# Limit - Squeeze Theorem or L'Hopital's Rule?

• Qube
In summary, there was a discussion about different methods to solve the limit of (1-cosx)/x as x approaches 0. L'Hopital's rule and the squeeze theorem were mentioned, with the correct application of L'Hopital's rule resulting in the limit being 1 and the application of the squeeze theorem not being successful. Another method using the limit of sinx/x was also mentioned.
Qube
Gold Member

## Homework Equations

Squeeze theorem: set up inequalities putting the function of interest between two integers.

L'Hopital's rule: when plugging in the number into the limit results in a specified indeterminate form such as 0/0 or infinity/infinity then take the derivative of the numerator and the derivative of the denominator and again plug-in to solve the limit.

The limit of 1-cosx/x as x --> 0 is 1.

## The Attempt at a Solution

I used L'Hopital's rule and applied two iterations of get

cos(x)/(4cosx+4cosx-4xsinx)

Plugging in 0 results in

1 / (4+4-0) = 1/8

However, we never learned L'Hopital's rule in class and the only thing we learned was the squeeze theorem. I attempted to use the squeeze theorem (without success):

0≤1-cosx≤2

I'm not sure where to proceed beyond the above step in using the squeeze theorem.

0/x≤(1-cosx)/x≤2/x
x--> 0 so infinity≤(1-cosx)/x≤infinity

Is the squeeze theorem the correct method to solve this problem? Or am I overlooking some common identity or method other than L'Hopital's rule to solve this problem?

Qube said:

## Homework Equations

Squeeze theorem: set up inequalities putting the function of interest between two integers.

Not generally between two integers, just two numbers.

L'Hopital's rule: when plugging in the number into the limit results in a specified indeterminate form such as 0/0 or infinity/infinity then take the derivative of the numerator and the derivative of the denominator and again plug-in to solve the limit.

The limit of 1-cosx/x as x --> 0 is 1.

I assume you mean ##\frac{1-\cos x}{x}## and not what you have written:##1 -\frac{\cos x}{x}##. Use parentheses! Also, you will have difficulty proving it since the limit isn't ##1##.

## The Attempt at a Solution

I used L'Hopital's rule and applied two iterations of get

cos(x)/(4cosx+4cosx-4xsinx)

How in the world did you get that with L'Hospital's rule?? Show your work.

Plugging in 0 results in

1 / (4+4-0) = 1/8

However, we never learned L'Hopital's rule in class and the only thing we learned was the squeeze theorem. I attempted to use the squeeze theorem (without success):

0≤1-cosx≤2

I'm not sure where to proceed beyond the above step in using the squeeze theorem.

0/x≤(1-cosx)/x≤2/x
x--> 0 so infinity≤(1-cosx)/x≤infinity

##\frac 0 x = 0## so you have ##0## on the left and infinity on the right. Nothing is being squeezed. Try doing L'Hospital's rule correctly.

If you use L'Hospital's rule you calculate the derivative of 1-cosx and x, which gives you sinx/1 or simply sinx. What's the limit as x approaches 0 of sinx?
Another method to evaluate this limit is using the fact that the limit as x approaches 0 of sinx/x is 1, and then multiplying your original problem ((1-cosx)/x) by the conjugate of the numerator. When you simplify this you'll see the limit I mentioned earlier pop up which helps you simplify and get to the correct answer.

I think you took the derivative of the whole functions, you're supposed to take the derivative of the top and bottom seperately. So the derivative of the top/derivative of the bottom. The bottom gives you 1 and the top gives you what?

## 1. What is the Limit - Squeeze Theorem?

The Limit - Squeeze Theorem, also known as the Sandwich Theorem, states that if two functions, g(x) and h(x), have the same limit as x approaches a, and if another function, f(x), is always between g(x) and h(x) near a, then f(x) also has the same limit as x approaches a.

## 2. When should I use the Limit - Squeeze Theorem?

The Limit - Squeeze Theorem is useful when you are trying to find the limit of a function that is difficult to evaluate directly. It can also be used to prove the existence of a limit.

## 3. What is L'Hopital's Rule?

L'Hopital's Rule is a technique used to evaluate limits of indeterminate forms, such as 0/0 or ∞/∞. It states that if the limit of the ratio of two functions, f(x) and g(x), as x approaches a is an indeterminate form, then the limit of the ratio of their derivatives, f'(x) and g'(x), is equal to the original limit.

## 4. When should I use L'Hopital's Rule?

L'Hopital's Rule is useful when you are trying to evaluate a limit that results in an indeterminate form. It can also be used to simplify complicated expressions and make them easier to evaluate.

## 5. Are there any limitations to using Limit - Squeeze Theorem or L'Hopital's Rule?

Yes, there are limitations to using these techniques. They can only be used to evaluate limits at specific points, and they do not work for all types of indeterminate forms. Additionally, L'Hopital's Rule can only be used if the limit of the ratio of the derivatives exists. It is important to use caution and check the assumptions before using these techniques.

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