Example 9: Continuity of $f(x)=\frac1x$⁠, $x\ne0$

Basics you need

Solution

1. The domain, and an arbitrary point in it

   The domain of $f$ is $\{x \in \R : x \ne 0\}$⁠. Fix any real number $c \ne 0$⁠. Then $c$ is in the domain and

   $$f(c) = \frac1c.$$
   Definition 2 only asks about points of the domain, so these are the only points we need to examine. Taking $c$ arbitrary (apart from $c \ne 0$⁠) handles all of them at once.
   $\frac1c$ is a real number exactly because $c \ne 0$⁠. Basics: functions

2. Near $c$⁠, every $x$ is in the domain

   If $|x - c| < \frac{|c|}{2}$⁠, then $x$ lies between $\frac{c}{2}$ and $\frac{3c}{2}$⁠, so $x$ has the same sign as $c$ and

   $$|x| > \frac{|c|}{2} > 0.$$

   So $f(x) = \frac1x$ is defined at every $x$ close to $c$⁠, on both sides.

   To talk about the limit of $f$ at $c$⁠, $f$ must be defined at the points near $c$ on both sides. Staying within half the distance to 0 keeps us safely away from the one excluded point.
   For $c = 2$⁠: every $x$ in $(1, 3)$ is positive. For $c = -0.2$⁠: every $x$ in $(-0.3, -0.1)$ is negative.

3. Left-hand and right-hand limits at $c$

   The formula $\frac1x$ is used on both sides of $c$⁠, and $\lim x = c \ne 0$⁠, so by the quotient rule (Basic 2)

   $$\begin{aligned} \lim_{x\to c^-}\frac1x &= \frac1c, \\ \lim_{x\to c^+}\frac1x &= \frac1c. \end{aligned}$$

   To see that this is forced, measure the distance:

   $$\left|\frac1x - \frac1c\right| = \frac{|x - c|}{|x|\,|c|} < \frac{2\,|x - c|}{c^2},$$

   using $|x| > \frac{|c|}{2}$ from Step 2. The right side shrinks to 0 as $x \to c$⁠.

   The quotient rule needs the limit of the denominator to be non-zero; that is where $c \ne 0$ is used. The inequality makes it concrete: at $c = 2$⁠, being within 0.001 of 2 keeps $\frac1x$ within $2(0.001)/4 = 0.0005$ of $\frac12$⁠.
   $\frac1x - \frac1c = \frac{c - x}{xc}$ (common denominator). Making the denominator smaller makes a fraction larger, so replacing $|x|$ by the smaller $\frac{|c|}{2}$ gives an upper bound. Basics: one-sided limits

4. Compare, and conclude

   $$\begin{aligned} \lim_{x\to c^-} f(x) &= \frac1c, \\ \lim_{x\to c^+} f(x) &= \frac1c, \\ f(c) &= \frac1c. \end{aligned}$$

   All three agree, so $f$ is continuous at $c$⁠. As $c$ was any point of the domain, $f$ is continuous at every point of its domain; $f$ is a continuous function (Definition 2).

   This is the textbook's conclusion: "$f$ is continuous at every point in the domain of $f$⁠. Thus $f$ is a continuous function."
   Lesson: continuity at a point

5. What about $x = 0$⁠?

   0 is not in the domain: $f(0)$ does not exist. Definition 1 is only about points of the domain, so 0 is neither a point of continuity nor a point of discontinuity of $f$ in this book's sense. The textbook still studies the behaviour near 0 (Tables 5.1 and 5.2):

   $x$	$f(x)$	side of 0
   $0.5$	$2$	right
   $0.1$	$10$	right
   $0.01$	$100$	right
   $0.001$	$1000$	right
   $-0.5$	$-2$	left
   $-0.1$	$-10$	left
   $-0.01$	$-100$	left
   $-0.001$	$-1000$	left

   As $x \to 0^+$⁠, $\frac1x$ exceeds any number you name (for any $N > 0$⁠, $\frac1x > N$ once $0 < x < \frac1N$⁠); the textbook writes $\lim_{x\to0^+} f(x) = +\infty$⁠. As $x \to 0^-$⁠, $\frac1x$ falls below any number: $\lim_{x\to0^-} f(x) = -\infty$⁠.

   $+\infty$ and $-\infty$ are not real numbers, so these one-sided limits do not exist as real numbers. This also shows that no choice of a value at 0 could make the extended function continuous there: there is no limit for the value to equal.
   Some books call 0 a discontinuity of $\frac1x$ (an "infinite discontinuity"). In this textbook, continuity and discontinuity are defined only at points of the domain, so the correct answer to this question is "continuous at every point of its domain". Basics: what a limit is

Every fact used, and where it comes from