Problem while calculating a limit with tangens I'm having trouble calculating this limit:
$$\lim _{x \rightarrow 0} (\frac{1}{2x^2} - \frac{1}{2x \tan x})$$
I've tried using the gact that $\lim_{x \rightarrow 0} \frac{\tan x}{x} = 1$, but that obviously doesn't work here.
Neither does the l'Hospital's rule.
Could you give me some idea how to solve this?
Thanks.

apply l'Hospital's rule:
$$\lim _{x \rightarrow 0} (\frac{1}{2x^2} - \frac{1}{2x \tan x}) = \lim _{x \rightarrow 0} (\frac{tanx - x}{2x^2 \tan x}) = \lim _{x \rightarrow 0} (\frac{\frac{1}{cos^2x} - 1}{4xtan x + \frac{2x^2}{cos^2x}})
= \lim _{x \rightarrow 0} (\frac{1 - cos^2x}{2x^2 + 4xsinxcosx}) = \lim _{x \rightarrow 0} (\frac{sin^2x}{2x^2 + 2xsin2x}) = \lim _{x \rightarrow 0} (\frac{sin2x}{4x+2sin2x+4xcos2x}) = \lim _{x \rightarrow 0} (\frac{2cos2x}{4 + 4cos2x + 4cos2x - 8xsin2x}) = \frac{2}{12} = \frac{1}{6} $$ 
