Properties of the limit of functions

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The limit of functions has properties that are similar to the limit of sequences.



Theorem (Properties of the limit of function)

Let \(f(x)\) and \(g(x)\) be functions such that \(\lim_{x\to a}f(x) = \alpha\) and \(\lim_{x\to a}g(x) = \beta\). The following hold.

  1. For any constants \(k, l\in\mathbb{R}\), \[\lim_{x\to a}(kf(x) + lg(x)) = k\alpha + l\beta.\]
  2. \[\lim_{x\to a}f(x)g(x) = \alpha\beta.\]
  3. If \(\beta \neq 0\), \[\lim_{x\to a}\frac{f(x)}{g(x)} = \frac{\alpha}{\beta}\]
Proof. In the following \(\varepsilon\) is always an arbitrary positive real number, and \(D_f\) and \(D_g\) are the domains of the functions \(f\) and \(g\), respectively.

1. Let \(M = \max\{|k|, |l|\} + 1\). Then \(\frac{\varepsilon}{2M}\) is also positive and real. Since \(\lim_{x\to a}f(x) = \alpha\), there exists \(\delta_1 > 0\) such that for any \(x\in D_f\) if \(0 < |x - a| < \delta_1\) then \(|f(x) - \alpha| < \frac{\varepsilon}{2M}\). Similarly, there exists \(\delta_2 > 0\) such that for any \(x \in D_g\) if \(0 < |x - a| < \delta_2\) then \(|g(x) - \beta| < \frac{\varepsilon}{2M}\). Let \(\delta = \min\{\delta_1, \delta_2\}\). If \(|x - a| < \delta\), then \[ \begin{eqnarray*} |(kf(x) + lg(x)) - (k\alpha + l\beta)| &=& |k(f(x) - \alpha) + l(g(x) - \beta)|\\ &\leq& |k||f(x) - \alpha| + |l||g(x) - \beta|\\ &< & M\cdot\frac{\varepsilon}{2M} + M\cdot\frac{\varepsilon}{2M}\\ &=& \varepsilon. \end{eqnarray*} \] Hence, \[\lim_{x\to a}(kf(x) + lg(x)) = k\alpha + l\beta.\]

2. Since \(\lim_{x\to a}f(x) = \alpha\), there exists \(\delta_0 > 0\) such that for any \(x\in D_f\), \(0 < |x - a| < \delta_0\) implies \(|f(x) - \alpha| < \varepsilon\). Then \(\alpha - \varepsilon < f(x) < \alpha + \varepsilon\), and hence
\[|f(x)| \leq \max\{|\alpha - \varepsilon|, |\alpha + \varepsilon|\}.\] Let us define \[M = \max\{|\alpha - \varepsilon|, |\alpha + \varepsilon|, |\beta|\}.\] Since \(\varepsilon > 0\), at least one of \(|\alpha - \varepsilon|\) or \(|\alpha + \varepsilon|\) is not 0 so that \(M > 0\). Again, we can find \(\delta_1 > 0\) such that \(|x - a| < \delta_1\) implies \(|f(x) - \alpha| < \frac{\varepsilon}{2M}\). Similarly, we can find \(\delta_2 > 0\) such that \(|x - a| < \delta_2\) implies \(|g(x) - \beta| < \frac{\varepsilon}{2M}\). Let \(\delta = \min\{\delta_0, \delta_1, \delta_2\}\). If \(0 < |x - a| < \delta\), we have \[ \begin{eqnarray*} |f(x)g(x) - \alpha\beta| &=& |f(x)g(x) -f(x)\beta + f(x)\beta - \alpha\beta|\\ &\leq& |f(x)g(x) -f(x)\beta| + |f(x)\beta - \alpha\beta|\\ &=& |f(x)||g(x) -\beta| + |f(x) - \alpha||\beta|\\ &<& M\cdot\frac{\varepsilon}{2M} + \frac{\varepsilon}{2M}\cdot M\\ &=& \varepsilon. \end{eqnarray*} \] Hence, \[\lim_{x\to a}f(x)g(x) = \alpha\beta.\]

3. It suffices to show that \[\lim_{x \to a}\frac{1}{g(x)} = \frac{1}{\beta}.\] We can find \(\delta_0 > 0\) such that \(0 < |x - a| < \delta_0\) implies \(|g(x) - \beta| < \frac{|\beta|}{2}\). By the triangle inequality, \[||g(x)| - |\beta|| \leq |g(x) - \beta| < \frac{|\beta|}{2},\] we have, in particular, \[|g(x)| > \frac{|\beta|}{2} > 0\] so that \[\frac{1}{|g(x)|} < \frac{2}{|\beta|}.\] Again, we can find \(\delta_1 > 0\) such that \(|x - a| < \delta_1\) implies \(|g(x) - \beta| < \frac{|\beta|^2\varepsilon}{2}\). Let \(\delta = \min\{\delta_0, \delta_1\}\). If \(0 < |x - a| < \delta\), we have \[\begin{eqnarray*} \left|\frac{1}{g(x)} - \frac{1}{\beta}\right| &=& \left|\frac{\beta - g(x)}{g(x)\beta}\right|\\ &=& \frac{|g(x) - \beta|}{|g(x)||\beta|}\\ &<& \frac{|\beta|^2\varepsilon}{2}\cdot\frac{2}{|\beta|}\cdot\frac{1}{|\beta|}\\ &=& \varepsilon. \end{eqnarray*} \]
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Remark. Part 1 of this theorem shows that
\[\lim_{x\to a}(kf(x) + lg(x)) = k\lim_{x\to a}f(x) + l\lim_{x\to a}g(x)\]
which demonstrates the linearity of the limit operation. □

Theorem (Limit of a composite function)

Let \(f(x)\) and \(g(x)\) be functions such that \(\lim_{x \to a}f(x) = b\) and \(\lim_{x\to b}g(x) = \alpha\). Then
\[\lim_{x\to a}(g\circ f)(x) = \alpha\]
where \(g\circ f\) is the function composition of \(g\) after \(f\) (or \(f\) then \(g\)), that is, \((g\circ f)(x) = g(f(x))\).
Proof. Let \(\varepsilon\) be an arbitrary positive real number. As usual, we can find \(\delta_1 > 0\) such that \(0 < |x - b| < \delta_1\) implies \(|g(x) - \alpha| < \varepsilon\). Similarly, we can find \(\delta > 0\) such that \(0 < |x - a| < \delta\) implies \(|f(x) - b| < \delta_1\). Then \(0 < |x - a| < \delta\) implies \(|g(f(x)) - \alpha| < \varepsilon\). Hence
\[\lim_{x \to a}g(f(x)) = \alpha.\]
■

Example. If \(x\) approaches \(a\), then \(x\) approaches \(a\) (tautology!). Therefore
\[\lim_{x\to a}x = a.\]
By the above theorems, we have, for example,
\[\begin{eqnarray} \lim_{x\to a}2x &=& 2a,\\ \lim_{x\to a}x^2 &=& a^2,\\ \lim_{x\to a}(x^2 + 2x) &=& a^2 + 2a. \end{eqnarray}\]
In general, if \(f(x)\) is a polynomial of \(x\) (such as above), then
\[\lim_{x \to a}f(x) = f(a).\]
□
Example. Consider
\[\lim_{x \to 1}\frac{x^3 - 1}{x - 1}.\]
Since we can factorize as \(x^3 - 1  = (x-1)(x^2 + x + 1)\), we have
\[\frac{x^3 - 1}{x - 1} = \frac{(x-1)(x^2 + x + 1)}{x-1} = x^2 + x + 1\]
which is a polynomial of \(x\). Therefore,
\[\lim_{x \to 1}\frac{x^3 - 1}{x - 1} = \lim_{x \to 1}(x^2 + x + 1) = 3.\]
□


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