The squeezing theorem, also called the sandwich theorem, is used to find limits.
In what follows, \( C \) represents the constant of integration where applicable.
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 \( a \) (except possibly at \( a \)), and if:
\[ \lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L \]then:
\[ \lim_{x \to a} g(x) = L \]Examples with Solutions
Click on each example to view its detailed step-by-step solution.
Example 1
Find the limit:
\[ \lim_{x \to 0} x^2 \cos\left(\dfrac{1}{x}\right) \]Show Solution to Example 1
As \( x \) approaches \( 0 \), \( \dfrac{1}{x} \) becomes very large in absolute value, and \( \cos\left(\dfrac{1}{x}\right) \) becomes highly oscillatory.
However, \( \cos\left(\dfrac{1}{x}\right) \) takes values within the interval \([-1, 1]\), which is the range of the cosine function. Hence:
\[ -1 \le \cos\left(\dfrac{1}{x}\right) \le 1 \]Multiply all terms of the above inequality by \( x^2 \) (noting that \( x^2 > 0 \) for \( x \neq 0 \)):
\[ -x^2 \le x^2 \cos\left(\dfrac{1}{x}\right) \le x^2 \]The above inequality holds for any value of \( x \) except \( 0 \), where \( x^2 \cos\left(\dfrac{1}{x}\right) \) is undefined. As \( x \) approaches \( 0 \), both \( -x^2 \) and \( x^2 \) approach \( 0 \):
\[ \lim_{x \to 0} (-x^2) = 0 \quad \text{and} \quad \lim_{x \to 0} x^2 = 0 \]Therefore, according to the Squeezing Theorem, we obtain:
\[ \lim_{x \to 0} x^2 \cos\left(\dfrac{1}{x}\right) = 0 \]Example 2
Find the limit:
\[ \lim_{x \to 0} \dfrac{\sin x}{x} \]Show Solution to Example 2
Assume that \( 0 < x < \pi/2 \) and consider the unit circle 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 \( A \), forming a right angle at this point.
Point \( C \) is a point on the unit circle (radius equal to 1) with coordinates \( (\cos x, \sin x) \). Let us find the areas of triangle \( OAC \), sector \( OAC \), and right triangle \( OAB \).
Using the sine formula for the area of a triangle:
\( \text{Area of triangle } OAC = \dfrac{1}{2} \cdot OA \cdot OC \cdot \sin x = \dfrac{1}{2} \cdot 1 \cdot 1 \cdot \sin x = \dfrac{1}{2} \sin x \)
Using the formula for the area of a circular sector:
\( \text{Area of sector } OAC = \dfrac{1}{2} \cdot x \cdot OC^2 = \dfrac{1}{2} x \)
Area of right triangle \( OAB \):
\( \text{Area of right triangle } OAB = \dfrac{1}{2} \cdot (\text{base}) \cdot (\text{height}) = \dfrac{1}{2} \cdot 1 \cdot \tan x = \dfrac{1}{2} \tan x \)
Comparing geometrically the three areas, we can write the inequality:
\[ \text{Area of triangle } OAC < \text{Area of sector } OAC < \text{Area of triangle } OAB \]Substitute the areas with their expressions:
\[ \dfrac{1}{2} \sin x < \dfrac{1}{2} x < \dfrac{1}{2} \tan x \]Multiply all terms by \( \dfrac{2}{\sin x} \) (since \( \sin x > 0 \) for \( 0 < x < \pi/2 \)):
\[ 1 < \dfrac{x}{\sin x} < \dfrac{1}{\cos x \]Take the reciprocal and reverse the inequality symbols:
\[ 1 > \dfrac{\sin x}{x} > \cos x \]Which is equivalent to:
\[ \cos x < \dfrac{\sin x}{x} < 1 \]It can be shown that this inequality also holds for \( -\pi/2 < x < 0 \), so the inequality holds for all \( x \) in an open interval around \( 0 \) except \( x = 0 \) where \( \dfrac{\sin x}{x} \) is undefined. Since:
\[ \lim_{x \to 0} \cos x = 1 \quad \text{and} \quad \lim_{x \to 0} 1 = 1 \]We can apply the Squeezing Theorem to obtain:
\[ \lim_{x \to 0} \dfrac{\sin x}{x} = 1 \]This fundamental limit is essential for evaluating other limits of trigonometric functions and calculating their derivatives.