Proving a Quadratic Limit at an Arbitrary Input

Why an Arbitrary Input Needs a Local Bound

A limit of \(x^2\) at zero is relatively direct: near zero, both factors in \(x^2=x\cdot x\) are small. At an arbitrary input \(a\), the expression \(x^2-a^2\) still becomes small as \(x\) approaches \(a\), but its factors do not both approach zero. One factor approaches \(a\), which need not be small.

The key is to factor the difference. One factor then records how close \(x\) is to \(a\); the other can be controlled by requiring \(x\) to remain in a suitable neighborhood of \(a\). This is a precise version of the idea that a small change in input produces a small change in its square, provided the input stays near the point under consideration.

The Claim and Its Conditions

There is no restriction on the real number \(a\), and the output tolerance \(\varepsilon\) may be any positive real number. The number \(\delta\) is allowed to depend on both \(a\) and \(\varepsilon\). The condition \(0<|x-a|\) describes inputs near \(a\) but not equal to \(a\), as required by the definition of a limit. The proof will in fact give the desired estimate even when \(x=a\).

The algebraic identity at the center of the proof is the difference-of-squares factorization:

The first factor, \(|x-a|\), is exactly the input distance controlled by \(\delta\). To control the second factor, we will also require \(|x-a|<1\). The triangle inequality then gives

This bound is local: it holds because \(x\) has been restricted to lie within distance \(1\) of \(a\). It does not claim that \(|x+a|\) is bounded by \(1+2|a|\) for every real \(x\).

A Complete Epsilon-Delta Proof

Fix an arbitrary \(a\in\mathbb{R}\) and an arbitrary \(\varepsilon>0\). Define

Both numbers inside the minimum are positive: \(1>0\), and \(\varepsilon/(2|a|+1)>0\) because \(\varepsilon>0\) and \(2|a|+1\ge 1\). Thus \(\delta>0\). Now take any \(x\in\mathbb{R}\) for which \(0<|x-a|<\delta\). Since \(\delta\le 1\), we have \(|x-a|<1\), and consequently \(|x+a|<2|a|+1\). Applying the factorization and the absolute-value product rule yields

The last inequality follows from \(\delta\le\varepsilon/(2|a|+1)\). We have therefore shown that every \(x\) in the required punctured neighborhood satisfies \(|x^2-a^2|<\varepsilon\). Since \(a\) and \(\varepsilon\) were arbitrary, the theorem holds for every real input \(a\).

Worked Applications

Common Mistakes and Their Corrections

A frequent error is to treat \(|x+a|\) as though it were automatically at most \(2|a|\). For example, with \(a=1\) and \(x=3/2\), this would claim

which is false. The missing condition is that \(x\) must be close to \(a\). A correct local statement is: if \(|x-a|<1\), then

Here the example value \(x=3/2\) has \(|x-1|=1/2<1\), and the corrected estimate gives \(|x+1|<3\), which is true. The bound need not be sharp; it only needs to be valid throughout the neighborhood.

Another error is to write \(x^2-a^2=(x-a)^2\). This identity is generally false: for example, when \(x=3\) and \(a=1\), the left side is \(9-1=8\), while the right side is \((3-1)^2=4\). The correct factorization is \(x^2-a^2=(x-a)(x+a)\), which retains the second factor that must be bounded.

Finally, a positive \(\delta\) must control both parts of the argument. Choosing only \(\delta=\varepsilon/(2|a|+1)\) would control the product if the factor bound held, but it would not ensure \(|x-a|<1\) when \(\delta>1\). Taking the minimum with \(1\) makes both requirements hold at once.

What the Proof Shows

The factorization explains why the limit is true: as \(x\) approaches \(a\), the factor \(|x-a|\) becomes as small as needed, while the factor \(|x+a|\) stays bounded near that fixed \(a\). Their product therefore becomes small. The point of choosing a neighborhood first is not to make both factors small; it is to keep one factor bounded so the other can control the product.

The particular radius above is a convenient sufficient choice, not the only possible choice. Any positive \(\delta\) that guarantees both \(|x-a|<1\) and \(|x-a|<\varepsilon/(2|a|+1)\) would work. The minimum enforces both conditions directly. The argument makes no assumption that \(a\) is positive: absolute values and the triangle inequality handle positive, negative, and zero inputs in the same proof.

This technique is useful beyond this one limit. Factoring an output difference can expose an input-distance factor, and a local restriction can control the remaining factor. The exact bounds may change in other problems, but the logical structure remains: identify the error, control all factors, and choose a positive neighborhood size that guarantees the required tolerance.