Exercise 5.1 · Q32: $|\cos x|$ is a continuous function

Basics you need

Solution

1. The domain is all of $\R$

   $\cos x$ is defined for every real $x$⁠, and the modulus of every real number is defined. So $f(x) = |\cos x|$ is defined for all real $x$⁠.

   "Continuous function" means continuous at every point of the domain (Definition 2), so we record the domain first. Nothing is excluded.

2. Write $f$ as a composite $g \circ h$

   Let

   $$h(x) = \cos x \ \ (\text{inner}), \qquad g(u) = |u| \ \ (\text{outer}).$$

   Then for every real $x$⁠,

   $$(g \circ h)(x) = g(\cos x) = |\cos x| = f(x).$$

   The range of $h$ is $[-1, 1]$⁠, which lies inside the domain $\R$ of $g$⁠, so the composite is defined everywhere.

   This is the order in which you would press calculator keys: cos first, then the absolute value. The other order, $h \circ g$⁠, would be $\cos|x|$⁠, which is a different function (although it happens to equal $\cos x$⁠, since cosine is even).
   Inner function first, outer function second (Basic 1). Basics: composite functions

3. The inner function $\cos x$ is continuous everywhere

   For every real $c$⁠, $\lim_{x\to c}\cos x = \cos c$⁠. With $x = c + t$⁠:

   $$\begin{aligned} \lim_{t\to0}\cos(c + t) &= \lim_{t\to0}(\cos c\cos t - \sin c\sin t)\\ &= \cos c\cdot1 - \sin c\cdot0 = \cos c. \end{aligned}$$
   This is the "similar proof" the textbook mentions after Example 17. It uses $\cos t \to 1$ and $\sin t \to 0$ as $t \to 0$⁠, limits from Class XI (Limits and Derivatives) that Example 17 uses without proof.
   Cosine is continuous (Basic 4). Basics: trigonometric identities

4. The outer function $|u|$ is continuous everywhere

   This is Example 7. In brief, for a point $d$⁠:

   • if $d < 0$⁠, near $d$ we have $|u| = -u$⁠, so $\lim_{u\to d}|u| = -d = |d|$⁠;
   • if $d > 0$⁠, near $d$ we have $|u| = u$⁠, so $\lim_{u\to d}|u| = d = |d|$⁠;
   • if $d = 0$⁠, the left-hand limit is $\lim(-u) = 0$⁠, the right-hand limit is $\lim u = 0$⁠, and $|0| = 0$ (Example 3).
   The only delicate point for $|u|$ is $u = 0$⁠, where the formula changes; the one-sided limits there both equal 0, which is the value. So $g$ is continuous at every real number.
   Modulus as a two-piece function (Basic 3). Basics: piecewise functions

5. Apply Theorem 2 at an arbitrary point

   Let $c$ be any real number. $h = \cos$ is continuous at $c$ (Step 3), and $g = |\cdot|$ is continuous at $h(c) = \cos c$ (Step 4, since $g$ is continuous at every real number). By Theorem 2, $g \circ h = f$ is continuous at $c$⁠:

   $$\lim_{x\to c} |\cos x| = |\cos c| = f(c).$$

   As $c$ was arbitrary, $f$ is continuous at every real number, so it is a continuous function.

   Theorem 2 asks for continuity of the outer function at the particular inner value $\cos c$⁠. Because $|\cdot|$ is continuous everywhere, that holds whatever $\cos c$ turns out to be, including $\cos c = 0$⁠.
   Theorem 2, Section 5.2.1; Example 20 uses $|\cdot|$ as an outer function in the same way. Lesson: algebra of continuous functions

6. A direct look at a corner: $x = \frac{\pi}{2}$

   The most suspicious points are where $\cos x = 0$⁠, because there the modulus switches from "leave alone" to "flip the sign". Take $x = \frac{\pi}{2}$⁠, where $f(\frac{\pi}{2}) = |0| = 0$⁠.

   • Just left of $\frac{\pi}{2}$⁠, $\cos x > 0$⁠, so $f(x) = \cos x$⁠, and $\lim_{x\to\pi/2^-} \cos x = \cos\frac{\pi}{2} = 0$⁠.
   • Just right of $\frac{\pi}{2}$⁠, $\cos x < 0$⁠, so $f(x) = -\cos x$⁠, and $\lim_{x\to\pi/2^+} (-\cos x) = 0$⁠.

   Left-hand limit $=$ right-hand limit $=$ value $= 0$⁠: continuous at $\frac{\pi}{2}$⁠, as Step 5 promised. The same happens at every odd multiple of $\frac{\pi}{2}$⁠.

   The general argument already covers these points; the direct check shows why the corners cause no trouble: both sides arrive at height 0, which is the value. A corner changes the direction of the graph, not its height. (Corners do matter for differentiability, Section 5.3.)
   One-sided limits (Basic 5). Basics: left-hand and right-hand limits

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