Question Number 30502 by abdo imad last updated on 22/Feb/18
$${find}\:\:{lim}_{{x}\rightarrow\mathrm{0}} \left({sinx}\:+{cosx}\right)^{\frac{\mathrm{1}}{{x}}} \:\:. \\ $$
Answered by sma3l2996 last updated on 24/Feb/18
$$\left({sinx}+{cosx}\right)^{\mathrm{1}/{x}} ={e}^{\frac{{ln}\left({sinx}+{cosx}\right)}{{x}}} \\ $$$${sin}\left({x}\right)\underset{\mathrm{0}} {\sim}{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}\:\:{and}\:\:{cos}\left({x}\right)\underset{\mathrm{0}} {\sim}\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${so}\:\:{sinx}+{cosx}\underset{\mathrm{0}} {\sim}\mathrm{1}+{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}} \\ $$$${ln}\left({cosx}+{sinx}\right)\underset{\mathrm{0}} {\sim}{ln}\left(\mathrm{1}+{x}−{x}^{\mathrm{2}} /\mathrm{2}−{x}^{\mathrm{3}} /\mathrm{6}\right)\underset{\mathrm{0}} {\sim}{x}−{x}^{\mathrm{2}} /\mathrm{2}−{x}^{\mathrm{3}} /\mathrm{6} \\ $$$$\frac{{ln}\left({sinx}+{cosx}\right)}{{x}}\underset{\mathrm{0}} {\sim}\mathrm{1} \\ $$$${so}\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left({sinx}+{cosx}\right)^{\mathrm{1}/{x}} ={e} \\ $$