Example 17: The sine function is continuous

Basics you need

Solution

1. The domain is all of $\R$

   $\sin x$ is defined for every real number $x$ (every real number is an angle in radians). So we must show continuity at every real $c$⁠. Fix an arbitrary real number $c$⁠.

   Definition 2 asks for continuity at every point of the domain. Since nothing is excluded, the argument must work for a general $c$⁠.

2. The two small-angle facts

   The book uses the fact

   $$\lim_{h\to 0} \sin h = 0.$$

   The last line of its working also needs $\cos h \to 1$⁠. This follows from the first fact through the identity $1 - \cos h = 2\sin^2 \frac{h}{2}$⁠, that is $\cos h = 1 - 2\sin^2 \frac{h}{2}$⁠. Put $t = \frac{h}{2}$⁠: as $h \to 0$⁠, also $t \to 0$⁠, so $\sin \frac{h}{2} = \sin t \to 0$ by the book's fact. Then, by the product and difference rules of limits,

   $$\begin{aligned} \lim_{h\to 0} \cos h &= 1 - 2 \cdot 0 \cdot 0 \\ &= 1. \end{aligned}$$
   The book says $\lim \sin x = 0$ "is intuitively clear from the graph". The circle shows it too: a small angle gives a point very close to $(1, 0)$⁠, with tiny height $\sin h$ and horizontal position $\cos h$ close to 1. Both facts are needed, because the expansion in Step 4 contains both $\sin h$ and $\cos h$⁠.
   $1 - \cos h = 2\sin^2\frac{h}{2}$ is the double-angle formula $\cos 2\theta = 1 - 2\sin^2\theta$ with $\theta = \frac{h}{2}$⁠. Going deeper: for $0 < |h| < \frac{\pi}{2}$ the height $|\sin h|$ is shorter than the arc length $|h|$⁠, so $|\sin h| < |h|$⁠, and the sandwich theorem (Class XI, Limits and Derivatives; not proved in this book) then squeezes $\sin h$ to 0. Basics: trigonometric identities

3. Substitute $x = c + h$

   Put $x = c + h$⁠. As $x \to c$⁠, $h = x - c \to 0$⁠, so

   $$\lim_{x\to c} \sin x = \lim_{h\to 0} \sin(c + h).$$
   This rewrites "$x$ close to $c$" as "$c$ plus a small $h$". The advantage is that the addition formula can now separate the fixed part $c$ from the small part $h$⁠.

4. Expand with the addition formula

   With $A = c$ and $B = h$⁠:

   $$\sin(c + h) = \sin c \cos h + \cos c \sin h.$$
   Now each term is a constant ($\sin c$ or $\cos c$⁠, which do not change as $h$ varies, because $c$ is fixed) times a function of $h$ whose limit we know from Step 2.
   Basics: trigonometric identities

5. Take the limit term by term

   $$\begin{aligned} \lim_{h\to 0} \sin(c + h) &= \lim_{h\to 0} [\sin c \cos h] \\ &\quad + \lim_{h\to 0} [\cos c \sin h] \\ &= \sin c \cdot 1 + \cos c \cdot 0 \\ &= \sin c = f(c). \end{aligned}$$
   The sum rule splits the limit into two limits, each of which exists by Step 2. In each, the constant factor comes outside: $\sin c \cdot \lim \cos h = \sin c \cdot 1$ and $\cos c \cdot \lim \sin h = \cos c \cdot 0$⁠.
   Basics: what a limit is

6. Conclude

   Steps 3 and 5 give $\lim_{x\to c} \sin x = \sin c = f(c)$⁠. By Definition 1, $\sin x$ is continuous at $c$⁠; since $c$ was any real number, the sine function is continuous at every real number, so it is a continuous function.

   This matches the book's conclusion "$f$ is a continuous function", now with the role of $\cos h \to 1$ made explicit.
   Lesson: continuity at a point

7. The cosine function (the book's Remark)

   The book remarks that a similar proof works for cosine. Using $\cos(c + h) =$ $\cos c \cos h - \sin c \sin h$⁠:

   $$\begin{aligned} \lim_{h\to 0} \cos(c + h) &= \cos c \cdot 1 - \sin c \cdot 0 \\ &= \cos c. \end{aligned}$$

   So $\cos x$ is also continuous at every real number.

   Exactly the same three moves: substitute $x = c + h$⁠, expand with an addition formula, and use $\cos h \to 1$⁠, $\sin h \to 0$⁠. Example 18 needs this result for $\tan x$⁠.
   $\cos(A + B) =$ $\cos A \cos B - \sin A \sin B$⁠. Basics: trigonometric identities

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