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{6.(\frac{cos6x}{6x})^2+(\frac{cosx}{x})^2}{11.(\frac{sin11x}{11x})^2}= \frac{6+1}{11}=\frac{7}{11} ( vì ta có \mathop {\lim }\limits_{x \to 0}\frac{sinx}{x}=1)
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}=$\mathop {\lim }\limits_{x \to 0}\frac{
36.(\frac{cos6x}{6x})^2+(\frac{cosx}{x})^2}{1
21.(\frac{sin11x}{11x})^2}
= \frac{
36+1}{1
21}
=\frac{
37}{1
21}
( vì ta có \mathop {\lim }\limits_{x \to 0}\frac{sinx}{x}=1$)