lim _{x rightarrow 0} frac{cos ax - cos bx}{x | vedclass

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(B) Using the trigonometric identity $\cos C - \cos D = -2 \sin \left( \frac{C+D}{2} \right) \sin \left( \frac{C-D}{2} \right)$,we have: $\lim _{x \ri

Vedclass · Accepted Answer

$\frac{b^{2} - a^{2}}{2}$

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(B) Using the trigonometric identity $\cos C - \cos D = -2 \sin \left( \frac{C+D}{2} ight) \sin \left( \frac{C-D}{2} ight)$,we have: $\lim _{x ightarrow 0} \frac{\cos ax - \cos bx}{x^{2}} = \lim _{x ightarrow 0} \frac{-2 \sin \left( \frac{ax+bx}{2} ight) \sin \left( \frac{ax-bx}{2} ight)}{x^{2}}$ $= -2 \lim _{x ightarrow 0} \left( \frac{\sin \left( \frac{a+b}{2} x ight)}{x} ight) \left( \frac{\sin \left( \frac{a-b}{2} x ight)}{x} ight)$ Using the standard limit $\lim _{	heta ightarrow 0} \frac{\sin k	heta}{	heta} = k$,we get: $= -2 \left( \frac{a+b}{2} ight) \left( \frac{a-b}{2} ight) = -2 \left( \frac{a^{2} - b^{2}}{4} ight) = \frac{b^{2} - a^{2}}{2}$

$\lim _{x \rightarrow 0} \frac{\cos ax - \cos bx}{x^{2}}$ is equal to

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