factored rational function. Sign is the only thing I care about now, so while I
could

pick a specific number for x (say 1.5) and get the value

for our function, that's way too much number crunching for me. All I care about
is

whether each quantity is positive or negative for x values in the chosen
interval:

That is, as x approaches 2 from the left, our rational function remains
positive|

which means it approaches positive ∞.

For x approaching 2 from the right, I think of x as belonging to the interval
(2,∞),

which lies on the right side of 2, and check the sign of the result:

As x approaches 2 from the right, we again have a limit of positive ∞. Since the

two one-sided limits agree, we say that

2. How do you find the limit, as x approaches ± ∞, of a
rational function?

In my opinion, this is an easier question to answer than the previous one .

Solution:

First - when I am considering a limit as x→ ± ∞, this is one of the few times when
I do not

try factor polynomials in the usual sense. The "factoring" we do is much easier,
once you get

the hang of it, you never need to think about FOIL or diamond problems or
divisibility or

any of the stuff that makes factoring hard.

We would like to find the limit as x approaches ∞ of some rational function. One
method

(with many variations) is to factor, out of top and bottom, the highest power of
x that appears

in each. Consider the problem:

Find

In this case, we identify 6x^{4} as the dominant term upstairs and 2x^{3} as the
dominant term

downstairs. That is, the highest power of x appearing is x^{4} on top and x^{3} on the
bottom

(at this point I don't worry too much about their coefficients, because I'm trying
to keep life

simple). Now I factor x^{4} out of the numerator and x^{3} out of the denominator, and
simplify. (If

you have trouble going left to right in the first line below, try multiplying out
the numerator

on the right side to confirm that it matches the numerator on the left, and same
for the

denominators.)

Now, as x grows, all of the little fractions shrink ( fixed numerator, big
denominator). That

is:

...and we are done. One shortcut, based on this method, says that if the degree
upstairs (in

this case 4) is greater than the degree downstairs (in this case 3), then the
limit will always

be either ∞ or -∞, and that you can decide which it is by comparing the leading coefficients

(in this case 6 and 2). If they have the same sign, as in this case, the limit
in ∞, if they

have different signs, the limit is -∞. I don't personally rely on this shortcut,
because the

kind of factoring we did is extremely fast when you get used to it-and because,
for limits as

x → -∞, the shortcut needs some careful buttressing. It gets a little too
confusing for my

poor brain. The method I use works to the right and the left, with no exceptions
or special

cases to worry about.

But I should give you a disclaimer about something I did during that method. Did
you notice

how I disposed of the little fractions in (1) above while keeping the front x in
the formula?

This amounts to sending some x's to infinity before I send
others. That is not always a safe

thing to do. For example, the same reasoning would show that:

The correct value of that limit is one, not zero!
Nevertheless, the simplifications in the method

are valid as demonstrated. I'm happy to talk about the theory with anyone who
wants to

know.

Let's see another example:

Find.

According to the method, we factor x^{5} from the top and x^{4} from the bottom,
simplify, discard

negligible fractions, and evaluate:

I have one more disclaimer, aimed at people who suffer from the same confusion I
had during

calculus (and for some time afterward). By reading the above sequence of
equations, you

could rightly conclude that

But it would not be correct to say that the functions y =
(x^{5} -1)/(x^{4} +x^{3} +x^{2} +x+1) and

y = x get arbitrarily close together as x heads for -∞. You may confirm (by long
division,

synthetic division, multiplication, extreme cleverness, or something else) that,
in actual fact,

for all x. So the two functions remain exactly one unit apart. If this kind of
thing interests

you, then by all means: find the paradox in what just happened, and then explain
why there's

no paradox at all.

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