# Use of Squeezing Theorem to Find Limits

The squeezing theorem, also called the sandwich theorem, is used to find limits.

## Squeezing Theorem.

If
are functions such that

for all values of x in some open interval containing a and if
then,

## Examples with Solutions

### Example 1:

Find the limit:__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

Multiply all terms of the above inequality by x

^{2}(x not equal to 0)

\( \) \( \)\( \)\( \)

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

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

### 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 in standard position. A is a point on the unit circle and AB is tangent to the circle at point A hence a right angle at this point.

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

Use the sine formula of the area of a triangle to obtain:

area of triangle \( OAC = (1/2) \sin x \cdot OA \cdot OC = (1/2) \sin x \cdot 1 \cdot 1 = (1/2) \sin x \)

Use the formula of the area of a sector to obtain:

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

area of right triangle \( OAB = (1/2) \cdot (base) \cdot (height) = (1/2) \cdot (1) \cdot (\tan x) = (1/2) \tan x \)

Comparing, geometrically, the three areas, we can write the inequality
\[ \text{area of triangle} \; OAC \; \lt \; \text{area of sector} \; OAC \; \lt \; \text{area of triangle} \; OAB \]
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
\[ 1 \lt \dfrac{x}{\sin x} \lt 1 / \cos x \]
Take the reciprocal and reverse the two inequality symbols in the double inequality
\[ 1 \gt \dfrac{\sin x} {x} \gt \cos x \]
Which is equivalent to
\[ \cos x \lt \dfrac{\sin x} {x} \lt 1 \]
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 \( \dfrac{\sin x} {x} \) is undefined. Since
\[ \lim_{x \to 0} \cos x = 1 \]
and
\[ \lim_{x \to 0} 1 = 1 \] ,
we can apply the squeezing theorem to obtain
\[ \lim_{x \to 0} \dfrac{\sin x}{x} = 1 \]
This result is very important and will be used to find other limits of trigonometric functions and derivatives

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