Sandwich Theorem

The Sandwich (Squeeze) Theorem works by "sandwiching" a given function between two other functions whose limits are easier to determine.

It states that let functions f(x), g(x), and h(x) be real-valued functions such that h(x) ≤ f(x) ≤ g(x) for all x in some neighborhood of a (except possibly at a).

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.

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).

Conditions

The Sandwich Theorem can be applied only when the following conditions are satisfied:

1. There exist functions h(x), f(x), and g(x) such that h(x) ≤ f(x) ≤ g(x)in a neighborhood of the point of interest.
2. The lower and upper bounding functions approach the same limit: \lim_{x\to a}h(x)=\lim_{x\to a}g(x)=L
3. The inequality remains valid near the point a.

When these conditions hold, the middle function must have the same limit L.

Proof

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 ∀ ε > 0, ∃ δ₁ > 0 such that |x - a| < δ₁ ⇒ |g(x) - L| < ε

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

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

|x - a| < δ₂ ⇒ - ε < h(x) - L < ε    ... (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}, Now if |x - a| < δ,

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

-ε < f(x) - L < ε

lim_x→a f(x) = L

Thus, this proved the Sandwich Theorem.

Important Limits Using the Sandwich Theorem

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

Proof of  lim_x→a (sin x / x) = 1

From the 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

Solved Examples

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

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)

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

Practice Problems

1. Evaluate: \lim_{x\to0} x\sin\left(\frac1x\right).

2. If 4-x^2\le f(x)\le 4+x^2, find \lim_{x\to0}f(x).

3. Evaluate: \lim_{x\to0} x^3\cos\left(\frac1x\right).

4. If 2-\sqrt{x}\le g(x)\le 2+\sqrt{x}, find \lim_{x\to0}g(x)

5. Evaluate: \lim_{x\to0}\frac{x^2}{1+|x|}

(Hint: Bound the expression between 0 and x²)