Exercise 5.1 · Q31: $\cos(x^2)$ is a continuous function

Basics you need

Solution

1. The domain is all of $\R$

   For any real $x$⁠, $x^2$ is a real number, and the cosine of any real number is defined. So $f(x) = \cos(x^2)$ is defined for every real $x$⁠: the domain is $\R$⁠.

   By Definition 2, "continuous function" means continuous at every point of the domain, so we first need to know which points those are. Here there are no exclusions.
   $\cos u$ is defined for every real $u$ (the point at angle $u$ on the unit circle always exists). Basics: trigonometric identities

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

   Let

   $$h(x) = x^2 \ \ (\text{inner}), \qquad g(u) = \cos u \ \ (\text{outer}).$$

   Then for every real $x$⁠,

   $$(g \circ h)(x) = g\big(h(x)\big) = g(x^2) = \cos(x^2) = f(x).$$

   So $f = g \circ h$⁠. The composite is defined at every $x$⁠, since the range of $h$ (the numbers $\ge 0$⁠) lies inside the domain of $g$ (all of $\R$⁠).

   Theorem 2 is about composites, so we must first exhibit $f$ as one. Pay attention to the order: $h \circ g$ would be $(\cos x)^2 = \cos^2 x$⁠, a different function. We square first, so squaring is the inner function.
   $g \circ h$ means "do $h$⁠, then $g$" (Basic 1). Basics: composite functions

3. The inner function $h(x) = x^2$ is continuous everywhere

   $h$ is a polynomial, so it is continuous at every real number (Example 14). Concretely, for any real $c$⁠,

   $$\lim_{x\to c} x^2 = c \cdot c = c^2 = h(c).$$
   The limit of a product is the product of the limits, and $\lim_{x\to c} x = c$ (Example 6). This is the step "as $x$ gets close to $c$⁠, $x^2$ gets close to $c^2$".
   Polynomials are continuous (Basic 3). Basics: what a limit is

4. The outer function $g(u) = \cos u$ is continuous everywhere

   Take any real $d$ and write $u = d + t$⁠, so $u \to d$ means $t \to 0$⁠. Then

   $$\begin{aligned} \lim_{u\to d} \cos u &= \lim_{t\to0} \cos(d + t)\\ &= \lim_{t\to0} (\cos d\cos t - \sin d\sin t)\\ &= \cos d\cdot 1 - \sin d\cdot 0 = \cos d. \end{aligned}$$

   So $\lim_{u\to d}\cos u = \cos d = g(d)$⁠: cosine is continuous at every real $d$⁠.

   The textbook proves this for sine in Example 17 and says "a similar proof" works for cosine; this is that proof. It needs only the expansion of $\cos(d + t)$ and the facts $\cos t \to 1$⁠, $\sin t \to 0$ as $t \to 0$⁠. Those two limits come from Class XI (Limits and Derivatives); the book's Example 17 uses them without proof (it says the sine limit is "intuitively clear from the graph").
   $\cos(A + B) = \cos A\cos B - \sin A\sin B$ (Basic 4). Basics: trigonometric identities

5. Apply Theorem 2 at an arbitrary point

   Let $c$ be any real number.

   • $h$ is continuous at $c$ (Step 3).
   • $g$ is continuous at $h(c) = c^2$⁠, because $g$ is continuous at every real number (Step 4 with $d = c^2$⁠).

   By Theorem 2, $g \circ h = f$ is continuous at $c$⁠. In limit form:

   $$\lim_{x\to c} \cos(x^2) = \cos(c^2) = f(c).$$

   Since $c$ was an arbitrary real number, $f$ is continuous at every point of its domain $\R$⁠, so $f$ is a continuous function.

   Theorem 2 needs two things at the point: the inner function continuous at $c$⁠, and the outer function continuous at the inner value $h(c)$⁠. Both hold for every $c$⁠, so the conclusion holds for every $c$⁠. Choosing $c$ arbitrary is what turns one check into a proof for all real numbers.
   Theorem 2, Section 5.2.1; the same pattern as Example 19 ($\sin(x^2)$⁠). Lesson: algebra of continuous functions

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