Box 3.0: Coached Exercise -- Continuity of f(x) = x3KCT logo

© 1996 by Karl Hahn

Here is the outline of what you will do. First, you will write the limit expression of what it means to say that  f(x) = x3  is continuous everywhere. Then you will apply the "contract" definition of limits to that limit expression to come up with a contract. Then you will make a subsitution for x using an error variable, the same way we did in 3.0-4. Then you do the algebra to expand the power, just as we did in 3.0-5, only this time it will be a little more complex because you are using a higher power. That means that you will end up with more than two terms in your sum. Each term in the sum will dictate a choice for how to compute error from e and a. The rule is to pick the least from among them. Again, you will have to make special arrangements for when  a = 0  in order to avoid dividing by zero. But once you have done that, you'll have a recipe for finding the worst that error can be, so you set d equal to that, and you're done.

Step 1: Write The Limit Expression. This is the easy part. If you have trouble here, go back to 3.0-2 and see if you can't figure out the expression by analogy. Once you have the limit expression, you job now is to show that it holds for any real number, a.

Step 2: Write The Contract that the Limit Expression Implies. Go back to 3.0-3 if you need help, and make the new contract by analogy. Then read the contract over and make sure you know what it means.

Step 3: Rewrite the Contract Using an error Variable that expresses how far x is from a. You will make precisely the same substitution as we did in the x2 example.

Step 4: Do the Algebra. Look at what you've written so far. You should see a sum raised to the 3rd power. You need to use your algebra skills to expand that. For reference,

  (u + v)3  =  u3 + 3u2v + 3uv2 + v3                             eq. 3.0a-1
but you ought to be able to multiply that one out for yourself. If you do this step properly, you will notice a cancellation you can make that gets rid of all terms in the sum that are not multiplied by some power of error. How many terms do you have left in the sum?

Step 5: You have how many terms? So at worst, you would like each term to account for what fraction of e? So, in separate equations, set each term in the sum equal to that fraction of e and solve separately for error. You now have as many candidates for error as you had terms in the sum. So state the rule of: you should pick candidate that is the least of these expressions. That choice will vary according to the values of a and e, but so what? You don't have to make the choice, you just have to state the rule by which to make that choice. Note that the chosen candidate for error is positive, and in each case, choosing a value for error that is even closer to zero than the candidate makes the sum even closer to zero as well. So any value even closer to zero than your chosen candidate satisfies your contract as well.

Step 6: Does a = 0 give you a problem in any of the candidates? Remember, you can't divide by zero. If it does, then the term it came from in the original sum probably contributes zero to the original sum. Does it? So state why you can ignore that candidate and pick from the candidate(s) that don't have zero in the denominator.

Step 7: Set d equal to the choice you made for error, and state why that finishes the proof and why it applies for all values of a.

Did you come through this exercise intact? If not, review some more about how a limit is like a contract. Then think hard about what it means to define continuity in terms of limits, as we do. Reread section 3.0. And then, run through these steps again.


As always, I am at hahn@netsrq.com

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