What a limit is
Tonight's idea
Put \(x = 0\) into \(\dfrac{\sin x}{x}\) and you get \(\dfrac{0}{0}\), which means nothing. But \(x = 0.1\) gives 0.99833 and \(x = 0.01\) gives 0.99998. No value at 0, yet a clear target. We write \(\lim_{x \to 0} \dfrac{\sin x}{x} = 1\) (read: the limit as \(x\) tends to 0). A limit asks where the outputs settle as the inputs crowd in, and never looks at the point itself.
Why 1? Take an angle \(x\) in radians, \(0 \lt x \lt \pi/2\), in a unit circle:
Triangle \(OAP\) has base 1 and height \(\sin x\); triangle \(OAT\) has height \(\tan x\). The sector is the fraction \(x/2\pi\) of a circle of area \(\pi\), so its area is \(x/2\). Multiply the caption's inequality by \(2/\sin x\) and flip:
\(\cos x \;\le\; \dfrac{\sin x}{x} \;\le\; 1\) for \(0 \lt x \lt \pi/2\), and also for negative \(x\), since both sides are even.
As \(x \to 0\), \(\cos x \to 1\). Pinned between two things that both go to 1, \(\sin x / x\) has no choice. That is the squeeze.
The \(0/0\) was a warning, not an answer: \(\dfrac{x^2}{x}\) and \(\dfrac{x}{x^2}\) also give \(0/0\) at zero, with limits 0 and nothing at all. Look at how the top and bottom shrink, by simplifying or by squeezing.
Watch one: \(\displaystyle\lim_{x \to 0} \frac{\sin 5x}{\sin 2x}\)
Again \(0/0\). Build the limit we know out of the pieces we have. Make each piece look like \(\sin(\text{thing})/\text{thing}\):
\[\frac{\sin 5x}{\sin 2x} = \frac{5}{2} \cdot \frac{\sin 5x}{5x} \cdot \frac{2x}{\sin 2x} \;\to\; \frac52 \cdot 1 \cdot 1 = \frac52.\]
As \(x \to 0\), so do \(5x\) and \(2x\), so both fractions tend to 1. The limit is also a derivative in disguise: \(\dfrac{\sin x}{x} = \dfrac{\sin x - \sin 0}{x - 0}\) is a secant slope, so \(\sin x / x \to 1\) says the slope of \(\sin x\) at 0 is exactly 1. Sheet 4 uses that.
Watch a secant slope settle onto a tangent slope as \(h \to 0\): mathdetours.com/demos/tangent.html.
Your turn
- Simplify first, then take the limit: (a) \(\lim_{x \to 2} \frac{x^2 - 4}{x - 2}\); (b) \(\lim_{h \to 0} \frac{(3 + h)^2 - 9}{h}\). What is (b) the slope of?
- Use \(\sin x / x \to 1\) to find (a) \(\lim_{x \to 0} \frac{\sin 3x}{x}\) and (b) \(\lim_{x \to 0} \frac{\tan x}{x}\). For (b), write \(\tan x\) as \(\sin x / \cos x\).
- Find \(\lim_{x \to 0} \frac{1 - \cos x}{x^2}\). Multiply top and bottom by \(1 + \cos x\), then use \(\sin^2 x + \cos^2 x = 1\). Deduce the value of \(\lim_{x \to 0} \frac{1 - \cos x}{x}\).
- (a) Explain why \(\lim_{x \to 0} \sin(1/x)\) does not exist. Consider \(x = 1/(2\pi n)\) and \(x = 1/(2\pi n + \pi/2)\) for large \(n\). (b) Show that \(\lim_{x \to 0} x \sin(1/x)\) does exist, by squeezing it between two things you can see.
Stretch: the polygon that becomes a circle
A regular \(n\)-gon is drawn inside a unit circle. Cut it into \(n\) triangles from the centre and show its area is \(\dfrac{n}{2}\sin\dfrac{2\pi}{n}\). Then use tonight's limit to find what the area tends to as \(n \to \infty\). Is that a new proof that the circle's area is \(\pi\)?
Answers
- (a) \(\dfrac{(x-2)(x+2)}{x-2} = x + 2 \to 4\). (b) \(\dfrac{6h + h^2}{h} = 6 + h \to 6\), the slope of \(y = x^2\) at \(x = 3\).
- (a) \(\dfrac{\sin 3x}{x} = 3 \cdot \dfrac{\sin 3x}{3x} \to 3\). (b) \(\dfrac{\tan x}{x} = \dfrac{\sin x}{x} \cdot \dfrac{1}{\cos x} \to 1 \cdot 1 = 1\).
- \(\dfrac{(1 - \cos x)(1 + \cos x)}{x^2 (1 + \cos x)} = \dfrac{\sin^2 x}{x^2} \cdot \dfrac{1}{1 + \cos x} \to 1 \cdot \dfrac12 = \dfrac12\). Then \(\dfrac{1 - \cos x}{x} = x \cdot \dfrac{1 - \cos x}{x^2} \to 0 \cdot \dfrac12 = 0\).
- (a) At \(x = 1/(2\pi n)\) the value is \(\sin 2\pi n = 0\); at \(x = 1/(2\pi n + \pi/2)\) it is 1. Both sequences of inputs tend to 0, but the outputs never settle: the function oscillates faster and faster, so there is no single target. (b) Since \(-1 \le \sin(1/x) \le 1\), we have \(-|x| \le x \sin(1/x) \le |x|\). Both \(-|x|\) and \(|x|\) tend to 0, so the squeeze gives a limit of 0. The wild factor is tamed by the small one.
Stretch. Each of the \(n\) triangles has two sides of length 1 meeting at angle \(2\pi/n\), so its area is \(\tfrac12 \sin(2\pi/n)\); the polygon has area \(\tfrac{n}{2}\sin\tfrac{2\pi}{n}\). Write \(t = 2\pi/n\), which tends to 0 as \(n\) grows. Then \(\tfrac{n}{2}\sin\tfrac{2\pi}{n} = \pi \cdot \dfrac{\sin t}{t} \to \pi\). Not a new proof: the squeeze used the circle's area to get the sector's \(x/2\). It is a consistency check. For \(n = 1000\) the polygon's area is 3.14157.