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Question Number 30502 by abdo imad last updated on 22/Feb/18
find  lim_(x→0) (sinx +cosx)^(1/x)   .
$${find}\:\:{lim}_{{x}\rightarrow\mathrm{0}} \left({sinx}\:+{cosx}\right)^{\frac{\mathrm{1}}{{x}}} \:\:. \\ $$
Answered by sma3l2996 last updated on 24/Feb/18
(sinx+cosx)^(1/x) =e^((ln(sinx+cosx))/x)   sin(x)∼_0 x−(x^3 /(3!))  and  cos(x)∼_0 1−(x^2 /2)  so  sinx+cosx∼_0 1+x−(x^2 /2)−(x^3 /6)  ln(cosx+sinx)∼_0 ln(1+x−x^2 /2−x^3 /6)∼_0 x−x^2 /2−x^3 /6  ((ln(sinx+cosx))/x)∼_0 1  so lim_(x→0) (sinx+cosx)^(1/x) =e
$$\left({sinx}+{cosx}\right)^{\mathrm{1}/{x}} ={e}^{\frac{{ln}\left({sinx}+{cosx}\right)}{{x}}} \\ $$$${sin}\left({x}\right)\underset{\mathrm{0}} {\sim}{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}\:\:{and}\:\:{cos}\left({x}\right)\underset{\mathrm{0}} {\sim}\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${so}\:\:{sinx}+{cosx}\underset{\mathrm{0}} {\sim}\mathrm{1}+{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}} \\ $$$${ln}\left({cosx}+{sinx}\right)\underset{\mathrm{0}} {\sim}{ln}\left(\mathrm{1}+{x}−{x}^{\mathrm{2}} /\mathrm{2}−{x}^{\mathrm{3}} /\mathrm{6}\right)\underset{\mathrm{0}} {\sim}{x}−{x}^{\mathrm{2}} /\mathrm{2}−{x}^{\mathrm{3}} /\mathrm{6} \\ $$$$\frac{{ln}\left({sinx}+{cosx}\right)}{{x}}\underset{\mathrm{0}} {\sim}\mathrm{1} \\ $$$${so}\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left({sinx}+{cosx}\right)^{\mathrm{1}/{x}} ={e} \\ $$

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