Hello, synx!
Proving limit using squeeze/sandwich theorem
x→0limx21−cosx=21
I too am having trouble finding the sandwiching functions.
I found two ways to prove it using the theorem:
x→0limxsinx=1
[1] Multiply top and bottom by
(1+cosx):
\(\displaystyle \;\;\frac{\1\,-\,\cos x}{x^2}\,\cdot\,\frac{1\,+\,\cos x}{1\,+\,\cos x}\;=\;\frac{1\,-\,\cos^2x}{x^2(1\,+\,\cos x)} \:=\:\frac{\sin^2x}{x^2(1\,+\,\cos x)} \;= \;\left(\frac{\sin x}{x}\right)^2\cdot\left(\frac{1}{1\,+\,\cos x}\right)\)
Then:
x→0lim(xsinx)2(1+cosx1)=(12)(21)=21
[2] Using that identify we have:
x22sin2(2x)=21⋅4x2sin2(2x)=21⋅(2x)2sin2(2x)
Then:
2x→0lim21[2xsin(2x)]2=21⋅12=21
Edit: Daon was too fast for me!