Solution to Problem 3.1-4

The problem is to prove that

             _____
   f(x)  =  √x - 3

is continuous for all x > 3.

Step 1: Set up the delta-epsilon contract for the limit. You need to show that

           _____      _____
    lim   √x - 3  =  √a - 3
   x -> a

whenever a > 3. In delta-epsilon language, that statement is that: for any ε > 0, no matter how small, you can always find a δ > 0 small enough so that

     _____     _____
   |√x - 3  - √a - 3|  < ε

whenever |x - a| < δ. The rest of the problem is cooking up a recipe to find δ for any given ε. See if you can use the difference of squares to find that recipe. Then click here to see the next step.

Step 2: Apply the difference of squares to simplify the algebra. We multiply both sides of the inequality by

    _____      _____
   √x - 3  +  √x - a

(which is always positive), and we get a real simplification of left side:

      _____      _____   _____      _____
   |(√x - 3  -  √a - 3)(√x - 3  +  √a - 3)|  =
                                       _____      _____
         |(x - 3)  -  (a - 3)|  <  ε (√x - 3  +  √a - 3)

Taking the cancellation, you get

                   _____      _____
   |x - a|  <  ε (√x - 3  +  √a - 3)

Now think about the |x - a| term above and how it relates to δ. The next step is to include δ in the inequality.

Step 3: Putting the delta in. Since we are presupposing that |x - a| < δ, you can just pop the δ in like this:

                         _____      _____
   |x - a|  <  δ  <  ε (√x - 3  +  √a - 3)

which give us a recipe, more or less, for finding δ.

             _____      _____
   δ  <  ε (√x - 3  +  √a - 3)

We still have that nasty multiply on the right, though. We'd like the right-hand side of the inequality to be in terms of ε and a alone. Recall that we postulated that

     _____      _____
   |√x - 3  -  √a - 3|  <  ε

Use that fact to eliminate x, which is the final step of the problem.

Step 4: Eliminate the x. We add

     _____      _____
   -√a - 3  +  √a - 3

which is, of course, zero, to the right-hand side of the inequality:

             _____      _____       _____
   δ  <  ε (√x - 3  -  √a - 3  +  2√a - 3)

Since

     _____      _____
   |√x - 3  -  √a - 3|  <  ε

the expression on the right of our inequality can be no less than

                 _____
   ε ((-ε)  +  2√a - 3)

Make sure you see why. The final result is that if for any a > 3 and ε > 0, you select δ such that

              _____
   δ  <  ε (2√a - 3 - ε)

then you will meet the delta-epsilon contract.

There is still a problem, though, if

            _____
   ε  >  2 √a - 3

because that would make δ have to be negative. But remember, if ε is too big, you are always allowed to use a smaller one. So if this happens, replace ε with any positive number that is smaller than

      _____
   2 √a - 3

and that becomes your recipe for δ.

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