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Question Number 137773 by Ajadiazeemolamilekan last updated on 06/Apr/21

$$\underset{x\to 0}{\lim }\frac{\cos px-\cos qx}{x^2}$$

Answered by bemath last updated on 06/Apr/21

$$\underset{x\to 0}{\lim }\frac{-2\sin \left(\frac{p+q}{2}x\right)\sin \left(\frac{p-q}{2}x\right)}{x^2}$$$$=-2\left(\frac{p+q}{2}\right)\left(\frac{p-q}{2}\right)=-\frac{p^2-q^2}{2}$$

Answered by mathmax by abdo last updated on 07/Apr/21

$$\cos \left(\mathrm{px}\right)\sim 1-\frac{{\mathrm{p}}^2{\mathrm{x}}^2}{2}\mathrm{and}\cos \left(\mathrm{qx}\right)\sim 1-\frac{{\mathrm{q}}^2{\mathrm{x}}^2}{2}\Rightarrow$$$$\frac{\cos \left(\mathrm{px}\right)-\cos \left(\mathrm{qx}\right)}{{\mathrm{x}}^2}\sim \frac{1-\frac{{\mathrm{p}}^2{\mathrm{x}}^2}{2}-1+\frac{{\mathrm{q}}^2{\mathrm{x}}^2}{2}}{{\mathrm{x}}^2}=\frac{{\mathrm{q}}^2-{\mathrm{p}}^2}{2}\Rightarrow$$$${\lim }_{\mathrm{x}\to 0}\frac{\cos \left(\mathrm{px}\right)-\cos \left(\mathrm{qx}\right)}{{\mathrm{x}}^2}=\frac{{\mathrm{q}}^2-{\mathrm{p}}^2}{2}$$

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