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Squeeze Theorem
The squeeze theorem is used to evaluate a kind of limits. This is also known as the sandwich theorem. To evaluate a limit lim ₓ → ₐ f(x), we usually substitute x = a into f(x) and if that leads to an indeterminate form, then we apply some algebraic methods. But neither of them may work in evaluating some kind of limits such as lim ₓ → ∞ (sin x / x). The squeeze theorem helps in this regard.
Let us see what does squeeze theorem states, how to prove it, and how to apply it along with various examples.
1.  What is Squeeze Theorem? 
2.  Squeeze Theorem Statement 
3.  Squeeze Theorem Proof 
4.  Important Limits Using Squeeze (Sandwich)Theorem 
5.  FAQs on Squeeze Theorem 
What is Squeeze Theorem?
The squeeze theorem (also known as sandwich theorem) states that if a function f(x) lies between two functions g(x) and h(x) and the limits of each of g(x) and h(x) at a particular point are equal (to L), then the limit of f(x) at that point is also equal to L. This looks something like what we know already in algebra. If a ≤ b ≤ c and a = c then b is also equal to c. The squeeze theorem says that this rule applies to limits as well.
Squeeze Theorem Statement
We define the squeeze theorem mathematically as follows:
"Let f(x), g(x), and h(x) are three functions that are defined over an interval I such that g(x) ≤ f(x) ≤ h(x) and suppose lim ₓ → ₐ g(x) = lim ₓ → ₐ h(x) = L, then lim ₓ → ₐ f(x) = L".
Here:
 The function f lies between g and h and hence they are lower and upper bounds of f respectively.
 'a' doesn't necessarily need to be within I.
This theorem looks so obvious given the following explanation. Since f(x) lies between g(x) and h(x), the curve of f(x) should lie between the curves of g(x) and h(x) graphically. Also, when both g(x) and h(x) tend to L as x → a, then f(x) cannot escape from having the same limit as L when x → a. This is because f(x) lies between the curves of g(x) and h(x). It can be understood from the following figure.
Squeeze Theorem Proof
We have already seen the proof of the squeeze theorem graphically. But we are going to prove it mathematically using the definition of limits. Assume that g(x) ≤ f(x) ≤ h(x) and lim ₓ → ₐ g(x) = lim ₓ → ₐ h(x) = L. By the definition of limits:
 lim ₓ → ₐ g(x) = L means that
∀ ∈ > 0, ∃ δ_{1} > 0 such that x  a < δ_{1} ⇒ g(x)  L < ∈
This gives x  a < δ_{1} ⇒ ∈ < g(x)  L < ∈... (1)  lim ₓ → ₐ h(x) = L means that
∀ ∈ > 0, ∃ δ_{2} > 0 such that x  a < δ_{2} ⇒ h(x)  L < ∈
This gives x  a < δ_{2} ⇒ ∈ < h(x)  L < ∈... (2)
Now, it is given that g(x) ≤ f(x) ≤ h(x). Subtracting L from each side,
g(x)  L ≤ f(x)  L ≤ h(x)  L
Choosing δ = minimum {δ_{1}, δ_{2}}, we have, whenever x  a < δ,
∈ < g(x)  L ≤ f(x)  L ≤ h(x)  L < ∈ (using (1) and (2))
∈ < f(x)  L < ∈
⇒ lim ₓ → ₐ f(x) = L.
Hence the squeeze theorem is proved.
Important Limits Using Squeeze (Sandwich) Theorem
There are two limits that occur most frequently while solving the problems: lim ₓ → ₀ (sin x / x) and lim ₓ → ₀ (1  cos x)/x. We can use the squeeze theorem to evaluate these two limits. After learning the process of evaluating these limits using the squeeze theorem, we can just memorize them so that we can use those values right away when solving other limits. We are going to prove:
 lim ₓ → ₀ (sin x / x) = 1
 lim ₓ → ₀ (1  cos x)/x = 0
To prove the first limit, we will prove an important relationship in trigonometry which says: cos x < (sin x)/x < 1. To prove this, consider a unit circle and let us say that 'x' is in the first quadrant (as sin(x) =  sin x and cos (x) = x, it is sufficient to prove the inequality in the first quadrant) such that ∠AOB = x (in radians). Draw a perpendicular from B to C. Draw a tangent AD at A (we know that tangent is perpendicular to radius). So both BC and AD are perpendicular to OA.
From the above figure,
Area of ΔOAB < Area of sector OAB < Area of ΔOAD
Using the formulas of the area of triangle and area of the sector:
(1/2) OA × BC < (1/2) (OA)^{2} × x < (1/2) OA × AD
Multiplying each side by 2,
OA × BC < (OA)^{2} × x < OA × AD
Since, it is a unit circle, its radius, OA = 1. So the above inequality becomes:
BC < x < AD ... (1)
From ΔOBC, sin x = BC/OB ⇒ BC = sin x (as OB = radius of unit circle = 1) ... (2)
From ΔOAD, tan x = AD/OA ⇒ AD = tan x (as OA = radius of unit circle = 1) ... (3)
Substituting the values of BC and AD from (2) and (3) in (1):
sin x < x < tan x
Since x lies in the first quadrant, sin x is positive. Thus, dividing throughout by sin x doesn't affect the signs of inequality.
(sin x)/(sin x) < x/(sin x) < (tan x)/(sin x)
1 < x/(sin x) < cos x (∵ (tan x)/(sin x) = (sin x/cos x)/(sin x/1) = cos x)
Taking reciprocals:
cos x < (sin x)/x < 1. Hence we proved the inequality. We will now use this to prove the first limit.
Proving lim ₓ → ₀ (sin x / x) = 1
Using the above inequality, cos x < (sin x)/x < 1. Also, it is very clear that lim ₓ → ₀ cos x = cos 0 = 1 and lim ₓ → ₀ 1 = 1. Hence, by squeeze theorem (sandwich theorem), lim ₓ → ₀ (sin x) / x = 1.
Hence, we proved that lim ₓ → ₀ (sin x) / x = 1.
Proving lim ₓ → ₀ (1  cos x)/x = 0
Using the halfangle formulas, 1  cos x = sin^{2} x/2. So the above limit becomes:
lim ₓ → ₀ (2sin^{2} x/2) / x = lim ₓ → ₀ 2 (sin x/2) (sin x/2) / x
As x → 0, x/2 → 0. So the above limit becomes:
= lim ₓ → ₀ (sin x/2) lim ₓ_{/2} → ₀ (sin x/2) / (x/2)
By the above limit, we have lim ₓ → ₀ (sin x / x) = 1. Thus, the value of the rightside limit of the above step is 1. Thus,
= (sin 0/2) (1)
= 0 (1)
= 0
Hence, we proved that lim ₓ → ₀ (1  cos x)/x = 0.
Important Notes on Squeeze Theorem:
When we have to evaluate a limit using squeeze theorem (or sandwich theorem), remember the following trig inequalities.
 cos x < (sin x)/x < 1
 1 ≤ sin x ≤ 1, for any x in the domain of sin x
 1 ≤ cos x ≤ 1, for any x in the domain of cos x
☛ Related Topics:
Squeeze Theorem Examples

Example 1: If 6  x^{2} ≤ f(x) ≤ 6 + x^{2 }then find the value of lim ₓ → ₀ f(x).
Solution:
We know that lim ₓ → ₀ 6  x^{2} = 6  0^{2} = 6 and
lim ₓ → ₀ 6 + x^{2} = 6 + 0^{2} = 6
Also, f(x) lies between 6  x^{2} and 6 + x^{2}. Thus, by squeeze theorem:
lim ₓ → ₀ f(x) = 6.
Answer: 6

Example 2: What is the value of the limit lim ₓ → ₀ x^{2} sin (1/x)?
Solution:
We know that 1 ≤ sin x ≤ 1 for any x. In the same way
1 ≤ sin (1/x) ≤ 1 (when x ≠ 0)
Multiplying each side by x^{2} (This does not change the sign of inequality as x^{2} ≥ 0):
x^{2} ≤ sin (1/x) ≤ x^{2}
Now, lim ₓ → ₀ x^{2} = 0^{2} = 0 and lim ₓ → ₀ x^{2} = lim ₓ → ₀ 0^{2} = 0.
So by sandwich theorem, lim ₓ → ₀ x^{2} sin (1/x) = 0.
Answer: 0

Example 3: Evaluate the limit lim ₓ → ₀ x e^{cos (1/x)}.
Solution:
We know that 1 ≤ cos x ≤ 1 for any x. In the same way
1 ≤ cos (1/x) ≤ 1 (when x ≠ 0)
Raising each side by "e",
e^{1 }≤ e^{cos (1/x)} ≤ e^{1}
Case 1: When x > 0
Multiply each side by x:
xe^{1 }≤ x e^{cos (1/x)} ≤ xe
Now, lim ₓ → ₀ xe^{1} =0/e = 0 and lim ₓ → ₀ xe = 0(e) = 0
So by sandwich theorem, lim ₓ → ₀ x e^{cos (1/x)} = 0.
Case 2: When x < 0
Multiply each side by x:
xe^{1 }≥ x e^{cos (1/x)} ≥ xe
xe^{ }≤ x e^{cos (1/x)} ≤ xe^{1}
Now, lim ₓ → ₀ xe =0/e = 0 and lim ₓ → ₀ xe^{1} = 0(e) = 0
So by sandwich theorem, lim ₓ → ₀ x e^{cos (1/x)} = 0.
Answer: In both the cases, the given limit = 0.
FAQs on Squeeze Theorem
Why is the Name Squeeze Theorem?
The squeeze theorem states that if a function f(x) is such that g(x) ≤ f(x) ≤ h(x) and suppose that the limits of g(x) and h(x) as x tends to a is equal to L then lim ₓ → ₐ f(x) = L. It is known as "squeeze" theorem because it talks about a function f(x) that is "squeezed" between g(x) and h(x).
What are the Other Names of Squeeze Theorem?
The squeeze theorem is also known with other names such as:
 sandwich theorem
 carabinieri theorem
 squeeze lemma
 police theorem
 pinching theorem
 between theorem
Can we Apply Sandwich Theorem for Infinite Limits?
Yes, the sandwich theorem can be applied for infinite limits as well. For example, to find the limit lim ₓ → ∞ (sin x) / x, we use the squeeze theorem as follows. We know that 1 ≤ sin x ≤ 1. Dividing by x, 1/x ≤ (sin x) / x ≤ 1/x. We know that lim ₓ → ∞ (1/x) = lim ₓ → ∞ (1/x) = 0 and hence by squeeze theorem, lim ₓ → ∞ (sin x) / x = 0.
How Do We Apply Squeeze Theorem?
To apply the squeeze theorem, first find between which two functions the given function lies. Then see whether the limits of those two functions at the given point are equal. If so, apply the squeeze theorem, by which the limit of the given function is also equal to the limit of each of the two functions. In this process, we can use one of the following popular inequalities:
 cos x < (sin x)/x < 1
 1 ≤ sin x ≤ 1
 1 ≤ cos x ≤ 1
What is the Name of the Mathematician Who Proposed Squeeze Theorem?
The squeeze theorem was proposed by a mathematician called Hugo Steinhaus. It was later proved by another mathematician called Stefan Banach.
Is Squeeze Theorem Only for Trig?
In fact, the squeeze function is applicable for any type of function. But while solving the problems, we mostly come across the ones that are with trig functions when we apply the squeeze theorem.
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