Question

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$\frac{lim}{x - > 0} \: \: \frac{cos \: ax \: - \: cos \: bx}{cos \: cx \: - \: cos \: dx}$
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Answer 1

$\sf \Large Solution!!$

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

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

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

$\sf \large So,$

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

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

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

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

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

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

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

Thanks for asking!! :)

Answer 2

$\sf \Large Solution!!$

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

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

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

$\sf \large So,$

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

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

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

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

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

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

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

Thanks for asking!! :)

Step-by-step explanation:

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