Question

꧁ HOLA BRAINLIANS ꧂
꧁ Qᴜᴇsᴛɪᴏɴ ꧂

\frac{lim}{x - > 0} \: \: \frac{cos \: ax \: - \: cos \: bx}{cos \: cx \: - \: cos \: dx}

​

__
SOLVE IT PLEASE ​

Answer 1

$\sf \Large Solution!!$

$\sf \to \displaystyle \lim _{x\to 0}\;\dfrac{\cos ax-\cos bx}{\cos cx-\cos dx}$

$\sf \large As\:we\:know$

$\sf \to \cos A-\cos B=-2\sin (\dfrac{A+B}{2})\sin (\dfrac{A-B}{2})$

$\sf \large So,$

$\sf \to \displaystyle \lim _{x\to 0}\;\dfrac{-2\sin (\dfrac{a+b}{2}x)\sin (\dfrac{a-b}{2}x)}{-2\sin (\dfrac{c+d}{2}x)\sin (\dfrac{c-d}{2}x)}$

$\sf \to \displaystyle \lim _{x\to 0}\;\dfrac{\dfrac{\sin (\dfrac{a+b}{2}x)}{(\dfrac{a+b}{2}x)}\times (\dfrac{a+b}{2}x)\times \dfrac{\sin (\dfrac{a-b}{2}x)}{(\dfrac{a-b}{2}x)}\times (\dfrac{a-b}{2}x)}{\dfrac{\sin (\dfrac{a+b}{2}x)}{(\dfrac{c+d}{2}x)}\times (\dfrac{c+d}{2}x)\times \dfrac{\sin (\dfrac{c-d}{2}x)}{(\dfrac{c-d}{2}x)}\times (\dfrac{c-d}{2}x)}$

$\sf \to \displaystyle \lim _{x\to 0}\;\dfrac{(1)(\dfrac{a+b}{2})(\dfrac{a-b}{2})(1)}{(1)(\dfrac{c+d}{2})(1)(\dfrac{c-d}{2})}$

$\sf \to \displaystyle \lim _{x\to 0}\;\dfrac{(a+b)(a-b)}{4}\times \dfrac{4}{(c+d)(c-d)}$

$\sf \to \displaystyle \lim _{x\to 0}\;\dfrac{(a+b)(a-b)}{(c+d)(c-d)}$

$\sf \to \displaystyle \lim _{x\to 0}\;\dfrac{a^{2}-b^{2}}{c^{2}-d^{2}}$

$\sf \therefore \displaystyle \lim _{x\to 0}\;\dfrac{\cos ax-\cos bx}{\cos cx-\cos dx}=\dfrac{a^{2}-b^{2}}{c^{2}-d^{2}}$

Thanks for asking!! :)

Answer 2

Step-by-step explanation:

\frac{lim}{x - > 0} \: \: \frac{cos \: ax \: - \: cos \: bx}{cos \: cx \: - \: cos \: dx}