Proving limit using squeeze/sandwich theorem

synx

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

1cosxx2=2(sin(x2)x)2\displaystyle \frac{1-cosx}{x^2} = 2(\frac{sin(\frac{x}{2})}{x})^2

=214(sin(x2)x2)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,

1cosxx2=214(1)2=12\displaystyle \frac{1-cosx}{x^2} = 2\frac{1}{4}(1)^2 = \frac{1}{2}

Hope that helps
 
Hello, synx!

I too am having trouble finding the sandwiching functions.

I found two ways to prove it using the theorem:   limx0sinxx  =  1\displaystyle \;\lim_{x\to0}\frac{\sin x}{x}\;=\;1


[1] Multiply top and bottom by (1+cosx):\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:   limx0(sinxx)2(11+cosx)  =  (12)(12)  =  12\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: 2sin2(x2)x2  =  12sin2(x2)x24  =  12sin2(x2)(x2)2\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: limx2012[sin(x2)x2]2  =  1212  =  12\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!
 
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!
 
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