Proving the Limit of x Squared at Zero

Why This Limit Is a Useful First Quadratic Proof

When the input \(x\) is close to zero, its square \(x^2\) is also close to zero. That observation is easy to see in a table or on a graph. An epsilon–delta proof makes the claim precise: it shows that every requested positive error tolerance can be met by requiring the input to be sufficiently close to zero.

The key feature here is that the input is squared, while the limiting input and limiting value are both zero. This makes it possible to read off a suitable distance directly from the tolerance. The proof will not depend on checking a few sample inputs; it will apply to every input satisfying the required condition.

The Definition Applied to This Limit

To prove that the limit of \(x^2\) is zero as \(x\) approaches zero, we use the epsilon–delta definition. In this case, it asks us to show that for every positive tolerance \(\varepsilon\), there is a positive distance \(\delta\) such that every \(x\) within that distance of zero, but not equal to zero, has its square within \(\varepsilon\) of zero.

The quantities \(|x-0|\) and \(|x^2-0|\) are the input's distance from zero and the output's distance from zero, respectively. Since \(x^2\geq 0\), the output distance simplifies to \(|x^2-0|=x^2\). Thus, the task is to make \(x^2<\varepsilon\) by restricting \(|x|\).

Finding the Distance from the Tolerance

Start from the inequality we need to establish: \(x^2<\varepsilon\). Because \(\varepsilon>0\), its positive square root exists. For a real number \(x\), the inequality \(x^2<\varepsilon\) holds exactly when \(|x|<\sqrt{\varepsilon}\). This suggests choosing the input distance to be \(\sqrt{\varepsilon}\).

This choice is positive whenever \(\varepsilon\) is positive, as required by the definition. Moreover, if \(0<|x|<\delta\), then \(|x|<\sqrt{\varepsilon}\). Squaring these nonnegative quantities gives \(x^2<\varepsilon\). The strict inequality matters: it produces the strict output bound required by the definition.

The proof has a fixed logical order. First, take an arbitrary positive \(\varepsilon\). Next, specify a positive \(\delta\) in terms of it. Finally, consider any \(x\) satisfying the input condition and show that it satisfies the output condition. The choice is not made separately for each \(x\); it works for all qualifying \(x\).

Complete Epsilon–Delta Proof

Proof. Let \(\varepsilon>0\) be arbitrary. Choose \(\delta=\sqrt{\varepsilon}\), which is positive because \(\varepsilon>0\). Now let \(x\) be any real number satisfying \(0<|x-0|<\delta\). Then

Thus every \(x\) with \(0<|x-0|<\delta\) satisfies \(|x^2-0|<\varepsilon\). Since \(\varepsilon>0\) was arbitrary, the epsilon–delta definition gives \(\lim_{x\to 0}x^2=0\). \(\square\)

The central implication is \(|x|<\delta\Rightarrow x^2<\delta^2\). It holds because both \(|x|\) and \(\delta\) are nonnegative, and squaring preserves a strict inequality between nonnegative numbers. With \(\delta^2=\varepsilon\), the desired output bound follows.

Worked Examples With Specific Tolerances

Why the Choice Works for Every Positive Tolerance

The choice \(\delta=\sqrt{\varepsilon}\) is obtained by matching the squared input distance to the allowed output error. If the input distance is less than \(\delta\), then its square is less than \(\delta^2\). Setting \(\delta^2\) equal to \(\varepsilon\) makes that upper bound exactly the tolerance.

The equality \(\delta^2=\varepsilon\) is convenient, but the definition does not demand the largest possible distance. Any positive \(\delta\) with \(\delta^2\leq\varepsilon\) will also work: from \(|x|<\delta\), it follows that \(x^2<\delta^2\leq\varepsilon\). The square-root choice gives a simple direct match.

It is also important that \(\delta\) is allowed to depend on \(\varepsilon\). A smaller requested output error generally calls for a smaller input distance. In this proof that relationship is explicit: halving the tolerance's square root halves the chosen distance, while the resulting squared distance matches the tolerance.

A Common Mistake: Using the Tolerance as the Distance

A tempting but incorrect choice is \(\delta=\varepsilon\), on the assumption that an input distance less than \(\varepsilon\) automatically makes the squared input less than \(\varepsilon\). That implication does not hold for all positive \(\varepsilon\).

For example, let \(\varepsilon=2\) and make the proposed choice \(\delta=2\). The input \(x=\frac{3}{2}\) satisfies \(0<|x|<\delta\), since \(0<\frac{3}{2}<2\). But its output error is

So \(\delta=\varepsilon\) fails for this tolerance. The mistaken reasoning overlooks that \(x^2<|x|\) is not true for every \(x\); for instance, when \(|x|>1\), squaring makes its value larger. The direct correction is to choose \(\delta=\sqrt{\varepsilon}\): with \(\varepsilon=2\), this gives \(\delta=\sqrt{2}\), and \(0<|x|<\sqrt{2}\) implies \(x^2<2\).