Sandwich Theorem

Sandwich Theorem, also known as the Squeeze Theorem, is a fundamental concept in calculus used to find the limit of a function. The theorem works by "sandwiching" a given function between two other functions whose limits are easier to determine. If the two bounding functions converge to the same limit at a certain point, then the sandwiched function also converges to that same limit.

Statement of Sandwich Theorem

Sandwich Theorem states that,

Let functions f(x), g(x), and h(x) be real values functions such that h(x) ≤ f(x) ≤ g(x). If \lim\limits_{x\rightarrow a}h(x)=\lim\limits_{x\rightarrow a}g(x)=L,     then  \lim\limits_{x\rightarrow a}f(x)=L    .

In the above situation, the function f(x) lies between h(x) and g(x), and hence h(x) and g(x) are called the lower bound and upper bound of f(x).

We can represent this condition easily using the graphs of the three functions h(x) ≤ f(x) ≤ g(x). The image added below shows the condition of the Sandwich Theorem.

Is Sandwich Theorem Results always in Zero?

The sandwich theorem states that for two functions f(x) and g(x), if both of them approach the same limit say L, when x approaches a certain point, and there is a third function, h(x), such that f(x) ≤ h(x) ≤ g(x) for all x in some interval containing the point, then h(x) also approaches L as x approaches that point.

No the value of the sandwiched function is not always zero the sandwich function can have any of the values that the other function approaches not necessarily the value of the function that is used in the sandwich theorem can be any value.

Example: If given g(x) =  (x³ + 2)/(x) and  3x < (x³ + 2)/(x-2) < 4x, where x = 0

Now the value of g(x) at x = 0 does not exist. But according to the Sandwich theorem,

3x  = 4x = 0 (at x = 0)
Then, g(x) = 0 is also 0 at x = 0.

Sandwich Theorem Proof

To prove the sandwich theorem we can use the epsilon-delta definition of limit, which is called algebraic proof. The following heading shows detailed proof of the sandwich theorem using limits.

Algebraic Proof Using Definition of Limit

Sandwich theorem can be easily proved using the definition of limit. Assume three real-valued functions g(x), f(x), and h(x) such that g(x) ≤ f(x) ≤ h(x) and lim_x→a g(x) = lim_x→a h(x) = L. 

Then by the definition of limits,

lim_x→a g(x) = L signifies ∀ ∈ > 0, ∃ δ₁ > 0 such that |x - a| < δ₁ ⇒ |g(x) - L| < ∈

|x - a| < δ₁ ⇒ -∈ < g(x) - L < ∈   ... (i)

lim_x→a h(x) = L signifies ∀ ∈ > 0, ∃ δ₂ > 0 such that |x - a| < δ₂ ⇒ |h(x) - L| < ∈

|x - a| < δ₂ ⇒ -∈ < h(x) - L < ∈    ... (ii)

Given, g(x) ≤ f(x) ≤ h(x)

Subtracting L from each side of the inequality

g(x) - L ≤ f(x) - L ≤ h(x) - L

Taking δ = minimum {δ₁, δ₂}, Nowr |x - a| < δ,

-∈ < g(x) - L ≤ f(x) - L ≤ h(x) - L < ∈                 [using (i) and (ii)]

-∈ < f(x) - L < ∈

lim_x→a f(x) = L

Thus, this proved the Sandwich Theorem.

Geometric Proof of cos x < sin x/x < 1

We can understand the geometric proof of the statement cos x < sin x/x < 1, as follows

To Prove:  cos x < sin x/x < 1, for 0 < |x| < π/2

Also, using trigonometric identites

• sin (– x) = – sin x
• cos( – x) = cos x

We are now proving the inequality in eq(i)

Draw a unit circle with centre O

Now if ∠COA is x radians and 0 < x < π/2

For the figure,

Area of ΔOAC < Area of sector OAC < Area of Δ OAB

1/2×OA×CD < x/2π × π(OA)² < 1/2×OA×AB

Cancelling OA from each side

CD < x.OA < AB...(i)

In ΔOCD

sin x = CD/OC = CD/OA

CD = OA sin x...(ii)

Similarly, In ΔOAB

tan x = AB/OA

AB = OA tan x...(iii)

Now, from eq (i), (ii) and (iii) we get

sin x < x < tan x  [given 0<x<π/2]

Dividing sin x from each side

sin x/ sin x < x/(sin x) < tan x/ (sin x)

1 < x/(sin x) < 1/ (cos x)

Taking reciprocal,

cos x < (sin x)/(x) < 1

Thus, Proved.

Sandwich Theorem Limits

Some important limits which can be proved using Sandwich theorems are,

• lim_x→a (sin x / x) = 1
• lim_x→a (1-cos x / x) = 0

The proof of these limits are discussed below,

Proof of  lim_x→a (sin x / x) = 1

From the above inequality, we proved that

\cos x\ <\ \dfrac{\sin x}{x}\ <\ 1

From this equation, we understand that (sin x/x) always lies between cos x and 1. So (sin x/x) is sandwiched between 1 and cos x. We know that

\text{If} \lim\limits_{x\rightarrow a}f(x)=\lim\limits_{x\rightarrow a}h(x)=L \text{, Then } \lim\limits_{x\rightarrow a}g(x)=L

\lim\limits_{x\rightarrow0}\cos\ x=1\ \text{ and }\ \lim\limits_{x\rightarrow0}1=1 \\ \Rightarrow \lim\limits_{x\rightarrow0}\dfrac{\sin\ x}{x}=1

Proof of lim_x→a (1 - cos x)/x = 0

To prove this limit we use the trigonometric identity

1-\cos x=2\sin^2(\dfrac{x}{2})\\\qquad\\ \lim\limits_{x\rightarrow0}\dfrac{1-\cos x}{x}\\=\lim\limits_{x\rightarrow0}\dfrac{2\sin^2\frac{x}{2}}{x}\\=\lim\limits_{x\rightarrow0}\dfrac{\sin \frac{x}{2}}{\frac{x}{2}}.\sin\frac{x}{2}\\\qquad\\=\lim\limits_{x\rightarrow0}\dfrac{sin\frac{x}{2}}{\frac{x}{2}}\lim\limits_{x\rightarrow0}\sin \dfrac{x}{2}\ \\= 1.0\\ =0\\\qquad\\ \implies 1-\cos x=2\sin^2(\dfrac{x}{2})\ =\ 0

Sandwich Theorem Examples

Example 1: Find the value of lim_{x→ 0} f(x) if 9 - x² ≤ f(x) ≤ 9 + x² 

Solution:

Now,

lim_{x→ 0} 9 - x² 
= 9 - 0² 
= 9

lim_{x→ 0} 9 + x² 
= 9 + 0² 
= 9

Now, if lim_{x→ 0} 9 - x² = lim_{x→ 0} 9 + x² = 9

Then using Sandwich Theorem we can say that,

lim_{x→ 0} 9 - x² = lim_{x→ 0} f(x) = lim_{x→ 0} 9 + x²

Thus the limit of lim_{x→ 0} f(x) = 9

Example 2: Find the limit lim_{x→ 0}  x² sin (1/x).

Solution:

As we know, the range of the sin function is [-1, 1] then,

-1 ≤ sin x ≤ 1
⇒ -1 ≤ sin (1/x) ≤ 1             (x ≠ 0)

Multiplying x² on each side

-x² ≤ x²sin (1/x) ≤ x²

Now, lim_{x→ 0} -x² = -0² = 0

and lim_{x→ 0} x² = -0² = 0

Now, if lim_{x→ 0} - x² = lim_{x→ 0} x² = 0

Then using Sandwich Theorem we can say that,

lim_{x→ 0} - x² = lim_{x→ 0} x² sin (1/x) = lim_{x→ 0} x²

Thus the limit of lim_{x→ 0}  x² sin (1/x) = 0