When Is the Squeeze(Sandwich) Theorem Used?

Question

I am beginning to learn Calculus 1, and I was taught about the squeeze (sandwich) theorem. It seems to me that all the problems given have $\sin$ involved. Is this true? When is the squeeze theorem applied?

Also, what are the standard "squeeze functions"? For example, I know that $-1\leq\sin(x)\leq 1$ and this is used in a $\sin$-involved problem. Are there other such functions to apply when using the squeeze theorem?

Answer 1

No it's not necessary that sine functions are involved in all Sandwich Theorem Problems.

1. Sandwich Theorem is commonly used in computing Integrals as a limit of a sum.

2. It is used in Limit Computations

3. It is used in proving convergence of many series by bounding it.

4. I found another interesting use with many applications in itself. The limit of the sinc function can be proved using the theorem which provides a first order approximation that is used in Physics. This function also shows up as the fourier transform of a rectangular wave. You can read more about this specific function in the top answer (answered anonymously) to this Quora question.

Examples of functions where squeeze theorem can be applied-

Limit of the function-

$f(x)=x^2 e^{\sin\frac{1}{x}}$

As x approaches 0

The limit is not normally defined because the function oscillates infinitely many times as it approaches zero

Answer 2

The squeeze theorem provides an intuitive rule for making statements about the convergence of a given series when it is bounded above and below ("squeezed") by 2 other series which are known to converge.

That you are seeing a lot of examples that use trig functions is purely coincidental. What is even more likely is that most of the examples you see also involve quotients, since the squeeze theorem is often used to prove the existence of a limit where the denominator of a function is zero. But agin, it is a general rule and is not restricted to any particular class of function.

Answer 3

A nice example for an application of the squeeze theorem which does not involve any trigonometric functions: Show that

$$\lim_{x\to 0} x \cdot \lfloor 1/x\rfloor =1$$

(where $\lfloor \rfloor$ is the floor function.)

Answer 4

Squeeze Theorem is used to find the limit of a function when other methods are failed to do that.

Now see this example: Show that $\lim _{t \to 2}g(t)=-1$ when $-\frac{1}{3}t^{3}+t^{2}-\frac{7}{3} \leq g\left ( t \right ) \leq \cos \left ( \frac{t \pi}{2} \right )$.

This is an example of the Squeeze theorem not involving sine function.

We can evaluate the limit using the squeeze theorem.

Solution:

Given that $-\frac{1}{3}t^{3}+t^{2}-\frac{7}{3} \leq g\left ( t \right ) \leq \cos \left ( \frac{t \pi}{2} \right )$

Now $\lim _{t \to 2}\left ( -\frac{1}{3}t^{3}+t^{2}-\frac{7}{3} \right )$

= $\lim _{t \to 2} -\frac{1}{3}t^{3}+ \lim_ {t _\to 2}t^{2}- \lim _{t _\to 2}\frac{7}{3}$

=$-\frac{1}{3}(2)^{3}+ (2)^{2}-\frac{7}{3}$

=$-\frac{8}{3}+ 4-\frac{7}{3}$

=$\frac{-8+12-7}{3}$

=$\frac{-3}{3}$

= $-1$

and $ \lim _{t \to 2} \cos \left ( \frac{t \pi}{2} \right ) = \cos \left ( \frac{2 \pi}{2} \right ) = \cos \pi = -1$

Since $-\frac{1}{3}t^{3}+t^{2}-\frac{7}{3} \leq g\left ( t \right ) \leq \cos \left ( \frac{t \pi}{2} \right )$ and $\lim _{t \to 2}\left (-\frac{1}{3}t^{3}+t^{2}-\frac{7}{3} \right )=-1=\cos \left ( \frac{t \pi}{2} \right )$

Therefore by Squeeze Theorem,

$$\lim _{t \to 2}g(t)=-1$$