A “rigorous proof” in mathematics is a proof based on a complete chain of logic without any gaps or ambiguity. The formal limit definition is a key ingredient of rigorous proofs in calculus. A few such proofs are included in Appendix D. More complete developments can be found in textbooks on the branch of mathematics called “analysis.”

Figure 2.48: A sequence of graphs of \(y=\tfrac{\sin x}{x}\), zoomed in at the point in question. To shrink the gap from \(0.2\) to \(0.004\), we require that \(x\) lie within \(0.15\) of \(0\).

The formal definition of a limit is often called the \(\epsilon-\delta\) definition. The tradition of using the symbols \(\epsilon\) and \(\delta\) originated in the writings of Augustin-Louis Cauchy on calculus and analysis in the 1820s.

FORMAL DEFINITION OF A LIMIT

Suppose that \(f(x)\) is defined for all \(x\) in an open interval containing \(c\) (but not necessarily at \(x = c\)). Then

if for all \(\epsilon > 0\), there exists \(\delta > 0\) such that

Example 1

Let \(f(x) = 8x + 3\).

(a) Prove that \(\lim\limits_{x\rightarrow 3} f(x) = 27\) using the formal definition of the limit.

(b) Find values of \(\delta\) that work for \(\epsilon = 0.2\) and \(0.001\).

Solution

(a) We break the proof into two steps.

Step 1. Relate the gap to \(|x - c|\).

We must find a relation between two absolute values: \(|f(x) - L|\) for \(L = 27\) and \(|x - c|\) for \(c = 3\). Observe that

Thus, the gap is 8 times as large as \(|x - 3|\).

Step 2. Choose \(\delta\) (in terms of \(\epsilon\)).

We can now see how to make the gap small: If \(|x-3|<\frac{\epsilon}{3}\), then the gap is less than \(8\left(\frac{\epsilon}{8}\right)\). Therefore, for any \(\epsilon > 0\), we choose \(\delta=\frac{\epsilon}{8}\). With this choice, the following statement holds:

Since we have specified \(\delta\) for all \(\epsilon > 0\), we have fulfilled the requirements of the formal definition, thus proving rigorously that \(\lim\limits_{x\rightarrow 3}(8x+3) = 27\).

(b) For the particular choice \(\epsilon = 0.2\), we may take \(\delta=\frac{\epsilon}{8} = \frac{0.2}{8} = 0.025\):

This statement is illustrated in Figure 2.49. But note that any positive \(\delta\) smaller than \( 0.025 \) will also work. For example, the following statement is also true, although it places an unnecessary restriction on \(x\):

Similarly, to make the gap less than \(\epsilon = 0.001\), we may take

Example 2

Prove that \(\lim\limits_{x\rightarrow 2}x^2 = 4\).

Solution Let \(f(x) = x^2\).

Step 1. Relate the gap to \(|x - c|\).

In this case, we must relate the gap \(|f(x) - 4|\) = \(|x^2 - 4|\) to the quantity \(|x - 2|\) (Figure 2.50). This is more difficult than in the previous example because the gap is not a constant multiple of \(|x - 2|\). To proceed, consider the factorization

Because we are going to require that \(|x - 2|\) be small, we may as well assume from the outset that \(|x - 2| < 1\), which means that \(1 < x < 3\). In this case, \(|x + 2|\) is less than 5 and the gap satisfies

Step 2. Choose \(\delta\) (in terms of \(\epsilon\)).

We see from Eq. (2) that if \(|x - 2|\) is smaller than both \(\frac{\epsilon}{5}\) and 1, then the gap satisfies

Therefore, the following statement holds for all \(\epsilon > 0\):

We have specified \(\delta\) for all \(\epsilon > 0\), so we have fulfilled the requirements of the formal limit definition, thus proving that \(\lim\limits_{x\rightarrow 2}x^2 = 4\).

Example 3

Prove that \(\lim\limits_{x\rightarrow 3}\frac{1}{x} = \frac{1}{3}\).

Solution

Step 1. Relate the gap to \(|x - c|\).

The gap is equal to

Because we are going to require that \(|x - 3|\) be small, we may as well assume from the outset that \(|x - 3| < 1\) —that is, that \(2 < x < 4\). Now observe that if \(x > 2\), then \(3x > 6\) and \(\frac{1}{3x}<\frac{1}{6}\), so the following inequality is valid if \(|x - 3| < 1\):

Step 2. Choose \(\delta\) (in terms of \(\epsilon\)).

By Eq. (3), if \(|x - 3| < 1\) and \(|x - 3| < 6\epsilon\), then

Therefore, given any \(\epsilon > 0\), we let \(\delta\) be the smaller of the numbers \(6\epsilon\) and 1. Then

Again, we have fulfilled the requirements of the formal limit definition, thus proving rigorously that \(\lim\limits_{x\rightarrow 3}\frac{1}{x} = \frac{1}{3}\).

GRAPHICAL INSIGHT

Keep the graphical interpretation of limits in mind. In Figure 2.51(A), \(f(x)\) approaches \(L\) as \(x \rightarrow c\) because for any \(\epsilon > 0\), we can make the gap less than \(\epsilon\) by taking \(\delta\) sufficiently small. By contrast, the function in Figure 2.51(B) has a jump discontinuity at \(x = c\). The gap cannot be made small, no matter how small \(\delta\) is taken. Therefore, the limit does not exist.

Proof of the Sum Law

Assume that

We must prove that \(\lim\limits_{x\rightarrow c}(f(x)+g(x)) = L+M\).

Apply the Triangle Inequality (see margin) with \(a = f(x)- L\) and \(b = g(x) - M\):

Each term on the right in (4) can be made small by the limit definition. More precisely, given \(\epsilon > 0\), we can choose \(\delta\) such that \(|f(x) - L|<\frac{\epsilon}{2}\) and \(|g(x) - M|<\frac{\epsilon}{2}\) if \(0 < |x-c| < \delta\) (in principle, we might choose different \(\delta\)’s for \(f\) and \(g\), but we may then use the smaller of the two \(\delta\)’s). Thus, Eq. (4) gives

This proves that

For a video of a proof of this theorem, click here.