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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.

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    $\begingroup$ $$\lim_{x \to 0} \frac{\tan x - x}{2x^2 \tan x}.$$ Now apply your favorite method: MacLaurin expansion, De l'Hospital... $\endgroup$ – Siminore Feb 4 '15 at 14:52
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apply l'Hospital's rule:

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

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Using a polynomial approximation at zero, $$ \frac{\tan x - x}{2x^2 \tan x} = \frac{x+\frac13 x^3 + \ldots -x}{2x^2 (x+\frac13 x^3 + \ldots)} = \frac{\frac13 x^3 + \ldots}{2x^3 + \ldots} \to \frac{1}{6}. $$

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