Question

302) по методу первого замечательного предела

Answer 1

$limlimits _{x to pi /6}, frac{1-2sinx}{pi /6-x}=limlimits _{x to pi /6}frac{2cdot (frac{1}{2}-sinx)}{pi /6-x}=limlimits _{x to pi /6}frac{2cdot (sinfrac{pi}{6}-sinx)}{pi /6-x}=\\=limlimits _{x to pi /6}frac{2cdot 2, sin(frac{pi}{12}-frac{x}{2})cdot cos(frac{pi}{12}+frac{x}{2})}{pi /6-x}=\\=Big[; sin(frac{pi}{12}-frac{x}{2})sim (frac{pi}{12}-frac{x}{2}); ,; tak; kak; ; (frac{pi}{12}-frac{x}{2})to 0; ,; esli; ; xto frac{pi}{6}; Big]=$

$=limlimits _{x to pi /6}frac{2cdot 2, (frac{pi}{12}-frac{x}{2})cdot cos(frac{pi}{12}+frac{x}{2})}{frac{pi}{6}-x}=limlimits _{x to pi /6}frac{2cdot (frac{pi}{6}-x)cdot cos(frac{pi}{12}+frac{x}{2})}{frac{pi}{6}-x}=\\=2cdot limlimits _{x to pi /6}; cos(frac{pi}{12}+frac{x}{2})=2cdot cos(frac{pi}{12}+frac{pi}{12})=2cdot cosfrac{pi}{6}=2cdot frac{sqrt3}{2}=sqrt3$

$P.S.; ; ; limlimits _{alpha (x) to 0}frac{sin,alpha (x)}{alpha (x)}=1; ; Rightarrow ; ; ; sin,alpha (x)sim alpha (x)$

Answer 2

спасибо большое за подробности)

Answer 3

Ответ: во вложении  Объяснение: