Якія вынікі мела барацьба супраць крыжакоў на беларускіх землях у XIV ст.?

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Ответ:

vladkoblov02

02.06.2020 05:09

$\lim_{x\to\frac{\pi}{3}}\frac{\tan^3 x-3\tan x}{\cos(x+\frac{\pi}{6})}=$

$=\lim_{x\to\frac{\pi}{3}}\frac{\tan x(\tan^2 x-3)}{\cos(x+\frac{\pi}{6})}=\lim_{x\to\frac{\pi}{3}}\frac{\tan x(\tan x+\sqrt{3})*(\tan x-\sqrt{3})}{\cos(x+\frac{\pi}{6})}=$

по свойству пределов $\lim_{x\to c}(a*b)=\lim_{x\to c}a*\lim_{x\to c}b$

$=\lim_{x\to\frac{\pi}{3}}\tan x(\tan x+\sqrt{3})*\lim_{x\to\frac{\pi}{3}}\frac{(\tan x-\sqrt{3})}{\cos(x+\frac{\pi}{6})}=$

$=\tan \frac{\pi}{3}(\tan \frac{\pi}{3}+\sqrt{3})*\lim_{x\to\frac{\pi}{3}}\frac{(\tan x-\sqrt{3})}{\cos(x+\frac{\pi}{6})}=$

$=\sqrt{3}(\sqrt{3}+\sqrt{3})*\lim_{x\to\frac{\pi}{3}}\frac{(\tan x-\sqrt{3})}{\cos(x+\frac{\pi}{6})}=$

$=\sqrt{3}*2\sqrt{3}*\lim_{x\to\frac{\pi}{3}}\frac{(\tan x-\sqrt{3})}{\cos(x+\frac{\pi}{6})}=$

$=6*\lim_{x\to\frac{\pi}{3}}\frac{(\tan x-\sqrt{3})}{\cos(x+\frac{\pi}{6})}=$

Замена $t=x-\frac{\pi}{3}$ при $x\to\frac{\pi}{3}$ . Тогда $t\to 0.$

$x=t+\frac{\pi}{3}$

$=6*\lim_{t\to 0}\frac{\tan (t+\frac{\pi}{3})-\sqrt{3}}{\cos(t+\frac{\pi}{3}+\frac{\pi}{6})}=$

$=6*\lim_{t\to 0}\frac{\tan (t+\frac{\pi}{3})-\sqrt{3}}{\cos(t+\frac{\pi}{2})}=$

Преобразуем знаменатель. Получим косинус суммы
cos (a+b)=cos a*cos b-sin a sin b
$\cos(t+\frac{\pi}{2})=\cos t\cos\frac{\pi}{2}-\sin t\sin\frac{\pi}{2}=\cos t*0-\sin t*1=-\sin t$

Перепишем предел в новом виде

$=6*\lim_{t\to 0}\frac{\tan (t+\frac{\pi}{3})-\sqrt{3}}{-\sin t}=6*\lim_{t\to 0}\frac{\tan (t+\frac{\pi}{3})-\tan\frac{\pi}{3}}{-\sin t}=$
Воспользуемся формулой разности тангенсов, чтобы преобразовать числитель

$\tan(a-b)=\frac{\sin(a-b)}{\cos a\cos b}$

$=6*\lim_{t\to 0}\frac{\tan (t+\frac{\pi}{3})-\tan\frac{\pi}{3}}{-\sin t}=$

$=6*\lim_{t\to 0}\frac{\tan (t+\frac{\pi}{3})-\tan\frac{\pi}{3}}{-\sin t}=6*\lim_{t\to 0}\frac{\sin(t+\frac{\pi}{3}-\frac{\pi}{3})}{\cos(t+\frac{\pi}{3})\cos\frac{\pi}{3}}*\frac{1}{-\sin t}=$

$=6*\lim_{t\to 0}\frac{\sin t}{\cos(t+\frac{\pi}{3})\cos\frac{\pi}{3}}*\frac{1}{-\sin t}=$

$=6*\lim_{t\to 0}\frac{\sin t}{\frac{1}{2}*\frac{1}{2}}*\frac{1}{-\sin t}=6*4\lim_{t\to 0}\frac{\sin t}{-\sin t}=-24.$