Tìm giới hạn
$d) \mathop {\lim }\limits_{x \to 0}\frac{\sin 3x}{1-2\cos x}  $
$e) \mathop {\lim }\limits_{x \to 0}\frac{1-\cos 5x \cos 7x}{\sin^2 11x}  $
$f) \mathop {\lim }\limits_{x \to 0}\frac{\cos 12x-\cos 10x}{\cos 8x-\cos 6x}  $
f) $\mathop {\lim }\limits_{x \to 0}\frac{cos12x-cos10x}{cos8x-cos6x}$

=$\mathop {\lim }\limits_{x \to 0}\frac{-2sin11x.sinx}{-2sin7x.sinx}$

=$\mathop {\lim }\limits_{x \to 0}\frac{11x.\frac{sin11x}{11x}}{7x.\frac{sin7x}{7x}}$

=$\frac{11}{7}$
e) $\mathop {\lim }\limits_{x \to 0}\frac{1-cos5x.cos7x}{sin^211x}$

=$\mathop {\lim }\limits_{x \to 0}\frac{1- cos12x + 1 - cos 2x}{2sin^211x}$

=$\mathop {\lim }\limits_{x \to 0}\frac{2cos^26x+2cos^2x}{2sin^211x}$

=$\mathop {\lim }\limits_{x \to 0}\frac{(6x)^2.(\frac{cos6x}{6x})^2+x^2.(\frac{cosx}{x})^2}{(11x)^2.(\frac{sin11x}{11x})^2}$

=$\mathop {\lim }\limits_{x \to 0}\frac{36.(\frac{cos6x}{6x})^2+(\frac{cosx}{x})^2}{121.(\frac{sin11x}{11x})^2}$

= $\frac{36+1}{121}$=$\frac{37}{121}$ ( vì ta có $\mathop {\lim }\limits_{x \to 0}\frac{sinx}{x}=1$)
ừm đúng rồi, bạn giải thích chỗ đó được ko? –  HọcTạiNhà 03-03-14 04:16 AM
sao mà lim[sinx/x]=1. Oa....Oa..... –  Laughing 02-03-14 05:19 PM

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