Proving limit using squeeze/sandwich theorem

[synx]

Proving limit using squeeze/sandwich theorem

Limit x->0

(1-cos(x))/x^2 = 1/2

Im having trouble finding out what to pull out to satisfy the squeeze theormen, everything i try doesn't seem to have the limit of 1/2. Also, 1-cos(x) also is = to 2sin^2(x/2) , if that helps.

 

[daon]

Using what you said, (without the squeeze theorem):

\(\displaystyle \frac{1-cosx}{x^2} = 2(\frac{sin(\frac{x}{2})}{x})^2\)

\(\displaystyle = 2\frac{1}{4}(\frac{sin(\frac{x}{2})}{\frac{x}{2}})^2\)

Since \(\displaystyle \L\\ Lim\limits_{x\rightarrow 0}\frac{Sin\frac{x}{2}}{\frac{x}{2}} = 1\)

So,

\(\displaystyle \frac{1-cosx}{x^2} = 2\frac{1}{4}(1)^2 = \frac{1}{2}\)

Hope that helps

 

[soroban]

Hello, synx!

Proving limit using squeeze/sandwich theorem

\(\displaystyle \;\;\lim_{x\to0}\frac{1\,-\,\cos x}{x^2}\;=\;\frac{1}{2}\)

I too am having trouble finding the sandwiching functions.

I found two ways to prove it using the theorem: \(\displaystyle \;\lim_{x\to0}\frac{\sin x}{x}\;=\;1\)


[1] Multiply top and bottom by \(\displaystyle (1\,+\,\cos x):\)

\(\displaystyle \;\;\frac{\1\,-\,\cos x}{x^2}\,\cdot\,\frac{1\,+\,\cos x}{1\,+\,\cos x}\;=\;\frac{1\,-\,\cos^2x}{x^2(1\,+\,\cos x)} \:=\:\frac{\sin^2x}{x^2(1\,+\,\cos x)} \;= \;\left(\frac{\sin x}{x}\right)^2\cdot\left(\frac{1}{1\,+\,\cos x}\right)\)

\(\displaystyle \;\;\)Then: \(\displaystyle \;\lim_{x\to0}\left(\frac{\sin x}{x}\right)^2\left(\frac{1}{1\,+\,\cos x}\right) \;= \;(1^2)\left(\frac{1}{2}\right) \;=\;\frac{1}{2}\)


[2] Using that identify we have: \(\displaystyle \:\frac{2\sin^2\left(\frac{x}{2}\right)}{x^2} \;=\;\frac{1}{2}\,\cdot\,\frac{\sin^2\left(\frac{x}{2}\right)}{\frac{x^2}{4}} \;= \;\frac{1}{2}\,\cdot\,\frac{\sin^2\left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)^2}\)

\(\displaystyle \;\;\)Then: \(\displaystyle \:\lim_{\frac{x}{2}\to0}\,\frac{1}{2}\,\left[\frac{sin(\frac{x}{2})}{\frac{x}{2}}\right]^2\;= \;\frac{1}{2}\cdot1^2\;=\;\frac{1}{2}\)

Edit: Daon was too fast for me!

 

[synx]

yeah i see how it can be solved that way, but unfortunately my instructor wants us to prove it using only the squeeze theorem thanks for the quick replys though!