Does that `delta`-`epsilon` method that we have been through
a bunch of times now seem like a pain in the butt? Mathematicians are,
by their nature, a lazy lot. If they can get away with doing some
pain-in-the-butt operation just once and thereby come up with some
easy and useful rule, then that's what they'll do. And once they know
the rule to be true, they'll use it every time instead of the
pain-in-the-butt method. Not only does the rule make life easier,
but without such rules, mathematics would be so thick with undergrowth
as to make it virtually impossible to understand.

We have already demonstrated some rules concerning the sums, products,
and quotients of limits. Now that you know them to be true, any time
you encounter a sum, product, or quotient and need to know the limit,
you can apply the appropriate
rule rather than struggling through the `delta`-`epsilon`
method.

In this section, we will be establishing a rule concerning *the limits
of rational functions as *`x`* goes to zero*. What is
a *rational function*? It is the quotient (or ratio -- that's
where the name comes from) of two polynomials. It's that simple.
So 2.3-1, which we studied earlier, is an example of a rational function.
It's graph is shown again here.

Notice from the graph that the limit as `x` goes to zero of this
function appears to be `-1`. Notice as well that `-1` is also
the ratio of the *constant* term of the two polynomials
(the constant term of a polynomial is the one that is not multiplied by
`x` or any power of `x`). So the
first rule that I'll propose is this: *if the denominator has a nonzero
constant term, then the limit of the quotient of the numerator and denominator
polynomials as *`x`* goes to zero is exactly the ratio of the
numerator's constant term over the denominator's constant term*.

This is very easy to prove. Separate each polynomial into its constant term
and everything else. In the example in 2.3-1, the numerator separates into
a constant term of `-1` and everything else of `x ^{2}`.
Likewise the denominator separates into a constant term of

As another example, if one of the polynomials were:

xThen the constant term is^{3}+ 6x^{2}+ 7x + 3 eq. 2.5-1

x^{3}+ 6x^{2}+ 7x eq. 2.5-2

Every term in the "everything else" part of each polynomial is multiplied
by at least one `x`. And what is the limit as of `x` as
`x` goes to zero? Surely you can do a quick
`delta`-`epsilon` proof to show that that limit is zero.

And since every term in the "everything else" is multiplied by at least
one `x` and the
limit of that `x` is zero, and since the limit of the product is the
product of the limits, what is the limit of each term of the "everything
else"? Hint: you are multiplying each term by zero.

And so if the limit of each term in the "everything else" is zero, what
is the limit of their sum? Zero, right? So, in the limit as `x`
goes to zero, the entire "everything else" has a limit of zero. All that's
left is the constant term. Can you show using a quick
`delta`-`epsilon` proof that the limit of a constant is
always itself?

The point is that the limit of a polynomial as `x` goes to zero
*is* its constant term. And the limit of the quotient
(we already said the constant term of the denominator is not zero) is
simply the quotient of the constant terms.

Clearly, if the denominator has a zero in the constant term and the numerator
does not, the quotient has no limit. As `x` goes to zero, the
numerator will approach its nonzero constant as a limit, but the denominator
will approach zero as a limit. As the denominator gets closer and closer
to zero, the absolute value of the quotient will increase without limit.

But what if *both* the numerator and the denominator have constant
terms that are zero? Can you find a limit then? Well, let's take a
very simple case:

x f(x) =Wherever~~eq. 2.5-3 x~~

I claim that

lim f(x) = 1 eq. 2.5-4 xTo prove that, I must demonstrate the contract that for any~~> 0~~

ç x ÷ çwhenever~~- 1 ÷ £ e eq. 2.5-5 ç x ÷~~

|1 - 1| £ e eq. 2.5-6Clearly the lefthand side of 2.5-6 is zero, which is always less than

So what happens when we use more complicated polynomials that have zero in their constant term? Here is an example:

xHow do we find^{3}+ 2x^{2}+ 7x f(x) =~~eq. 2.5-7 8x~~^{2}- x

lim f(x) eq. 2.5-8 xfor this~~> 0~~

Well, we can factor an `x` out of both the top and bottom of this
to get:

xYou will agree that 2.5-9 represents precisely the same function as 2.5-7, won't you? So now we have the product of two things, each of which we know how to take the limit of. The lefthand factor is simply the quotient of two polynomials, both of which have nonzero constant terms. We already know that the limit of that as^{2}+ 2x + 7 x f(x) =~~eq. 2.5-9 8x - 1 x~~

If the polynomials are such that dividing out `x/x` still results in
neither of them having a nonzero constant term, then do it again.
Then you'll have two `x/x`'s in the product. But both of their
limits are still `1`, so multiplying by them does not effect the
result. In fact, you can divide out `x/x` as many times as
it takes until one of the polynomials has a constant term, and still
not effect the result. Once one of polynomials does have a constant
term, you can take the limit. Of course if you divide out `x/x`
some number of times and end up with the numerator's polynomial having
a nonzero constant term but the denominator still has a zero constant
term (that is no constant term), then the original function has no
limit.

From that, we can derive a general rule for quotients of polynomials.
*The limit of the quotient as *`x`* goes to zero is
exactly the quotient of the lowest power coefficients from the numerator
and denominator that are not both zero* (and remember that when no
term is shown for a particular power, the coefficient for that power
is zero). Note that the two coefficients
must come from the same order term. You can't, for example, take the
`x ^{3}` coefficient from the numerator and the

Use the rule discussed above to find the limits as `x` goes toward
zero of the following functions (or determine if no limit exists):

x^{2}- 7x + 12 1)~~x + 6 x~~^{4}- x^{2}2)~~x~~^{3}+ 2x^{2}x^{4}- x^{2}3)~~x~~^{4}+ 2x^{3}x^{4}- x^{2}4)~~x~~^{3}+ 2x

email me at *hahn@netsrq.com*