Sandwich Theorem (Squeeze Theorem): Statement, Proof …

Sandwich Theorem (Squeeze Theorem)

The Sandwich Theorem (also known as the Squeeze Theorem) is a fundamental result in calculus that allows us to find the limit of a function by “squeezing” it between two other functions whose limits are known and are equal at a particular point.

Statement:

Let f(x)f(x), g(x)g(x), and h(x)h(x) be functions defined on an interval around aa (except possibly at aa). If the following three conditions hold:

1. g(x)≤f(x)≤h(x)g(x) \leq f(x) \leq h(x) for all xx in some interval around aa (except possibly at a),
2. lim⁡x→ag(x)=L
3. Then lim⁡x→af(x)=L

In simpler terms, if f(x)f(x) is squeezed between two functions g(x)g(x) and h(x)h(x) which both approach the same limit LL as x→ax \to a, then f(x)f(x) must also approach L.

Proof:

Step 1: Set Up the Squeeze Inequality

We are given that for all xx sufficiently close to aa, except possibly at aa,

g(x)≤f(x)≤h(x).g(x) 

Also, we know that

lim⁡x→ag(x)=L

Step 2: Use the Definition of Limit

Since lim⁡x→ag(x)=L there exists a δ1>0\delta_1 > 0 such that for all xx in the interval (a−δ1,a+δ1) (a – + \delta_1),

Similarly, since lim⁡x→ah(x)=L

Step 3: Combine the Results

Let δ=min⁡(δ1,δ2)

This implies:

L−ϵ<g(x)<L+ϵ and L−ϵ<h(x)<L+ϵ.L – 

Step 4: Apply the Squeeze Inequality

Since g(x)≤f(x)≤h(x)g(x) we can conclude:

L−ϵ<f(x)<L+ϵ.L – \epsilon < f(x) < L + \epsilon.

Step 5: Conclusion

Thus, for all xx sufficiently close to aa, we have

∣f(x)−L∣<ϵ.|f(x) – L| < \epsilon.

By the definition of the limit, this implies that

lim⁡x→af(x)=L.\lim_{x \to a} f(x) = L.

This completes the proof of the Sandwich (Squeeze) Theorem.

Example:

Consider the following example:

Let f(x)=x2sin⁡(1x)f(x) = x^2 .

1. We know that −1≤sin⁡(1x)≤1-1 
2. Thus, we have: −x2≤x2sin⁡(1x)≤x2.-x^2 
3. Taking the limit of the bounds as x→0x \to 0: lim⁡x→0−x2=0andlim⁡x→0x2=0
4. By the Squeeze Theorem, we conclude that lim⁡x→0x2sin⁡(1x)=0.