# A Closer Look at a Limit Proof

#### (A new question of the week)

A recent question asked about one of our explanations of the limit of x2 (which we have discussed at least five times).  This led to a deeper examination of what was said; and as I have looked through this and other pages, I have realized that it would be worthwhile to look at the details more carefully.

A Limit Proof Using EstimationHow do I show that the limit of x^2 as x->(-2) is 4, using the delta-epsilon definition of a limit?

Here is the first part of Doctor Rob’s reply:

The definition says that for any epsilon > 0, no matter how small, you can find a delta > 0 such that:   |x-(-2)| = |x+2| < delta implies that |x^2-4| < epsilonThe idea is to start with what you want to show, that |x^2-4| < epsilon, and to manipulate this until you can get it into the form |x+2| < some expression in epsilon. Then picking delta to be this expression in epsilon will do, and the proof is to work backwards through the steps of the manipulation. In this case:   |x^2-4| < epsilon    <==> |(x+2)^2 - 4*(x+2)| < epsilon    <==  |x+2|^2 + 4*|x+2| < epsilon (by the triangle inequality)    <==> |x+2|^2 + 4*|x+2| - epsilon < 0Now you can use the Quadratic Formula to solve for |x+2|, and thus find an upper bound on |x+2| in terms of epsilon. That will be what you choose for delta. One tricky part is that each step needs an implication arrow in one direction (<==) but not necessarily in the other.

Before I look at the new question, we should read between the lines to see if we can follow this, which is not the method I am most familiar with.

First, note the structure of what he is doing: The goal is to show that, given a fact about delta, we can make a conclusion about epsilon; but first we have to find the appropriate delta. So the proof actually has two parts, which shouldn’t be confused. The search (exploration) part looks as if we were solving an inequality; but because of the ultimate goal, we need each step not to imply the next, but to be implied by it. Note the arrows, <==> and <==, the latter meaning that the new line will imply the one before.

Now, how does he get $$|x^2-4| < \epsilon \Leftrightarrow |(x+2)^2 – 4(x+2)| < \epsilon$$? This is not obvious from what was said; but we can check it by expanding $$(x+2)^2 – 4(x+2)$$ as $$x^2 + 4x + 4 – 4x – 8 = x^2 – 4$$ as claimed. One way to obtain that result (without just guessing) is to make a substitution: We want an expression in terms of $$x+2$$, so we let $$u = x+2$$, and replace $$x$$ with $$u-2$$:

$$x^2 – 4 = (u – 2)^2 – 4 = u^2 – 4u + 4 – 4 = u^2 – 4u = (x+2)^2 – 4(x + 2)$$.

The benefit of this is that we can use our knowledge of $$|x+2$$ in subsequent steps.

How about the next line, which refers to the triangle inequality? This is the fact that, for any a and b, $$|a – b| \le |a| + |b|$$, just as any side of a triangle is less than the sum of the other two sides. Specifically, it tells us that
$$|(x+2)^2 – 4(x+2)| \le |(x+2)^2| + |4(x+2)|$$. As a result, if the right side is less than epsilon, we can conclude that the left side is also less than epsilon (since it is even smaller), which is what we will want to conclude when this chain of statements is reversed to make the final proof. This is why we were looking for an expression greater than the expression we had.

Now we have a quadratic inequality, $$|x+2|^2 + 4|x+2| – \epsilon < 0$$, that we can solve. Doctor Rob suggests using the quadratic formula to do this, and this is where the recent question came in.

Juares wrote,

In this link, http://mathforum.org/library/drmath/view/53357.html , have example:

$$\displaystyle\lim_{x\rightarrow -2}x^2 = 4$$:

$$|x+2|^2 + 4|x+2| < \epsilon$$

$$|x+2|^2 + 4|x+2| – \epsilon < 0$$

and solve by quadratic formula for any ε

$$\displaystyle\frac{-b \pm\sqrt{b^2-4ac}}{2a}$$

$$\displaystyle\frac{-4|x+2| \pm\sqrt{(4|x+2|)^2-4(|x+2|)(-\epsilon)}}{2|x+2|}$$

He showed more work, but this is enough to see the error. Doctor Rick responded:

Hi, Juares. It looks like you haven’t understood what Doctor Rob meant when he said:

”  |x^2-4| < epsilon

<==> |(x+2)^2 – 4(x+2)| < epsilon

<== |x+2|^2 + 4|x+2| < epsilon (by the triangle inequality)

<==> |x+2|^2 + 4*|x+2| – epsilon < 0

Now you can use the Quadratic Formula to solve for |x+2|, and thus find an upper bound on |x+2| in terms of epsilon.”

In your work, you are taking a = |x + 2|, b = 4|x + 2|, and c = – ε, which implies that you are trying to solve the “quadratic equation” |x + 2|x2 + 4|x + 2|x – ε = 0. That’s not a quadratic equation!

When Doctor Rob said to solve for |x + 2|, he meant that you treat |x + 2| as the variable. If we let u = |x + 2|, then the inequality becomes

u2 + 4u – ε < 0

Now solve this inequality for u. That is, solve the related equation u2 + 4u – ε = 0 for u, which gives the endpoints of the solution interval(s), and decide whether the solution set for the inequality lies between these points or outside them.

Now Juares correctly applied the quadratic formula to solve the equation:

$$\displaystyle\frac{-4\pm\sqrt{16+4\epsilon}}{2} = -2+\sqrt{4+\epsilon}$$

This is δ. Let ε = 0.01. Then $$\displaystyle\delta =-2+\sqrt{4,1} = 0,025$$

This was followed by some long calculations presented as an image, so I will not attempt to reproduce it all. The important thing is that he has found a value for delta in general, and then is observing what happens for a specific value of epsilon, which is an excellent way to get a better feel for what is happening.

Hi again, Juares. Let me see whether I understand what you have done here.

We are working on solving the inequality u2 + 4u – ε < 0, where u = |x + 2|. You apply the Quadratic Formula to solve the related equation u2 + 4u – ε = 0, and this gives you the result

-2 – √(4 + ε) < |x + 2| < -2 + √(4 + ε)

Now, the quantity on the left is clearly negative (for any positive value of ε), whereas an absolute value (or modulus, as you probably call it), |x + 2|, must be non-negative, so we can further restrict the possible values for |x + 2|:

0 ≤ |x + 2| < -2 + √(4 + ε)

As Doctor Rob said, “The definition says that for any ε > 0, no matter how small, you can find a delta > 0 such that: |x-(-2)| = |x+2| < δ implies that |x2 – 4| < ε.” We have in fact found that:

For any ε > 0, we can choose δ = -2 + √(4 + ε)

and then |x – (-2)| < δ implies that |x2 – 4| < ε.

So we have found what is needed for the proof.

The rest of what you have written appears to be a check of this result by testing one possible value for ε, namely, ε = 0.1. In this case, we choose

δ = -2 + √(4 + 0.1) = 0.024845…

0 < |x – 2| < 0.024845…You have rounded that last number to 0.025; I have just kept more digits.
What I find, next, is
2 – 0.025 < x < 2 + 0.025
1.975 < x < 2.025
(I used the idea that |x – a| < b means that the distance between x and a is less than b, that is, x is between a-b and a+b.)

We want to demonstrate that, for any x in this interval, |x2 – 4| < 0.1. I suppose that you may be trying to use the fact that |x2 – 4| = |x + 2| |x – 2| to prove this. I would prefer to note that f(x) = |x2 – 4| is an increasing function on [1.975, 2.025], so that it obtains its greatest value at x = 2.025; and in that case, we find

If x = 2.025 then |x2 – 4| = 0.10065

If we hadn’t rounded our δ up, we would find that when x = 2 + δ, |x2 – 4| = 0.1; and for any x between 2 – δ and 2 + δ, |x2 – 4| < 0.1. This is what we expected to find, if our formula for δ in terms of ε is good.

So you have found a valid proof, and I hope I have helped you see that it is valid.

Note that since delta is an upper bound, we don’t want to increase it! So rather than rounding up, it would have been appropriate to round down, to ensure the required conclusion.

Now I want to continue and look at the last part of Doctor Rob’s answer, where he gives an alternative approach (which is what I am familiar with). It is important to observe that there is not just one valid delta in these proofs; not only can Doctor Rob’s first delta be replaced by anything smaller (since that will still imply the required conclusion), but he hasn’t determined that his delta is the largest or best possible number to use. In the next part, he obtains an entirely different formula for delta:

Another tricky part is that there isn't necessarily a unique answer. In this case, you could have proceeded like this instead:   |x^2-4| < epsilon    <==> |(x+2)*(x-2)| < epsilon    <==> |x+2|*|x-2| < epsilon    <==  |x+2|*(4 + |x+2|) < epsilon (by the triangle inequality)    <==  5*|x+2| < epsilon (provided |x+2| <= 1)and so on. At the end, you pick delta to be the minimum of 1 and the expression involving epsilon, and this ensures the "provided ..." part.Other expressions for delta in terms of epsilon may also work.

Finishing this work, we see that if $$|x+2| < \epsilon/5$$, then $$|x^2-4| < \epsilon$$, so we can take delta to be $$\epsilon/5$$ or anything smaller.

So this version of the proof takes $$\delta=min(\epsilon/5, 1)$$. In Juares’ example with ε = 0.01, we got δ = 0.024845… . This new version finds $$\delta = min(0.01/5, 1) = 0.002$$. This is a much smaller value (and therefore also works). The first method obtained a less restrictive value for delta, but both obtain values that provably lead to the required limit.

This same method (though for x approaching 2 rather than -2) is discussed in detail by two Math Doctors here:

Epsilon/Delta Definition of Limits

We also saw Doctor Fenton’s version of the same approach on Monday, as part of this answer:

Formal Definition of a Limit

In the following page, I looked at the same problem with a numerical value of delta (similar to Juares’ check):

Definition of the Limit

Finally, I examined a specific feature of this proof here:

Delta-Epsilon Proofs and Arbitrary Epsilon Choice

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