Question Number 169479 by cortano1 last updated on 01/May/22
$$\:\:\:\underset{{x}\rightarrow\mathrm{4}} {\mathrm{lim}}\:\frac{\left(\mathrm{cos}\:{a}\right)^{{x}} −\left(\mathrm{sin}\:{a}\right)^{{x}} −\mathrm{cos}\:\mathrm{2}{a}}{{x}−\mathrm{4}}\:=? \\ $$
Commented by infinityaction last updated on 01/May/22
$${use}\:{l}\:{hospital}\:{rule} \\ $$$$\:\:\:\:\:{p}\left({let}\right)\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{cos}\:{a}\right)^{{x}} \mathrm{log}\:_{{e}} \mathrm{cos}\:{a}−\left(\mathrm{sin}\:{a}\right)^{{x}} \mathrm{log}\:_{{e}} \mathrm{sin}\:{a}}{\mathrm{1}} \\ $$$$\:\:\:\:\:\:{p}\:\:\:=\:\:\left(\mathrm{cos}\:^{\mathrm{4}} {a}\:\mathrm{log}\:\mathrm{cos}\:{a}\:−\:\mathrm{sin}\:^{\mathrm{4}} {a}\mathrm{log}\:\mathrm{sin}\:{a}\right) \\ $$$$\:\:\:\: \\ $$
Commented by cortano1 last updated on 01/May/22
$${not}\:{correct} \\ $$
Answered by greougoury555 last updated on 01/May/22
![let h(x)= (cos α)^x −(sin x)^x h(4)= (cos α)^4 −(sin α)^4 = cos 2α L= lim_(x→4) (((cos α)^x −(sin α)^x −cos 2α)/(x−4)) L= lim_(x→4) ((h(x)−h(4))/(x−4)) = h ′(4) h′(x)= (d/dx) [ (cos α)^x −(sin α)^x ] h′(x) = (cos α)^x ln (cos α)−(sin α)^x ln (sin α) L= h′(4)=(cos α)^4 ln (cos α) −(sin α)^4 ln (sin α)](https://www.tinkutara.com/question/Q169500.png)
$$\:{let}\:{h}\left({x}\right)=\:\left(\mathrm{cos}\:\alpha\right)^{{x}} −\left(\mathrm{sin}\:{x}\right)^{{x}} \\ $$$$\:{h}\left(\mathrm{4}\right)=\:\left(\mathrm{cos}\:\alpha\right)^{\mathrm{4}} −\left(\mathrm{sin}\:\alpha\right)^{\mathrm{4}} =\:\mathrm{cos}\:\mathrm{2}\alpha \\ $$$$\:{L}=\:\underset{{x}\rightarrow\mathrm{4}} {\mathrm{lim}}\:\frac{\left(\mathrm{cos}\:\alpha\right)^{{x}} −\left(\mathrm{sin}\:\alpha\right)^{{x}} −\mathrm{cos}\:\mathrm{2}\alpha}{{x}−\mathrm{4}} \\ $$$$\:{L}=\:\underset{{x}\rightarrow\mathrm{4}} {\mathrm{lim}}\:\frac{{h}\left({x}\right)−{h}\left(\mathrm{4}\right)}{{x}−\mathrm{4}}\:=\:{h}\:'\left(\mathrm{4}\right) \\ $$$$\:{h}'\left({x}\right)=\:\frac{{d}}{{dx}}\:\left[\:\left(\mathrm{cos}\:\alpha\right)^{{x}} −\left(\mathrm{sin}\:\alpha\right)^{{x}} \:\right] \\ $$$$\:{h}'\left({x}\right)\:=\:\left(\mathrm{cos}\:\alpha\right)^{{x}} \:\mathrm{ln}\:\left(\mathrm{cos}\:\alpha\right)−\left(\mathrm{sin}\:\alpha\right)^{{x}} \:\mathrm{ln}\:\left(\mathrm{sin}\:\alpha\right) \\ $$$$\:{L}=\:{h}'\left(\mathrm{4}\right)=\left(\mathrm{cos}\:\alpha\right)^{\mathrm{4}} \:\mathrm{ln}\:\left(\mathrm{cos}\:\alpha\right)\:−\left(\mathrm{sin}\:\alpha\right)^{\mathrm{4}} \:\mathrm{ln}\:\left(\mathrm{sin}\:\alpha\right)\: \\ $$
Answered by som(math1967) last updated on 01/May/22
$$\:\underset{{x}\rightarrow\mathrm{4}} {{lim}}\frac{\left({cosa}\right)^{{x}} −\left({sina}\right)^{{x}} −{cos}^{\mathrm{4}} {a}+{sin}^{\mathrm{4}} {a}}{\left({x}−\mathrm{4}\right)} \\ $$$$\underset{{x}\rightarrow\mathrm{4}} {{lim}}\frac{{cos}^{\mathrm{4}} {a}\left\{\left({cosa}\right)^{{x}−\mathrm{4}} −\mathrm{1}\right\}}{\left({x}−\mathrm{4}\right)}\: \\ $$$$\:\:\:\:\:\:\:\:\:−\:\:\:\:\:\underset{{x}\rightarrow\mathrm{4}} {{lim}}\frac{{sin}^{\mathrm{4}} {a}\left\{\left({sina}\right)^{{x}−\mathrm{4}} −\mathrm{1}\right\}}{\left({x}−\mathrm{4}\right)} \\ $$$${let}\:{x}−\mathrm{4}={z}\:\Rightarrow{x}\rightarrow\mathrm{4}\therefore\left({x}−\mathrm{4}\right)\rightarrow\mathrm{0} \\ $$$$\:\therefore\:{cos}^{\mathrm{4}} {a}\:\underset{{z}\rightarrow\mathrm{0}} {{lim}}\frac{\left({cosa}\right)^{{z}} −\mathrm{1}}{{z}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:−{sin}^{\mathrm{4}} {a}\:\underset{{z}\rightarrow\mathrm{0}} {{lim}}\frac{\left({sina}\right)^{{z}} −\mathrm{1}}{{z}} \\ $$$${cos}^{\mathrm{4}} {alncosa}\:−{sin}^{\mathrm{4}} {alnsina} \\ $$