Proving the Squeeze Theorem - Finding Limits Using Squeeze Theorem

  • MHB
  • Thread starter nycmathdad
  • Start date
  • Tags
    Theorem
In summary: I learn best by going step by step. So, I will work on understanding the concepts first and then use the derivative when I need to. I thank you for the insight. I learn best by going step by step. So, I will work on understanding the concepts first and then use the derivative when I need to.
  • #1
nycmathdad
74
0
Use the Squeeze Theorem to find the limit.

lim [x^2 • (1 - cos(1/x)]
x--> 0

Let me see.

-1 ≤ cos (1/x) ≤ 1

-x^2 ≤ x^2 • [1 - cos(1/x)] ≤ x^2

-|x^2| ≤ x^2 • [1 - cos(1/x)] ≤ |x^2|

lim -|x^2| as x tends to 0 = 0.

lim |x^2| as x tends to 0 = 0.
.
By the Squeeze Theorem, [x^2 • (1 - cos(1/x)] was squeezed between the limit of -|x^2| as x tends to 0 and the limit of |x^2| as x tends to 0.

Conclusion:

lim [x^2 •(1 - cos(1/x)] = 0
x--> 0

The limit is 0.

Correct?
 
Physics news on Phys.org
  • #2
Beer soaked ramblings follow.
nycmathdad said:
...
-1 ≤ cos (1/x) ≤ 1

-x^2 ≤ x^2 • [1 - cos(1/x)] ≤ x^2
...
Does not follow.

Multiplying -1 ≤ cos (1/x) ≤ 1 by x^2 yields -x^2 ≤ x^2[cos (1/x)] ≤ x^2
not -x^2 ≤ x^2 • [1 - cos(1/x)] ≤ x^2
 
Last edited:
  • #3
nycmathdad said:
Use the Squeeze Theorem to find the limit.

lim [x^2 • (1 - cos(1/x)]
x--> 0

Let me see.

-1 ≤ cos (1/x) ≤ 1

-x^2 ≤ x^2 • [1 - cos(1/x)] ≤ x^2

-|x^2| ≤ x^2 • [1 - cos(1/x)] ≤ |x^2|

lim -|x^2| as x tends to 0 = 0.

lim |x^2| as x tends to 0 = 0.
.
By the Squeeze Theorem, [x^2 • (1 - cos(1/x)] was squeezed between the limit of -|x^2| as x tends to 0 and the limit of |x^2| as x tends to 0.

Conclusion:

lim [x^2 •(1 - cos(1/x)] = 0
x--> 0

The limit is 0.

Correct?

There's no need to use the Squeeze Theorem here at all. The product of a bounded function and a function going to 0 has a limit of 0.
 
  • #4
Prove It said:
There's no need to use the Squeeze Theorem here at all. The product of a bounded function and a function going to 0 has a limit of 0.

Brother, using the Squeeze Theorem here makes sense because I am learning one chapter, one section at a time. Understand? I can use the derivative to solve many of the limit problems but I am currently at the end of Chapter 1, which is all about limits at the Calculus 1 level. Chapter 2 introduces the derivative idea at its basic level. Work with me here. Travel with me one chapter, one section at a time.
 
  • #5
nycmathdad said:
Brother, using the Squeeze Theorem here makes sense because I am learning one chapter, one section at a time. Understand? I can use the derivative to solve many of the limit problems but I am currently at the end of Chapter 1, which is all about limits at the Calculus 1 level. Chapter 2 introduces the derivative idea at its basic level. Work with me here. Travel with me one chapter, one section at a time.

Then you can take what I said as something new you have learnt. Be grateful.
 
  • #6
Also, there is NOTHING here about using a derivative.

A limit is a statement of the behaviour of a function in a local neighbourhood of a specific point.

A derivative is an instantaneous rate of change of a function.

Mathematics is not something to learn in a linear fashion, it's a connected network of concepts that link together. Understand the concepts and then the mathematics will be manageable.
 
  • #7
Prove It said:
Also, there is NOTHING here about using a derivative.

A limit is a statement of the behaviour of a function in a local neighbourhood of a specific point.

A derivative is an instantaneous rate of change of a function.

Mathematics is not something to learn in a linear fashion, it's a connected network of concepts that link together. Understand the concepts and then the mathematics will be manageable.

I thank you for the insight.
 

1. What is the Squeeze Theorem?

The Squeeze Theorem, also known as the Sandwich Theorem, is a mathematical theorem used to prove the limit of a function. It states that if two functions, g(x) and h(x), are both approaching the same limit as x approaches a certain value, and another function f(x) is always between g(x) and h(x), then f(x) must also approach the same limit.

2. How is the Squeeze Theorem used to prove a limit?

To prove a limit using the Squeeze Theorem, you must first identify two functions, g(x) and h(x), that are approaching the same limit as x approaches a certain value. Then, you must show that another function, f(x), is always between g(x) and h(x). Finally, you can conclude that since g(x) and h(x) both approach the same limit, f(x) must also approach the same limit.

3. What is the importance of the Squeeze Theorem in calculus?

The Squeeze Theorem is an important tool in calculus because it allows us to prove the limit of a function without using the formal definition of a limit. It is often used in situations where the direct evaluation of a limit is difficult or impossible.

4. Can the Squeeze Theorem be used to prove the limit of a function at infinity?

Yes, the Squeeze Theorem can be used to prove the limit of a function at infinity. In this case, g(x) and h(x) would both approach infinity, and f(x) would be sandwiched between them. This allows us to conclude that f(x) also approaches infinity.

5. Are there any limitations to the Squeeze Theorem?

Yes, there are limitations to the Squeeze Theorem. It can only be used to prove the limit of a function if the two limiting functions, g(x) and h(x), are approaching the same limit. Additionally, f(x) must always be between g(x) and h(x) for the theorem to hold. If these conditions are not met, then the Squeeze Theorem cannot be used to prove the limit.

Similar threads

  • Calculus
Replies
2
Views
796
Replies
8
Views
376
Replies
2
Views
241
  • Calculus
Replies
6
Views
1K
Replies
5
Views
1K
Replies
3
Views
991
Replies
4
Views
286
Replies
20
Views
2K
  • Calculus
Replies
2
Views
839
Back
Top