The squeeze theorem, and why sin x / x tends to 1

The Squeeze Theorem

Suppose that for every $x$ near $a$, except possibly at $a$ itself,

and that $g(x)$ and $h(x)$ both approach one number $L$ as $x\to a$. Then $f(x)$ approaches $L$ too: ${lim}_{x\to a}f(x)=L$.

The picture: $f$ is caught between two functions that close in on the same height, so it has nowhere else to go. Both hypotheses matter. The inequality only has to hold near $a$, but it has to hold on both sides of $a$. And the two bounds must have the same limit; bounds that stay apart trap nothing.

Warm-up: $x^2\sin \left(\frac{1}{x}\right)$ as $x\to 0$

The limit laws cannot do this one. As $x\to 0$, the input $\frac{1}{x}$ runs off without bound, so $\sin \left(\frac{1}{x}\right)$ passes through every value from $-1$ to $1$ again and again on each side of $0$ and never settles. With no limit for that factor, the limit of the product cannot be split into a product of limits. So stop asking where the sine is going, and ask only how far it can go.

1. Bound the bad factor. The sine of any number lies between $-1$ and $1$, so $-1\le \sin \left(\frac{1}{x}\right)\le 1$ whenever $x\ne 0$.
2. Multiply by the good factor. Multiplying all three parts by $x^2$ leaves their order alone, since $x^2>0$ whenever $x\ne 0$: $$-x^2\le x^2\sin \left(\frac{1}{x}\right)\le x^2$$
3. Check the bounds. $-x^2\to 0$ and $x^2\to 0$. Same limit, so the Squeeze Theorem gives $$\underset{x\to 0}{lim}{x}^2\sin \left(\frac{1}{x}\right)=0$$

Notice what made it work: one factor that can never leave $[-1,1]$, multiplied by one that shrinks to $0$. The same three lines handle any product built that way. Without the $x^2$, the bounds would be $-1$ and $1$, which never meet, and the theorem rightly says nothing.

Why $\frac{\sin x}{x}\to 1$: three areas on the unit circle

Here the limit laws fail differently: the top and the bottom both go to $0$, and $\sin x$ is not a product, so there is nothing to factor or cancel. The bounds come from geometry instead. Take $0<x<\frac{\pi }{2}$, measured in radians.

1. Compare three areas. Triangle $OAP$ has base $OA=1$ and height $\sin x$. The sector $OAP$ has radius $1$ and angle $x$, so its area is $\frac{1}{2}·1^2·x$. Triangle $OAT$ has base $1$ and height $\tan x$. Each region sits inside the next, so $$\frac{1}{2}\sin x<\frac{1}{2}x<\frac{1}{2}\tan x$$
2. Divide by $\frac{1}{2}\sin x$, which is positive here, so the inequalities keep their direction: $$1<\frac{x}{\sin x}<\frac{1}{\cos x}$$
3. Take reciprocals. All three quantities are positive, and taking reciprocals of positive numbers reverses the inequalities: $$\cos x<\frac{\sin x}{x}<1$$
4. Cover negative $x$. Replacing $x$ by $-x$ changes neither $\cos x$, which is an even function, nor $\frac{\sin x}{x}$, whose top and bottom both change sign. If $-\frac{\pi }{2}<x<0$, then $-x$ lies in $(0,\frac{\pi }{2})$, where step 3 holds, so the same inequality holds for $-\frac{\pi }{2}<x<0$ as well.
5. Squeeze. Now let $x\to 0$ from either side. The ceiling never moves from $1$, and the floor $\cos x$ rises to $\cos 0=1$, because the cosine is continuous at $0$. Trapped between a floor and a ceiling that meet at $1$, the quotient has nowhere else to go: $$\underset{x\to 0}{lim}\frac{\sin x}{x}=1$$

The radians are doing real work in step 1: a sector of angle $x$ has area $\frac{x}{2}$ only when $x$ is measured in radians. In degrees the same quotient tends to a different number, and question 4 below asks which.

What it is for: the derivative of $\sin x$

The derivative of $\sin x$ is the limit of a difference quotient. The addition formula $\sin (x+h)=\sin x\cos h+\cos x\sin h$ splits it into two pieces:

As $h\to 0$, the second fraction tends to $1$ by the limit just proved, and the first tends to $0$, which question 7 derives from the same limit. So the derivative of $\sin x$ is $\cos x$. The derivative of $\cos x$ comes the same way, and the other trigonometric derivatives are built from those two, so every one of them rests on $\frac{\sin x}{x}\to 1$.

Check your understanding

No calculator needed.

1. Find $\underset{x\to 0}{lim}\frac{\sin (3x)}{x}$.
2. Find $\underset{x\to \infty }{lim}\frac{\sin x}{x}$.
3. True or false: if $g(x)\le f(x)\le h(x)$ for every $x$ near $a$, with ${lim}_{x\to a}g(x)=1$ and ${lim}_{x\to a}h(x)=3$, then ${lim}_{x\to a}f(x)$ exists and lies between $1$ and $3$. Give a reason.
4. If $x$ were measured in degrees, what would $\frac{\sin (x^{\circ })}{x}$ tend to as $x\to 0$?
5. Find the error. “Since $-1\le \sin \left(\frac{1}{x}\right)\le 1$, multiplying by $x$ gives $-x\le x\sin \left(\frac{1}{x}\right)\le x$. Both bounds tend to $0$, so ${lim}_{x\to 0}x\sin \left(\frac{1}{x}\right)=0$.”
6. Find $\underset{x\to 0}{lim}\frac{\sin (x^2)}{x}$.
7. Use $\frac{\sin x}{x}\to 1$ to show that $\frac{1-\cos x}{x}\to 0$ as $x\to 0$. (Hint: multiply top and bottom by $1+\cos x$.)

Practise it

The free CEGEP Calculus I worksheet on limits with trig, exponential and log functions has Squeeze Theorem limits to do, alongside trigonometric limits settled by an identity. Once the derivatives arrive, the derivatives of transcendental functions worksheet puts the derivative of $\sin x$ to work. For every squeeze, write the inequality, say why it holds, and check that both bounds have the same limit.

Free worksheets for this

• CEGEP Calculus I: Limits with Trig, Exponential and Log Functions worksheet
• CEGEP Calculus I: Derivatives of Transcendental Functions worksheet
• AP Calculus AB Unit 1: Limits and Continuity worksheet