lim _{x rightarrow 0} frac{cos (m x)-cos (n x | vedclass

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Question

(C) We use the trigonometric identity $\cos A - \cos B = -2 \sin \frac{A+B}{2} \sin \frac{A-B}{2}$.$\lim _{x \rightarrow 0} \frac{\cos (m x)-\cos (n x

Vedclass · Accepted Answer

$\frac{n^2-m^2}{2}$

Vedclass · Answer

(C) We use the trigonometric identity $\cos A - \cos B = -2 \sin \frac{A+B}{2} \sin \frac{A-B}{2}$.$\lim _{x \rightarrow 0} \frac{\cos (m x)-\cos (n x)}{x^2} = \lim _{x \rightarrow 0} \frac{-2 \sin \frac{(m+n) x}{2} \sin \frac{(m-n) x}{2}}{x^2}$$= -2 \lim _{x \rightarrow 0} \left( \frac{\sin \frac{(m+n)x}{2}}{x} \right) \left( \frac{\sin \frac{(m-n)x}{2}}{x} \right)$Multiplying and dividing by $\frac{m+n}{2}$ and $\frac{m-n}{2}$ respectively:$= -2 \left( \frac{m+n}{2} \right) \left( \frac{m-n}{2} \right) \lim _{x \rightarrow 0} \left( \frac{\sin \frac{(m+n)x}{2}}{\frac{(m+n)x}{2}} \right) \left( \frac{\sin \frac{(m-n)x}{2}}{\frac{(m-n)x}{2}} \right)$$= -2 \left( \frac{m^2-n^2}{4} \right) (1)(1) = \frac{n^2-m^2}{2}$

$\lim _{x \rightarrow 0} \frac{\cos (m x)-\cos (n x)}{x^2} =$

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