Limit of a univariate function

Let \(f(x)\) be a function. Suppose we move \(x\in\mathbb{R}\) towards \(a\) while keeping \(x \neq a\). If in this case \(f(x)\) approaches a constant value \(\alpha\) irrespective of the way how \(x\) approaches \(a\), we say that \(f(x)\) converges to \(\alpha\) as \(x \to a\) and write \[\lim_{x\to a}f(x) = \alpha\] or \[f(x) \to \alpha \text{ as \(x \to a\)}.\]

Remark. \(a\) needs not belong to \(\text{dom}(f)\) (the domain of \(f\)) as long as \(x\) can approach \(a\) arbitrarily closely. □

But what does this mean exactly? Here's a rigorous definition in terms of what is called the \(\varepsilon-\delta\) argument.

Definition (Limit of a function)

We say that the function \(f(x)\) converges to \(\alpha\) as \(x \to a\) and write

\[\lim_{x\to a}f(x) = \alpha\]

if the following condition is satisfied.

• For any \(\varepsilon > 0\),  there exists \(\delta > 0\) such that, for all \(x\in \text{dom}(f)\), if \(0 < |x - a| < \delta\) then \(|f(x) - \alpha| < \varepsilon\).

In a logical form,

\[\forall \varepsilon > 0, \exists \delta > 0, \forall x\in \text{dom}(f) ~ (0 < |x-a| < \delta \implies |f(x) - \alpha| < \varepsilon).\]

Remark. We are implicitly assuming \(\varepsilon, \delta \in\mathbb{R}\). □

Here's how this definition works. Suppose \(f(x) \to \alpha\) as \(x \to a\). Let us pick any positive real number \(\varepsilon\). However small this \(\varepsilon\) may be, if \(x\) is sufficiently close to \(a\), we can always have \(|f(x) - \alpha| < \varepsilon\). Here ``sufficiently close'' means that we can pick some sufficiently small positive real number \(\delta\) such that \(|x - a| < \delta\) implies \(|f(x) - \alpha| < \varepsilon\). In other words, we move \(x\) closer and closer to \(a\) until \(|f(x) - \alpha| < \varepsilon\) holds. Conversely, if this operation is possible for any \(\varepsilon\), it makes sense to say that \(f(x)\) converges to \(\alpha\) as \(x \to a\).

Example. Consider \(f(x) = x^2  + 1\). We have

\[\lim_{x \to 1}f(x) = 2.\]

Let \(\varepsilon = 0.1\). Let us find \(\delta > 0\) such that \(0 < |x - 1| < \delta\) implies \(|f(x) - 2| < \varepsilon\). Suppose

\[|f(x) - 2| < 0.1.\]

Then, since \(f(x) - 2 = (x^2 + 1) - 2 = x^2 - 1\),

\[-0.1 < x^2 - 1 < 0.1\]

so

\[0.9 < x^2 < 1.1.\]

If \(x > 0\), then

\[\sqrt{0.9}  < x < \sqrt{1.1}.\]

Since \(\sqrt{0.9} - 1 = -0.0513\cdots\) and \(\sqrt{1.1} - 1 = 0.0488\cdots\), if

\[|x - 1| < \sqrt{1.1} - 1,\]

then we have

\[|f(x) - 2| < 0.1 = \varepsilon.\]

So \(\delta\) can be any positive number less than \(\sqrt{1.1} - 1\). For example, let \(\delta = 0.04\). Then \(|x - 1| < 0.04\) implies \(0.96 < x < 1.04\), which implies \(-0.0784 < x^2 - 1 < 0.0816\) so

\[|f(x) - 2| = |x^2 - 1| < 0.0816 < 0.1 = \varepsilon.\]

□

Example. Let us show that \(\lim_{x\to 1}2x = 2\) by using the \(\epsilon-\delta\) argument.

Let \(\varepsilon > 0\) be any positive real number. Define \(\delta = \frac{\varepsilon}{2}\). Then, if \(0 < |x - 1| < \delta\), then we have \(|f(x) - 2| = |2x - 2| = 2|x - 1| < 2\delta = \varepsilon\). Therefore \(\lim_{x\to 1}f(x) = 2\). □