Squeeze a Function Between Two Others and Its Limit Has Nowhere to Go

Substituting $x=0$ in $\frac{\sin x}{x}$ gives $\frac{0}{0}$, but there is no common factor to cancel — $\sin x$ is not a polynomial.

A calculator in radian mode hints at the answer: $\frac{\sin 0.1}{0.1}=0.9983$ and $\frac{\sin 0.01}{0.01}$ is closer still to $1$.

To prove it, we need a new tool — the Sandwich Theorem — and a little geometry.

This part covers that theorem, the two basic trigonometric limits, composite trigonometric limits, and exponential and logarithmic limits.

**If $f(x)\le g(x)\le h(x)$ near $a$ and both $f$ and $h$ tend to the same limit $l$, then $g$ is forced to tend to $l$ as well; trapping $\frac{\sin x}{x}$ between $\cos x$ and $1$ shows its limit is $1$.

The geometry.** In a circle of radius $1$ with angle $x$ radians at the centre, $0<x<\frac{\pi }{2}$:

$$\text{area of small triangle}<\text{area of sector}<\text{area of large triangle}$$

$$\frac{1}{2}\sin x<\frac{1}{2}x<\frac{1}{2}\tan x$$

Dividing by $\frac{1}{2}\sin x$ and taking reciprocals:

$$\cos x<\frac{\sin x}{x}<1$$

As $x\to 0$, $\cos x\to 1$, so **${\lim }_{x\to 0}\frac{\sin x}{x}=1$.** Both sides are even functions, so negative $x$ behaves the same way.

A second use. Since $-x^2\le x^2\sin \frac{1}{x}\le x^2$, and both bounds tend to $0$:

$$\underset{x\to 0}{\lim }{x}^2\sin \frac{1}{x}=0$$

An everyday example. If two friends walk on either side of you down a narrow lane and both reach the chai stall, you have no choice but to reach the stall too.

The substance. The angle must be in radians — the sector area $\frac{1}{2}x$ uses radian measure.

**Writing $1-\cos x=2{\sin }^2\frac{x}{2}$ turns the expression into $\sin \frac{x}{2}$ times a $\frac{\sin \theta }{\theta }$ form, which is $0\times 1=0$.**

$$\frac{1-\cos x}{x}=\frac{2{\sin }^2(x/2)}{x}=\sin \frac{x}{2}\cdot \frac{\sin (x/2)}{x/2}$$

As $x\to 0$, $\sin \frac{x}{2}\to 0$ and the second factor $\to 1$, so

$$\underset{x\to 0}{\lim }\frac{1-\cos x}{x}=0$$

**Worked example — dividing by $x^2$ instead.**

$$\underset{x\to 0}{\lim }\frac{1-\cos 4x}{x^2}=\underset{x\to 0}{\lim }\frac{2{\sin }^{2}2x}{x^2}=2\times 4\times {\left(\frac{\sin 2x}{2x}\right)}^2=8$$

In general, ${\lim }_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2}$.

An everyday example. Tilt a mobile phone slightly from flat: its height above the table changes far less than the tilt angle, which is why $1-\cos x$ is tiny compared with $x$.

The substance. The denominator decides the answer — dividing by $x$ gives $0$, but dividing by $x^2$ gives $\frac{1}{2}$.

**Rewrite the expression so that every sine or tangent is divided by exactly its own angle, using ${\lim }_{\theta \to 0}\frac{\sin \theta }{\theta }={\lim }_{\theta \to 0}\frac{\tan \theta }{\theta }=1$, and shift the variable when $x$ does not tend to $0$.

Worked example 1.**

$$\underset{x\to 0}{\lim }\frac{\sin 4x}{\sin 2x}=\underset{x\to 0}{\lim }\frac{\frac{\sin 4x}{4x}\cdot 4x}{\frac{\sin 2x}{2x}\cdot 2x}=\frac{4}{2}=2$$

Worked example 2. Since $\frac{\tan x}{x}=\frac{\sin x}{x}\cdot \frac{1}{\cos x}$, it tends to $1$, and

$$\underset{x\to 0}{\lim }\frac{\tan 3x}{5x}=\frac{3}{5}$$

Worked example 3 — shifting. For ${\lim }_{x\to \pi /2}\frac{\tan 2x}{x-\frac{\pi }{2}}$, put $y=x-\frac{\pi }{2}$, so $y\to 0$ and $\tan 2x=\tan (2y+\pi )=\tan 2y$:

$$\underset{y\to 0}{\lim }\frac{\tan 2y}{y}=2$$

Worked example 4.

$$\underset{x\to 0}{\lim }\frac{\cos 2x-1}{\cos x-1}=\underset{x\to 0}{\lim }\frac{2{\sin }^{2}x}{2{\sin }^2(x/2)}=\frac{x^2}{x^2/4}=4$$

An everyday example. A railway engineer marking a gentle curve treats a very short arc and its chord as equal in length, which is this limit at work.

The substance. Only the matching pair tends to 1 — $\frac{\sin 4x}{x}$ tends to $4$, not $1$.

**From the series $e^x=1+x+\frac{x^2}{2!}+\dots$, the quotient $\frac{e^x-1}{x}=1+\frac{x}{2!}+\dots$ tends to $1$; putting $y={\log }_e(1+x)$ turns $\frac{{\log }_e(1+x)}{x}$ into $\frac{y}{e^y-1}$, which also tends to $1$.

A numerical check.** $\frac{e^{0.01}-1}{0.01}\approx 1.005$.

Worked example 1.

$$\underset{x\to 0}{\lim }\frac{e^{3x}-1}{x}=3\times \underset{x\to 0}{\lim }\frac{e^{3x}-1}{3x}=3$$

Worked example 2 — split into two.

$$\underset{x\to 0}{\lim }\frac{e^{2x}-e^x}{x}=\underset{x\to 0}{\lim }\frac{e^{2x}-1}{x}-\underset{x\to 0}{\lim }\frac{e^x-1}{x}=2-1=1$$

Worked example 3.

$$\underset{x\to 0}{\lim }\frac{{\log }_e(1+4x)}{x}=4$$

Worked example 4 — any base. Since $a^x=e^{x{\log }_{e}a}$,

$$\underset{x\to 0}{\lim }\frac{a^x-1}{x}={\log }_{e}a\underset{x\to 0}{\lim }\frac{2^x-1}{x}={\log }_{e}2\approx 0.693$$

An everyday example. **Money growing continuously at a small rate $r$** becomes $e^r$ times itself in a year; for $r=0.05$, $e^{0.05}-1\approx 0.0513$, very close to $0.05$.

The substance. These results use natural logarithms — with base $10$, $\frac{{\log }_{10}(1+x)}{x}$ tends to ${\log }_{10}e$, not $1$.