In almost any calculus or analysis textbook, in the chapter on continuity of functions, you’ll encounter four theorems about the operations on functions that preserve continuity: multiplying a continuous function by a scalar (real number), adding two continuous functions, multiplying two continuous functions, and dividing two continuous functions. But no textbooks that I have seen, nor websites nor lecture notes nor study guides nor homework solutions nor Stack Exchange questions nor anything else I’ve found online, have the proof to the last one! On the chance that I can add something new to the internet for the first time, here is the proof that Professor Yuri Ledyaev did in my Advanced Calculus (introductory real analysis) class at Western Michigan University:

Let \(f, g\) be two continuous functions with domain \(A \subset \mathbb{R}\) and let \(a \in A\). Let \(g(a) \neq 0\). Then \(f/g\) is also continuous at point \(a\). The proof is nearly identical if \(A \subset \mathbb{R}^n\) —which is in fact the way we did the proof, in the first chapter on functions of multiple variables—but there’s no way I’m typing every single \(x\) and \(a\) in vector form in Latex. Just imagine they’re all bold or they have a little line over them, and imagine that the \(|x-a|<\delta\)'s are all \(\vec{x} \in B_\delta (\vec{a})\)'s.

** Proof:** For the function \(f/g\) to be continuous, this means

$$

\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{f(a)}{g(a)}.

$$

The way to prove this equality is to apply the Cauchy definition of continuity, i.e., \(\forall \epsilon > 0\), \(\exists \delta > 0\) such that if \(|x – a| < \delta\), then \(\left|\frac{f(x)}{g(x)} - \frac{f(a)}{g(a)}\right| < \epsilon\). To obtain this inequality, we start with the latter absolute-value expression, get a common denominator, use the ol' add-and-subtract-the-same-thing trick, and apply the fact that both \(f\) and \(g\) are individually continuous. Observe:

$$

\begin{eqnarray}

\left|\frac{f(x)}{g(x)} - \frac{f(a)}{g(a)}\right| &=& \frac{|f(x)g(a) - f(a)g(x)|}{|g(x)||g(a)|} \nonumber\\[13pt]

&=& \frac{|f(x)g(a) - f(a)g(a) + f(a)g(a) - f(a)g(x)|}{|g(x)||g(a)|} \nonumber \\[13pt]

&\leq& \frac{|f(x)g(a) - f(a)g(a)| + |f(a)g(a) - f(a)g(x)|}{|g(x)||g(a)|} \nonumber \\[13pt]

&=& \frac{|g(a)||f(x) - f(a)| + |f(a)||g(x) - g(a)|}{|g(x)||g(a)|} \nonumber \\

\end{eqnarray}

$$

(The \(\leq\) comes from the Triangle Inequality.)

Now, we have to interrupt our equation display to make use of the fact that \(g\) is continuous. Since \(g\) is continuous, we can let \(\epsilon_{\small{1}} = \frac{1}{2} |g(a)| > 0\). Then there exists \(\delta_{\small{1}}\) such that whenever \(|x-a|<\delta_{\small{1}}\),

$$

\begin{eqnarray}

|g(a)| - |g(x)| &\leq& |g(x) - g(a)| < \epsilon_{\small{1}} \nonumber \\[13pt]

|g(a)| - \epsilon_{\small{1}} &<& |g(x)| \nonumber \\[13pt]

\frac{1}{2} |g(a)| &<& |g(x)| \nonumber \\[13pt]

\frac{2}{|g(a)|} &>& \frac{1}{|g(x)|}

\end{eqnarray}

$$

(Again the \(\leq\) in this equation display is due to (one form of) the Triangle Inequality.)

Applying this inequality to the \(|g(x)|\) in the denominator up above gives us

$$

\begin{eqnarray}

\frac{|g(a)||f(x) – f(a)| + |f(a)||g(x) – g(a)|}{|g(x)||g(a)|} &<& \frac{2}{|g(a)|^2} \frac{|g(a)||f(x) - f(a)| + |f(a)||g(x) - g(a)|}{|g(x)||g(a)|} \nonumber \\[13pt]

&=& \frac{2}{|g(a)|} |f(x) - f(a)| + \frac{2|f(a)|}{|g(a)|^2} |g(x) - g(a)| \nonumber \\[13pt]

\end{eqnarray}

$$

Now, since \(f\) is continuous, we know that for any \(\frac{\epsilon}{2}\), there exists \(\delta_2\) such that making \(|x - a| < \delta_2\) will make \(|f(x) - f(a)| < \frac{\epsilon}{2}\cdot\frac{|g(a)|}{2}\).

Similarly, since \(g\) is continuous, there exists \(\delta_3\) such that making \(|x - a| < \delta_3\) will make \(|g(x) - g(a)| < \frac{\epsilon}{2}\cdot\frac{|g(a)|^2}{2|f(a)| + 1}\). (The \(+1\) must be added in case \(f(a) = 0\).)

Finally, we simply choose \(\delta = \min{\{\delta_1, \delta_2, \delta_3\}}\), and this will give us

$$

\begin{eqnarray}

\left|\frac{f(x)}{g(x)} - \frac{f(a)}{g(a)}\right| &<& \frac{2}{|g(a)|} |f(x) - f(a)| + \frac{2|f(a)|}{|g(a)|^2} |g(x) - g(a)| \nonumber\\[13pt]

&<& \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon

\end{eqnarray}

$$

To summarize, this proof showed that if \(f\) and \(g\) are continuous, then for any \(\epsilon > 0\), letting \(|x-a|<\delta\) will make \(\left|\frac{f(x)}{g(x)} – \frac{f(a)}{g(a)}\right| < \epsilon\), meaning \(\frac{f(x)}{g(x)}\) is continuous at \(x=a\).