Use of Squeezing Theorem to Find Limits

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

Squeezing Theorem.

If Functions f, g anf h are functions such that
Inequalities Between Functions f, g anf h
for all values of x in some open interval containing a and if Squeezing Therem Condition then,
Squeezing Therem

Examples with Solutions

Example 1:

Find the limit: Limit of x^2 1/cos 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
Prepare Inequality to Apply Squeezing Theorem
Multiply all terms of the above inequality by x2 (x not equal to 0)
Inequality to Apply Squeezing Theorem \( \) \( \)\( \)\( \)
The above inequality holds for any value of x except 0 where \( x^2 \cos (1/x) \) in undefined. As x approaches 0 both - x2 and x2 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.

Geometrical Computation of Limit of sin x/x
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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