Use of Squeezing Theorem to Find Limits

The squeezing theorem, also called the sandwich theorem, is used 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

\[ \lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L \]

then,

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

Examples with Solutions

Example 1:

Find the limit:

\[ \lim_{x \to 0} x^2 \cos\left(\frac{1}{x}\right) \]
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,

\[ -1 \le \cos\left(\frac{1}{x}\right) \le 1 \]

Multiply all terms of the above inequality by \( x^2 \) (where \( x \ne 0 \)):

\[ - x^2 \le x^2 \cos\left(\frac{1}{x}\right) \le x^2 \] The above inequality holds for any value of x except 0 where \( x^2 \cos (1/x) \) in undefined. As x approaches 0 both - 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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