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

#### Qube

Gold Member
1. The problem statement, all variables and given/known data

2. Relevant 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.

3. 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?

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#### LCKurtz

Homework Helper
Gold Member
1. The problem statement, all variables and given/known data

2. Relevant 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 $1$.

3. 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.

#### milkedpig

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.

#### iRaid

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?

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