The Squeeze Theorem

Some limits cannot be found with the algebra methods from computing limits. For example, look at

$\displaystyle\lim_{x\to 0}x^2\sin\left(\dfrac{1}{x}\right)$.

As $x\to 0$, the factor $\sin\left(\dfrac{1}{x}\right)$ keeps oscillating and has no limit. So we cannot use the product law. The Squeeze Theorem solves problems like this one. It is also called the Sandwich Theorem.

The theorem

The idea is simple. Suppose a function is trapped between two other functions. If the two outer functions approach the same number, the function in the middle must approach that number too. It has nowhere else to go.

The Squeeze Theorem: Suppose that

$g(x)\le f(x)\le h(x)$

for all $x$ near $a$ (except possibly at $a$ itself), and suppose that

$\displaystyle\lim_{x\to a}g(x)=L$ and $\displaystyle\lim_{x\to a}h(x)=L$.

Then

$\displaystyle\lim_{x\to a}f(x)=L$.

Here the symbol $\le$ means "is less than or equal to." The theorem also works for one-sided limits, and as $x\to\infty$ or $x\to -\infty$.

To use the theorem, follow these steps:

1. Find a lower function $g(x)$ and an upper function $h(x)$, with $g(x)\le f(x)\le h(x)$.
2. Check that $g(x)$ and $h(x)$ have the same limit $L$.
3. Conclude that the limit of $f(x)$ is also $L$.

Most often, the bounds come from the sine and cosine. For every real number $\theta$,

$-1\le \sin\theta\le 1$ and $-1\le \cos\theta\le 1$.

Examples

The limit of sin x / x as x approaches 0

In introduction to limits, a table suggested that

$\displaystyle\lim_{x\to 0}\dfrac{\sin x}{x}=1$.

Now we can prove it with the Squeeze Theorem. This limit is used to find the derivatives of the trigonometric functions. In this proof, the angle $x$ is in radians.

Step 1: Compare three areas

Take $x$ between $0$ and $\dfrac{\pi}{2}$. Draw a circle with radius $1$ and center $O$. Let $A$ be the point $(1,0)$, and let $P$ be the point on the circle at angle $x$. Draw a vertical line through $A$. It meets the line $OP$ (extended) at the point $T$.

The three shaded regions fit inside each other. So their areas are in order: (a) is less than (b), and (b) is less than (c).

• Triangle $OAP$: The base $OA$ is $1$. The height is the $y$-coordinate of $P$, which is $\sin x$. So the area is $\dfrac{1}{2}\sin x$.
• Sector $OAP$: A sector is the part of a circle between two radii, like a slice of pie. With radius $1$ and angle $x$ (in radians), its area is $\dfrac{1}{2}x$.
• Triangle $OAT$: The base $OA$ is $1$. The height $AT$ is $\tan x$. So the area is $\dfrac{1}{2}\tan x$.

So,

$\dfrac{1}{2}\sin x<\dfrac{1}{2}x<\dfrac{1}{2}\tan x$.

Multiply all three parts by $2$:

$\sin x<x<\tan x$.

Step 2: Rearrange

Divide all three parts by $\sin x$. It is positive, so the inequality signs do not change. Also, $\dfrac{\tan x}{\sin x}=\dfrac{1}{\cos x}$:

$1<\dfrac{x}{\sin x}<\dfrac{1}{\cos x}$.

Now take the reciprocal (flip each fraction). For positive numbers, taking reciprocals reverses the inequality signs:

$\cos x<\dfrac{\sin x}{x}<1$.

Step 3: Squeeze

As $x\to 0^+$, the lower function $\cos x$ approaches $\cos 0=1$. The upper function is $1$. Both approach $1$. So, by the Squeeze Theorem,

$\displaystyle\lim_{x\to 0^+}\dfrac{\sin x}{x}=1$.

For $x<0$, use $\dfrac{\sin(-x)}{-x}=\dfrac{\sin x}{x}$. The function has the same values on both sides of $0$, so the left-hand limit is also $1$. Therefore,

$\displaystyle\lim_{x\to 0}\dfrac{\sin x}{x}=1$.

Remember: This is true only when $x$ is in radians. In degrees, the limit would be $\dfrac{\pi}{180}$.

Using the limit of sin x / x

Many other limits can be found from this one.

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