# Use of Squeezing Theorem to Find Limits

## Squeezing Theorem.

If $f$, $g$ and $h$ are functions such that

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

for all values of $x$ in some open interval containing $a$ and if $\lim_{\to a} f(x) = \lim_{\to a} h(x) = L$ then

$\lim_{\to a} g(x) = L$

How to use the squeezing theorem?

Example 1: Find the limit $\lim_{\to 0} x^2 \cos(\dfrac{1}{x})$

Solution to Example 1:

As $x$ approaches $0$, $1 / x$ becomes very large in absolute value and $\cos(1 / x)$ becomes highly oscillatory. However $\cos(1 / x)$ takes values within the interval $[-1,1]$ which is the range of $\cos x$, hence

$-1 \le \cos (1/x) \le 1$

Multiply all terms of the above inequality by $x^2$ ($x$ not equal to $0$)

$- x^2 \le x^2 \cos (1/x) \le x^2$

The above inequality holds for any value of $x$ except $0$ where $x^2 \cos (1/x)$ in undefined. As $x$ approaches $0$ both $- x^2$ and $x^2$ approach $0$ and according to the squeezing theorem we obtain

$\lim_{\to 0} x^2 cos(1/x) = 0$

Example 2: Find the limit $\lim_{\to 0} \dfrac{\sin x}{x}$

Solution to Example 2:

Assume that $0 \lt x \lt \pi / 2$ and let us us consider the unit circle, shown below, and a sector $OAC$ with central angle $x$ where $x$ is in standard position. $A$ is a point on the unit circle and $AB$ is tangent to the circle at $A$.

Point $C$ is a point on the unit circle (radius = $1$)and has coordinates $(\cos x, \sin x)$. Let us find the areas of triangle $OAC$, sector $OAC$ and triangle

area of triangle $OAC = (1/2) \cdot (base) \cdot (height)$

$= (1/2) \cdot (1) \cdot$ ($y$ coordinate of point $C$)$= (1/2)(\sin x)$

Note: we have used base = radius = $1$

area of sector $OAC = (1/2)\cdot(x)\cdot(radius)^2$

$= (1/2) \cdot x \cdot (1)^2= (1/2) x$

area of triangle $OAB = (1/2) \cdot (base) \cdot (height)$

$= (1/2) \cdot (1) \cdot (\tan x) = (1/2) \tan x$

Comparing the three areas, we can write the inequality

area of triangle $OAC \lt$ area of sector $OAC \lt$ area of triangle $OAB$

Substitute the areas in the above inequality by their expressions obtained above.

$(1/2)(\sin x) \lt (1/2) x \lt (1/2) \tan x$

Multiply all terms by $\dfrac{2}{\sin x}$ gives

$1 \lt x / \sin x \lt 1 / \cos x$

Take the reciprocal and reverse the two inequality symbols in the double inequality

$1 \gt \sin x / x \gt \cos x$

Which is equivalent to

$\cos x \lt \sin x / x \lt 1$

It can be shown that the above inequality hols for $-\pi/ 2 \lt x \lt 0$ so the the above inequality hold for all $x$ except $x = 0$ where $\sin x / x$ is undefined. Since

$\lim_{\to 0} \cos x = 1$

and

$\lim_{\to 0} 1 = 1$ ,

we can apply the squeezing theorem to obtain

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

This result is very important and will be used to find other limits of trigonometric functions and derivatives

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Updated: 2 April 2013