Exercise 5.1 · Q28: Find $k$⁠: $kx+1$ for $x\le\pi$⁠, $\cos x$ for $x>\pi$

Basics you need

Solution

1. The value at the point: $f(\pi) = k\pi + 1$

   Every real $x$ satisfies exactly one of $x \le \pi$⁠, $x > \pi$⁠, so $f$ is defined everywhere, and $\pi$ is in the domain. Since $\pi \le \pi$⁠, the first line applies:

   $$f(\pi) = k\pi + 1.$$
   The value is what the limits must match. It comes from the line whose condition includes $\pi$⁠; the cosine line is only for $x$ strictly greater than $\pi$⁠.
   "$\le$" includes equality; "$>$" does not (Basic 2).

2. Left-hand limit: follow the line $kx + 1$

   For $x < \pi$⁠, $f(x) = kx + 1$⁠. Put $x = \pi - h$⁠, $h > 0$⁠, $h \to 0$⁠:

   $$\begin{aligned} \lim_{x\to\pi^-} f(x) &= \lim_{h\to0^+} \big(k(\pi - h) + 1\big)\\ &= \lim_{h\to0^+} (k\pi + 1 - kh)\\ &= k\pi + 1. \end{aligned}$$
   $kx + 1$ is a polynomial (degree 1, or a constant if $k = 0$⁠), so its limit is its value at $\pi$ (Example 14). The term $-kh$ is a fixed number $k$ times something shrinking to 0, so it vanishes.
   Polynomials are continuous: substitute to get the limit. Basics: what a limit is

3. Right-hand limit: follow $\cos x$

   For $x > \pi$⁠, $f(x) = \cos x$⁠. Put $x = \pi + h$⁠, $h > 0$⁠, $h \to 0$⁠:

   $$\begin{aligned} \lim_{x\to\pi^+} f(x) &= \lim_{h\to0^+} \cos(\pi + h)\\ &= \lim_{h\to0^+} (-\cos h)\\ &= -1. \end{aligned}$$
   By the formula for $\cos(A + B)$⁠, $$\begin{aligned} \cos(\pi + h) &= \cos\pi\cos h - \sin\pi\sin h\\ &= (-1)\cos h - 0 = -\cos h, \end{aligned}$$ and $\cos h \to \cos 0 = 1$⁠. This is the same as saying: cosine is continuous at $\pi$⁠, so its limit there is $\cos\pi = -1$ (Basic 4).
   $\cos\pi = -1$⁠, $\sin\pi = 0$⁠. Basics: trigonometric identities

4. Set them equal: $k\pi + 1 = -1$

   Continuity at $\pi$ needs

   $$\lim_{x\to\pi^-} f(x) = \lim_{x\to\pi^+} f(x) = f(\pi),$$

   that is, $k\pi + 1 = -1 = k\pi + 1$⁠. The left-hand limit and the value are the same expression, so the condition is

   $$k\pi + 1 = -1 \;\Longleftrightarrow\; k\pi = -2 \;\Longleftrightarrow\; k = -\frac{2}{\pi}.$$

   Dividing by $\pi$ is allowed because $\pi \ne 0$⁠. For any other $k$⁠, the line arrives at a height $k\pi + 1 \ne -1$ and the graph jumps. So $k = -\frac{2}{\pi}$ is the only value.

   Again the piece that owns the point makes the value agree with the left-hand limit for free; the equation forces the right-hand limit to agree as well. The answer is negative: the line must slope downwards from $(0, 1)$ to reach $-1$ at $x = \pi$⁠.
   Definition 1: left-hand limit = right-hand limit = value. Lesson: continuity at a point

5. Confirm with $k = -\frac{2}{\pi}$

   Now the first piece is $1 - \frac{2}{\pi}x$⁠. At $x = \pi$ it gives $1 - \frac{2}{\pi}\cdot\pi = 1 - 2 = -1$⁠. So

   $$\begin{aligned} \lim_{x\to\pi^-} f(x) &= -1,\\ \lim_{x\to\pi^+} f(x) &= \cos\pi = -1,\\ f(\pi) &= -1. \end{aligned}$$

   All three agree, so $f$ is continuous at $x = \pi$⁠. ✓

   The graph shows the meaning of the answer: the line must be aimed exactly at the bottom of the cosine wave at $x = \pi$⁠.

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