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Question Number 168576 by cortano1 last updated on 13/Apr/22

$$If\underset{x\to 0}{\lim }\frac{\cos {}^m\left(mx\right)-\cos {}^n\left(nx\right)}{(m^2+n^2+mn)x^2}=1$$$$find\frac{m^2+n^2-4}{mn}.$$

Commented by blackmamba last updated on 17/Apr/22

$$m^2+n^2+mn=\underset{x\to 0}{\lim }\frac{\cos {}^m\left(mx\right)-\cos {}^n\left(nx\right)}{x^2}$$$$m^2+n^2+mn=\frac{n^3-m^3}{2}$$$$\frac{n^3-m^3}{(n-m)}=\frac{n^3-m^3}{2}$$$$(n^3-m^3)(\frac{1}{n-m}-\frac{1}{2})=0$$$$n-m=2\Rightarrow n^2-2mn+m^2=4$$$$\Rightarrow m^2+n^2-4=2mn$$$$\Rightarrow \frac{m^2+n^2-4}{mn}=2$$

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