Combination Theorem for Continuous Functions/Quotient Rule

Theorem

Let $X$ be one of the standard number fields $\Q, \R, \C$.

Let $f$ and $g$ be functions which are continuous on an open subset $S \subseteq X$.

Then:

$\dfrac f g$ is continuous on $S \setminus \left\{{x \in S: g \left({x}\right) = 0}\right\}$

that is, on all the points $x$ of $S$ where $g \left({x}\right) \ne 0$.

Proof

By definition of continuous, we have that

$\forall c \in S: \displaystyle \lim_{x \to c} \ f \left({x}\right) = f \left({c}\right)$

$\forall c \in S: \displaystyle \lim_{x \to c} \ g \left({x}\right) = g \left({c}\right)$

Let $f$ and $g$ tend to the following limits:

$\displaystyle \lim_{x \to c} \ f \left({x}\right) = l$

$\displaystyle \lim_{x \to c} \ g \left({x}\right) = m$

From the Quotient Rule for Limits of Functions, we have that:

$\displaystyle \lim_{x \to c} \ \frac {f \left({x}\right)} {g \left({x}\right)} = \frac l m$

wherever $m \ne 0$.

So, by definition of continuous again, we have that $\dfrac f g$ is continuous on all points $x$ of $S$ where $g \left({x}\right) \ne 0$.

$\blacksquare$

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 9.4 \ \text{(iii)}$