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Trigonometric Limits

Posted November 27th, 2009 by alpha
in

$ \mathop{\lim}\limits_{x \to 0}\frac{tan x -sinx}{x^3} $

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answer

On November 30th, 2009 Structure says:
$$ \mathop{\lim}\limits_{x \to 0}\frac{tan x -sinx}{x^3} = \mathop{\lim}\limits_{x \to 0}\frac{tan x -x}{x^3} + \mathop{\lim}\limits_{x \to 0}\frac{ x -sinx}{x^3} =\frac{1}{3}+\frac{1}{6}=\frac{1}{2}$$

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