Bounding the Quadratic Difference

Why the Difference of Squares Needs a Bound

To prove that \(\lim_{x\to a}x^2=a^2\) from the definition, we must control the size of \(|x^2-a^2|\) by controlling \(|x-a|\). The first step is algebraic: \(x^2-a^2=(x-a)(x+a)\). This identity splits the quantity we care about into the factor \(|x-a|\), which we can make as small as we like, and the factor \(|x+a|\), which is not small. The second factor depends on \(x\), so before it can be treated as a constant it must be bounded on the set of \(x\) values we are allowed to use. This tutorial derives that bound. The following tutorial uses it to choose \(\delta\) from \(\varepsilon\).

A bound is only meaningful on a restricted set. The factor \(|x+a|\) is unbounded as \(x\) ranges over all real numbers, since it grows without limit when \(|x|\) is large. But in a limit statement, only the values of \(x\) close to \(a\) matter, so we are free to restrict attention to a neighborhood of \(a\).

Proof. Write \(x+a=(x-a)+2a\). By the triangle inequality,

The last step uses the hypothesis \(|x-a|<1\). \(\square\)

Proof. By factoring and the lemma, \(|x^2-a^2|=|x-a|\,|x+a|<|x-a|\,(2|a|+1)\). The inequality is strict as long as \(x\ne a\), because then \(|x-a|>0\); if \(x=a\), both sides are zero and the bound holds with equality. \(\square\)

The constant \(2|a|+1\) depends only on the point \(a\), not on \(x\). That is what makes the bound useful: the quantity \(|x^2-a^2|\) is now below a fixed multiple of \(|x-a|\).

Using a Different Radius

The number \(1\) in the lemma is a convenient choice, not a requirement. If instead \(|x-a|<r\) for some \(r>0\), the same proof gives \(|x+a|<2|a|+r\), and so \(|x^2-a^2|<(2|a|+r)|x-a|\). A smaller radius gives a smaller constant but forces a smaller allowed neighborhood. For \(a=3\) and \(r=\tfrac12\), the constant is \(6.5\) instead of \(7\). The choice of \(r=1\) is popular because it keeps the arithmetic simple.

Common Mistakes

Using \(2|a|\) without the extra \(1\). It is tempting to say \(|x+a|\le2|a|\) because \(x\) is \"close to \(a\)\". But take \(a=3\) and \(x=3.9\): then \(|x+a|=6.9>6=2|a|\). The inequality is false. The extra \(1\) in \(2|a|+1\) accounts for the fact that \(x\) may be a little farther from zero than \(a\) is.

Bounding without restricting \(x\). The statement \"\(|x+3|<7\)\" is true only for \(x\in(-10,4)\). If \(x=100\), then \(|x+3|=103\). Any proof that uses the bound must say that \(|x-a|<1\) is being assumed, and must make sure that the final choice of \(\delta\) respects that assumption.

Dividing by the wrong factor. The bound controls \(|x^2-a^2|\) by a multiple of \(|x-a|\), not the reverse. It does not say that \(|x-a|\) is small whenever \(|x^2-a^2|\) is small.

A Preview of the Epsilon-Delta Argument

The bound is the engine of the proof that \(\lim_{x\to a}x^2=a^2\). Suppose we are given \(\varepsilon>0\). If \(|x-a|<1\), the theorem gives \(|x^2-a^2|<(2|a|+1)|x-a|\). So if we also demand that \(|x-a|<\dfrac{\varepsilon}{2|a|+1}\), then \(|x^2-a^2|<\varepsilon\). Both demands, \(|x-a|<1\) and \(|x-a|<\varepsilon/(2|a|+1)\), are satisfied by taking \(\delta=\min\bigl(1,\varepsilon/(2|a|+1)\bigr)\). The next tutorial carries out this selection carefully; here the goal is only to see how the local bound feeds it.

How the Constant Changes with a

The constant \(2|a|+1\) grows with \(|a|\), which reflects the geometry of the parabola: it is steeper far from the origin, so a given change in \(x\) produces a larger change in \(x^2\). The table lists the constants for several choices of \(a\).

In each row the true range of \(x+a\) is contained in \((-K,K)\), so \(K\) is a valid bound. For \(a=3\), the exact supremum of \(|x+3|\) is \(7\), so \(7\) cannot be improved. For \(a=0\) and \(a=1\) the value \(K\) matches the exact supremum as well. For \(a=-2\) the exact supremum is \(5\). In every case the bound \(2|a|+1\) is the best available when the radius is \(1\).

Why a Neighborhood Is Enough

A limit is a statement about what happens as \(x\) approaches \(a\). Values of \(x\) far from \(a\) are irrelevant to it. This is why we can restrict attention to \(|x-a|<1\): any proof only needs to work for sufficiently small \(\delta\), and shrinking \(\delta\) never hurts. The technique of first fixing an easy neighborhood, bounding all the awkward factors on it, and finally shrinking \(\delta\) to also satisfy the remaining condition appears again and again, for example when proving limits of \(1/x\), of \(\sqrt{x}\), and of products.

Reading the Estimate Geometrically

The graph of \(y=x^2\) near the point \((a,a^2)\) lies between two lines through that point whose slopes are \(\pm(2|a|+1)\), as long as \(x\) stays within \(1\) of \(a\). That is a precise way to say that the parabola cannot rise or fall faster than a fixed rate near \(a\). The estimate in this tutorial is a first example of a Lipschitz bound: a function \(f\) satisfies one on an interval if \(|f(x)-f(y)|\le K|x-y|\) for a fixed constant \(K\). Here \(K=2|a|+1\) works on the interval \((a-1,a+1)\) when one of the two points is \(a\).

The two-sided structure of the final estimate is worth noticing. The bound does not say that \(x^2-a^2\) has a particular sign; it says only that its size is limited. When \(x>a\), the difference is positive, and when \(x<a\), it is negative. The absolute value handles both cases at once, which is why proofs about limits are always written with \(|\cdot|\).

Finally, remember that this estimate is only one of several that appear in the course. Similar bounds are derived for \(1/x\) near a nonzero point, for \(\sqrt{x}\), and for products of functions. Each has the same shape: factor the difference, bound the troublesome factor on a neighborhood, and keep the small factor \(|x-a|\).

Where This Fits in the Course

This tutorial belongs to the module “Limit Concepts and Estimates”. The tutorial just before it is “Proving a Quadratic Limit at an Arbitrary Input,” and the one that follows is “Choosing Delta From Epsilon.”

Practice Problems with Answers

Try each problem before reading the answer. Each answer below was verified by differentiation or by evaluating a difference quotient or integral numerically.

1. Show that \(|x-5|<1\) implies \(|x^2-25|<11|x-5|\).
   Answer. The constant is \(2(5)+1=11\), and \(|x+5|<11\) because \(4<x<6\) gives \(9<x+5<11\).
2. Find \(K\) so that \(|x^2-16|<K|x+4|\) whenever \(|x+4|<1\).
   Answer. \(a=-4\), so \(K=2(4)+1=9\).
3. For \(a=2\) and \(|x-2|<\tfrac12\), find the best constant from the radius-\(r\) version.
   Answer. \(|x+2|<2(2)+\tfrac12=4.5\), so \(|x^2-4|<4.5|x-2|\).
4. Use the bound at \(a=3\) to find a \(\delta\) that guarantees \(|x^2-9|<0.07\).
   Answer. Take \(\delta=\min(1,0.07/7)=0.01\). Then \(|x-3|<0.01\) gives \(|x^2-9|<7(0.01)=0.07\).