How to solve lim Without L'Hospital's Rule?
you can use trigonometric identities and inequalities, but you can't use series or more advanced stuff.
Answer
The given expression is odd; therefore it is enough to consider x>0. We then have
0<{x-\sin x\over x^2}<{\tan x -\sin x\over x^2}=\tan x\ {1-\cos x\over x^2}={\tan x\over2}\ \Bigl({\sin(x/2)\over x/2}\Bigr)^2\ ,
and right side obviously converges to 0 when x\to0+.
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