# Squeeze Theorem Proof....3

• MHB
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?

jonah1
Beer soaked ramblings follow.
...
-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:
Gold Member
MHB
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.

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.

Gold Member
MHB
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.

Gold Member
MHB
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.