Exercise 5.1 · Q26: Find $k$⁠: $\frac{k\cos x}{\pi-2x}$ continuous at $\frac{\pi}{2}$

Basics you need

Solution

1. The point $\frac{\pi}{2}$ is in the domain, and $f(\frac{\pi}{2}) = 3$

   The top rule needs $\pi - 2x \ne 0$⁠, that is $x \ne \frac{\pi}{2}$⁠, which is exactly where it is used. The bottom rule covers the single point $x = \frac{\pi}{2}$⁠. So $f$ is defined at every real number, and in particular

   $$f\!\left(\tfrac{\pi}{2}\right) = 3.$$
   Continuity is only asked at points of the domain (Definition 1), so first we confirm $\frac{\pi}{2}$ is one, and we record the value that the limits will have to match. Note that the value 3 is given; it does not depend on $k$⁠.
   A piecewise rule: decide which line applies to the input, then use only that line. At exactly $x = \frac{\pi}{2}$ the second line applies. Basics: piecewise functions

2. Why we cannot simply substitute $x = \frac{\pi}{2}$ in the top rule

   At $x = \frac{\pi}{2}$⁠: the numerator is $k\cos\frac{\pi}{2} = k \cdot 0 = 0$ and the denominator is $\pi - 2\cdot\frac{\pi}{2} = 0$⁠. The top rule would give $\frac{0}{0}$⁠, which is not a number.

   So the limit has to be found by letting $x$ get close to $\frac{\pi}{2}$ without being equal to it, from each side separately.

   A limit only looks at values of $x$ near $c$⁠, never at $x = c$ itself. That is why the top rule (valid for $x \ne \frac{\pi}{2}$⁠) is the one that decides the limits, while the bottom rule only supplies $f(\frac{\pi}{2})$⁠.
   "$\frac{0}{0}$" means "both parts shrink to 0"; their ratio can tend to any number, or to none. Rewriting is needed (Basic 5).

3. Left-hand limit: put $x = \frac{\pi}{2} - h$

   For $x$ just below $\frac{\pi}{2}$⁠, write $x = \frac{\pi}{2} - h$ with $h > 0$ and $h \to 0$⁠. Then

   $$\begin{aligned} \cos x &= \cos\!\left(\tfrac{\pi}{2} - h\right) = \sin h,\\ \pi - 2x &= \pi - \pi + 2h = 2h. \end{aligned}$$

   So

   $$\begin{aligned} \lim_{x\to \frac{\pi}{2}^-} f(x) &= \lim_{h\to0^+} \frac{k\sin h}{2h}\\ &= \frac{k}{2}\lim_{h\to0^+} \frac{\sin h}{h}\\ &= \frac{k}{2}\cdot 1 = \frac{k}{2}. \end{aligned}$$
   The substitution moves the problem to $h \to 0$⁠, where we know a standard limit. After the substitution the troublesome $\cos x$ becomes $\sin h$ and the denominator becomes $2h$⁠, so the expression is a constant times $\frac{\sin h}{h}$⁠. A constant factor $\frac{k}{2}$ can be taken outside the limit.
   By the formula for $\cos(A - B)$ (Basic 3), $$\begin{aligned} \cos\!\left(\tfrac{\pi}{2} - h\right) &= \cos\tfrac{\pi}{2}\cos h + \sin\tfrac{\pi}{2}\sin h\\ &= 0\cdot\cos h + 1\cdot\sin h = \sin h, \end{aligned}$$ and $\frac{\sin h}{h} \to 1$ (Basic 4). Basics: trigonometric identities

4. Right-hand limit: put $x = \frac{\pi}{2} + h$

   For $x$ just above $\frac{\pi}{2}$⁠, write $x = \frac{\pi}{2} + h$ with $h > 0$ and $h \to 0$⁠. Then

   $$\begin{aligned} \cos x &= \cos\!\left(\tfrac{\pi}{2} + h\right) = -\sin h,\\ \pi - 2x &= \pi - \pi - 2h = -2h. \end{aligned}$$

   Both numerator and denominator are negative, and the two minus signs cancel:

   $$\begin{aligned} \lim_{x\to \frac{\pi}{2}^+} f(x) &= \lim_{h\to0^+} \frac{-k\sin h}{-2h}\\ &= \frac{k}{2}\lim_{h\to0^+} \frac{\sin h}{h}\\ &= \frac{k}{2}. \end{aligned}$$
   The right-hand side has to be checked on its own: a function can approach different numbers from the two sides. Here the sign changes in numerator and denominator cancel, so the right-hand limit is also $\frac{k}{2}$⁠. Hence the (two-sided) limit at $\frac{\pi}{2}$ exists and equals $\frac{k}{2}$⁠, whatever $k$ is.
   By the formula for $\cos(A + B)$⁠, $$\begin{aligned} \cos\!\left(\tfrac{\pi}{2} + h\right) &= \cos\tfrac{\pi}{2}\cos h - \sin\tfrac{\pi}{2}\sin h\\ &= 0\cdot\cos h - 1\cdot\sin h = -\sin h. \end{aligned}$$ Basics: left-hand and right-hand limits

5. Make the limit equal the value: $\frac{k}{2} = 3$

   By Definition 1, $f$ is continuous at $\frac{\pi}{2}$ exactly when

   $$\lim_{x\to \frac{\pi}{2}^-} f(x) = \lim_{x\to \frac{\pi}{2}^+} f(x) = f\!\left(\tfrac{\pi}{2}\right),$$

   that is, $\frac{k}{2} = \frac{k}{2} = 3$⁠. The two limits already agree (Steps 3 and 4), so the only condition is

   $$\frac{k}{2} = 3 \quad\Longleftrightarrow\quad k = 6.$$

   For any other $k$ the limit $\frac{k}{2}$ differs from 3 and $f$ is discontinuous at $\frac{\pi}{2}$⁠. So $k = 6$ is the only value.

   "Find the values of $k$" asks for all of them, so we also note that no other value works: the condition $\frac{k}{2} = 3$ is a linear equation with exactly one solution. The picture shows what goes wrong otherwise: the limit is pulled to $\frac{k}{2}$ while the point stays at 3.
   Continuity at $c$⁠: left-hand limit = right-hand limit = $f(c)$ (Basic 1). Lesson: continuity at a point

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