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 x2. Likewise the denominator separates into a constant term of 1 and everything else of x2.
As another example, if one of the polynomials were:
x3 + 6x2 + 7x + 3 eq. 2.5-1Then the constant term is 3 and the "everything else" is
x3 + 6x2 + 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 x is nonzero, we have a cancellation, and
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
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 e regardless of what I choose for d. So my recipe for picking d is, "pick any
So what happens when we use more complicated polynomials that have zero in their constant term? Here is an example:
x3 + 2x2 + 7x f(x) =How do we find
eq. 2.5-7 8x2 - x
lim f(x) eq. 2.5-8 xfor this f(x)?
Well, we can factor an x out of both the top and bottom of this to get:
x2 + 2x + 7 x f(x) =You 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 x goes to zero is the quotient of the constant terms. We know the limit of the product is the product of the limits. Since the righthand factor is x/x and that has a limit of 1, the limit of the whole thing is simply the quotient of the new constant terms.
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 x3 coefficient from the numerator and the x2 coefficient from the denominator. Both powers must be the same. If it turns out that that gives you a nonzero coefficient from the numerator but a zero coefficient from the denominator, then the quotient has no limit.
Use the rule discussed above to find the limits as x goes toward zero of the following functions (or determine if no limit exists):
x2 - 7x + 12 1)
x + 6 x4 - x2 2) x3 + 2x2 x4 - x2 3) x4 + 2x3 x4 - x2 4) x3 + 2x
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