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Question Number 169479 by cortano1 last updated on 01/May/22

$$\underset{x\to 4}{\lim }\frac{{\left(\cos a\right)}^x-{\left(\sin a\right)}^x-\cos 2a}{x-4}=?$$

Commented by infinityaction last updated on 01/May/22

$$uselhospitalrule$$$$p\left(let\right)=\underset{x\to 0}{\lim }\frac{{\left(\cos a\right)}^x\log {}_e\cos a-{\left(\sin a\right)}^x\log {}_e\sin a}{1}$$$$p=(\cos {}^{4}a\log \cos a-\sin {}^{4}a\log \sin a)$$$$$$

Commented by cortano1 last updated on 01/May/22

$$notcorrect$$

Answered by greougoury555 last updated on 01/May/22

$$leth\left(x\right)={\left(\cos \alpha \right)}^x-{\left(\sin x\right)}^x$$$$h\left(4\right)={\left(\cos \alpha \right)}^4-{\left(\sin \alpha \right)}^4=\cos 2\alpha$$$$L=\underset{x\to 4}{\lim }\frac{{\left(\cos \alpha \right)}^x-{\left(\sin \alpha \right)}^x-\cos 2\alpha }{x-4}$$$$L=\underset{x\to 4}{\lim }\frac{h\left(x\right)-h\left(4\right)}{x-4}=h{}^{\prime }\left(4\right)$$$$h^{\prime }\left(x\right)=\frac{d}{dx}[{\left(\cos \alpha \right)}^x-{\left(\sin \alpha \right)}^x]$$$$h^{\prime }\left(x\right)={\left(\cos \alpha \right)}^x\ln \left(\cos \alpha \right)-{\left(\sin \alpha \right)}^x\ln \left(\sin \alpha \right)$$$$L=h^{\prime }\left(4\right)={\left(\cos \alpha \right)}^4\ln \left(\cos \alpha \right)-{\left(\sin \alpha \right)}^4\ln \left(\sin \alpha \right)$$

Answered by som(math1967) last updated on 01/May/22

$$\underset{x\to 4}{lim}\frac{{\left(cosa\right)}^x-{\left(sina\right)}^x-{cos}^{4}a+{sin}^{4}a}{(x-4)}$$$$\underset{x\to 4}{lim}\frac{{cos}^{4}a\{{\left(cosa\right)}^{x-4}-1\}}{(x-4)}$$$$-\underset{x\to 4}{lim}\frac{{sin}^{4}a\{{\left(sina\right)}^{x-4}-1\}}{(x-4)}$$$$letx-4=z\Rightarrow x\to 4\therefore (x-4)\to 0$$$$\therefore {cos}^{4}a\underset{z\to 0}{lim}\frac{{\left(cosa\right)}^z-1}{z}$$$$-{sin}^{4}a\underset{z\to 0}{lim}\frac{{\left(sina\right)}^z-1}{z}$$$${cos}^{4}alncosa-{sin}^{4}alnsina$$

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