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?

Related Calculus and Beyond Homework News on Phys.org

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?

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

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