Tìm giới hạn
$a)\mathop {\lim }\limits_{x \to 0}\frac{\tan x-\sin x}{x^3}  $
$b) \mathop {\lim }\limits_{x \to 0}\frac{1-\sin 2x-\cos 2x}{1+\sin 2x- \cos 2x}  $
$c) \mathop {\lim }\limits_{x \to 0} \frac{1-\cos^2 2x}{x\sin x} $
a) $\mathop {\lim }\limits_{x \to 0}\frac{tanx-sinx}{x^3}$

=$\mathop {\lim }\limits_{x \to 0}\frac{\frac{sinx-sinx.cosx}{cosx}}{x^3}$

=$\mathop {\lim }\limits_{x \to 0}\frac{sinx(\frac{1-cosx}{cosx})}{x^3}$

=$\mathop {\lim }\limits_{x \to 0}\frac{sinx.(2\frac{sin^2\frac{x}{2}}{cosx})}{x^3}$

=$\mathop {\lim }\limits_{x \to 0}\frac{x.\frac{sinx}{x}.\frac{2}{cosx}(\frac{x}{2})^2.(\frac{sin\frac{x}{2}}{\frac{x}{2}})^2}{x^3}$

=$\mathop {\lim }\limits_{x \to 0}\frac{x^3\frac{sinx}{x}.\frac{1}{2.cosx}.(\frac{sin\frac{x}{2}}{\frac{x}{2}})^2}{x^3}$

=$\frac{1}{2cosx}=\frac{1}{2}$
từ dòng 1 xuống dòng 2 là sao ta? –  HọcTạiNhà 03-03-14 06:48 PM
c) $\mathop {\lim }\limits_{x \to 0}\frac{1-cos^22x}{x.sinx}$

= $\mathop {\lim }\limits_{x \to 0}\frac{sin^22x}{x.sinx}$

=$\mathop {\lim }\limits_{x \to 0}\frac{(2x)^2.(\frac{sin2x}{2x})^2}{x^2.\frac{sinx}{x}}$

= 4
tai sao lim[sin(x)/x]=1??? –  Laughing 02-03-14 04:43 PM

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