Alternative Proof for Product of Limits © 2002 by Karl Hahn

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Again we start with a lemma, but a different one this time. The lemma this time is that if

    lim   f(x)  =  L
   x -> a

then, for any constant, C, then it must follow that

   lim   C × f(x)  =  C × L
  x -> a

We shall assume that C is not zero, since it is very easy to prove this for when C is zero.

Start with the delta-epsilon contract for the premise limit: For any  ε > 0  that you might name, then I can name a  δ > 0  small enough so that

   |f(x) - L|  <  ε

whenever  |x - a| < δ.  By saying that the limit of  f(x)  as x goes to a is equal to L, we are asserting this exact contract.

If you multiply the epsilon inequality by |C|, you find that

   |C| × |f(x) - L|  <  |C| × ε

Since the product of absolute values is equal to the absolute value of the product, you get

  |(C × f(x)) - (C × L)|  < |C| × ε

This is a delta-epsilon contract for the limit of  C × f(x)  being equal to  C × L,  except for one thing. This contract only guarantees the delta for  |C| × ε,  rather than for ε.

So if you said, "Find the delta that works for ε," I would have to say: Let ε^* be equal to  ε/|C|. The original contract stipulates that there is a delta that makes

   |f(x) - L|  <  ε*

whenever  |x - a| < δ.  Clearly that delta will also satisfy the contract

   |(C × f(x)) - (C × L)|  <  |C| × ε*  =  ε

That completes the recipe for finding the delta. The recipe is, divide the epsilon that they give you by  |C|.  Then, to that result, apply whatever recipe worked for the premise limit to get its delta. Because we assume that the premise limit is true, that recipe must exist.

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Now we move on to the main theorem, that the limit of a product is the product of the limits. If you have two limits:

    lim   f(x)  =  Lf
   x -> a

and

    lim   g(x)  =  Lg
   x -> a

then you have two delta-epsilon contracts, that for any  ε > 0,  no matter how small, you can always find a  δ_f > 0  to make:

   |f(x) - Lf|  <  ε

whenever  |x - a| < δ_f,  and you can always find a  δ_g > 0  to make:

   |g(x) - Lg|  <  ε

whenever  |x - a| < δ_g. 

Let δ be the smaller of δ_f and δ_g. It should be clear that δ satisfies both contracts. Now simply multiply the two contracts together:

   |f(x) - Lf| × |g(x) - Lg|  <  ε2

whenever  |x - a| < δ.  Recall that the product of the absolute values is the absolute value of the product. So, multiplying out the above we have:

   |f(x)g(x) - Lfg(x) - Lgf(x) + LfLg|  <  ε2

whenever  |x - a| < δ.  But we know that L_f and L_g are constants. And by the lemma, we know that a constant times a limit of a function is the limit of that constant times the function. That means that  L_fg(x)  will differ from  L_fL_g  by no more than  L_fε. Likewise  L_gf(x)  will differ from  L_fL_g  by no more than  L_gε. So the worst our product contract can possibly be is:

   |f(x)g(x) - LfLg - LfLg + LfLg|  <  ε2 + |Lf|ε + |Lg|ε

We get some cancellation in this contract:

   |f(x)g(x) - LfLg|  <  ε2 + |Lf|ε + |Lg|ε  =  ε(ε + |Lf| + |Lg|)

Once again we have arrived at a contract that involves something other than just a pure ε on the right. But suppose that  ε < 1.  Since the whole idea is that we can make it as small as we like, let it be less than one. Then if you let

ε* =	ε 1 + |Lf| + |Lg|

and use ε^* to generate the lesser of the two delta's from the two factor limits using their delta-epsilon contracts, you find that

   |f(x)g(x) - LfLg|  <  ε*(ε* + |Lf| + |Lg|)  <  ε

and you're done.

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