Question

Lim x-> 0 (1-cos7x)/(x*sin7/2* (x))

Answer 1

$lim_{x\to 0}\, \frac{1-cos7x}{x\cdot sin\frac{7}{2}x}=lim_{x\to 0} \, \frac{2sin^2\frac{7x}{2}}{x\cdot sin\frac{7x}{2}} =lim_{x\to 0}\, \frac{2sin\frac{7x}{2}}{x}=\\\\=lim_{x\to 0}\, \frac{2sin\frac{7x}{2}}{\frac{7x}{2}}\cdot \frac{2}{7}=lim_{x\to 0}\frac{4}{7}\cdot \frac{sin\frac{7x}{2}}{\frac{7x}{2}} =\frac{4}{7}\\\\P.S.\; \; 1-cos \alpha =2sin^2\frac{ \alpha }{2}\\$

Answer 2

$\lim_{x \to 0} \frac{(1-cos7x)}{x*sin\frac{7}2x}= \lim_{x \to 0} \frac{(1-cos7x^{\to1})}{(x^{\to0}*sin\frac{7}2x^{\to0})}=\frac{1-1}{0*0}=\frac{0}0=(*)$
Попробуем разложить знаменатель по формуле синус половинного угла
$\lim_{x \to 0} \frac{(1-cos7x)}{x*sin\frac{7}2x}=\lim_{x \to 0} \frac{(1-cos7x)}{x*\sqrt{\frac{1-cos7x}{2}}}=\lim_{x \to 0} \frac{(1-cos7x)}{x*\sqrt{1-cos7x}*\sqrt{\frac{1}{2}}}=\\ =\lim_{x \to 0} \frac{\sqrt{1-cos7x}}{\sqrt{\frac{1}{2}}x}=(*)\\ t=cos7x\\ x \to 0; \ \ \ t \to1\\ arccos \ t=arccos(cos7x)=7x\\ x=\frac{arccos \ t}7\\ (*)=lim_{t \to 1}\frac{\sqrt t^{\to1}}{\sqrt{\frac{1}2}\frac{arccos \ t^{\to0}}7}=\frac{1}0=\infty$