Exercise 5.1 · Q33: $\sin|x|$ is a continuous function

Basics you need

Solution

1. The domain is all of $\R$

   $|x|$ is a real number for every real $x$⁠, and sine is defined for every real number, so $\sin|x|$ is defined for all real $x$⁠.

   Continuity of a function is checked at every point of its domain (Definition 2); here that is every real number.

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

   Let

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

   Then $(g \circ h)(x) = g(|x|) = \sin|x| = f(x)$ for every real $x$⁠. The range of $h$⁠, $[0, \infty)$⁠, lies inside the domain $\R$ of $g$⁠.

   The modulus acts on $x$ first, so it is the inner layer. The other order, $h \circ g$⁠, would be $|\sin x|$⁠, a different function. For example, at $x = \frac{3\pi}{2}$ we get $\sin|\frac{3\pi}{2}| = -1$ but $|\sin\frac{3\pi}{2}| = 1$⁠.
   Inner first, outer second (Basic 1). Basics: composite functions

3. The inner function $|x|$ is continuous everywhere

   This is Example 7: for $c < 0$⁠, near $c$ we have $|x| = -x$ and the limit is $-c = |c|$⁠; for $c > 0$⁠, $|x| = x$ near $c$ and the limit is $c = |c|$⁠; at $c = 0$ both one-sided limits are 0 $= |0|$ (Example 3).

   The modulus is a two-piece function, and the only joint is 0. Away from 0 it is a polynomial near the point; at 0 the one-sided limits agree with the value.
   Modulus (Basic 3). Basics: piecewise functions

4. The outer function $\sin u$ is continuous everywhere

   Example 17 proves it: for any real $d$⁠, with $u = d + t$⁠,

   $$\begin{aligned} \lim_{t\to0}\sin(d + t) &= \lim_{t\to0}(\sin d\cos t + \cos d\sin t)\\ &= \sin d\cdot1 + \cos d\cdot0 = \sin d. \end{aligned}$$
   This is exactly the textbook's argument in Example 17, written with $t$ in place of $h$ (we are using $h$ for the inner function). 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.
   $\sin(A + B) = \sin A\cos B + \cos A\sin B$ (Basic 4). Basics: trigonometric identities

5. Apply Theorem 2 at an arbitrary point

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

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

   Since $c$ is arbitrary, $\sin|x|$ is continuous at every real number: it is a continuous function.

   The two conditions of Theorem 2 hold at every $c$⁠, so the conclusion holds at every $c$⁠. Arguing at an arbitrary $c$ is what makes this a proof for all real numbers at once.
   Theorem 2, Section 5.2.1. Lesson: algebra of continuous functions

6. A direct look at $x = 0$

   At 0 the inner function changes formula, so it is worth seeing the limits explicitly. $f(0) = \sin|0| = \sin 0 = 0$⁠.

   • For $x < 0$⁠: $|x| = -x$⁠, so $f(x) = \sin(-x)$⁠, and $\lim_{x\to0^-}\sin(-x) = \sin 0 = 0$⁠.
   • For $x > 0$⁠: $|x| = x$⁠, so $f(x) = \sin x$⁠, and $\lim_{x\to0^+}\sin x = \sin 0 = 0$⁠.

   Left-hand limit $=$ right-hand limit $=$ value $= 0$⁠, so $f$ is continuous at 0, in agreement with Step 5.

   Because $\sin|x|$ takes the same value at $x$ and $-x$⁠, its graph is symmetric about the $y$-axis. At 0 the two halves meet with slopes $-1$ and $+1$ (the slope of $\sin x$ at 0 is $\cos 0 = 1$⁠, a Class XI derivative fact, and the mirror image reverses its sign): a corner, which does not affect continuity.
   One-sided limits (Basic 5). Basics: left-hand and right-hand limits

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