BeMath-1-1-1-2-1-3-1-4-1-5-1-6-2-lim-x-0-1-sin-x-1-x-

Question Number 107352 by bemath last updated on 10/Aug/20

$$\:\:\:\:\:\:\:\:\circledcirc\mathcal{B}{e}\mathcal{M}{ath}\circledcirc \\ $$$$\left(\mathrm{1}\right)\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}}+…=? \\ $$$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+\mathrm{sin}\:{x}\right)^{\frac{\mathrm{1}}{{x}}} \:? \\ $$

Answered by Dwaipayan Shikari last updated on 10/Aug/20

2)lim_(x→0) (1+sinx)^(1/x) =y (1/x)log(1+sinx)=logy ((sinx)/x) ((log(1+sinx))/(sinx))=logy (x/x).1=logy y=e

$$\left.\mathrm{2}\right)\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+\mathrm{sinx}\right)^{\frac{\mathrm{1}}{\mathrm{x}}} =\mathrm{y} \\ $$$$\frac{\mathrm{1}}{\mathrm{x}}\mathrm{log}\left(\mathrm{1}+\mathrm{sinx}\right)=\mathrm{logy} \\ $$$$\frac{\mathrm{sinx}}{\mathrm{x}}\:\frac{\mathrm{log}\left(\mathrm{1}+\mathrm{sinx}\right)}{\mathrm{sinx}}=\mathrm{logy} \\ $$$$\frac{\mathrm{x}}{\mathrm{x}}.\mathrm{1}=\mathrm{logy} \\ $$$$\mathrm{y}=\mathrm{e} \\ $$

Answered by john santu last updated on 10/Aug/20

⋇JS⋇ (2) lim_(x→0) (1+sin x)^(1/x) = e^(lim_(x→0) (1+sin x−1).(1/x)) = e^(lim_(x→0) (((sin x)/x))) = e^1 = e

$$\:\:\:\:\:\:\:\:\divideontimes\mathcal{JS}\divideontimes \\ $$$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+\mathrm{sin}\:\mathrm{x}\right)^{\frac{\mathrm{1}}{\mathrm{x}}} =\:\mathrm{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+\mathrm{sin}\:\mathrm{x}−\mathrm{1}\right).\frac{\mathrm{1}}{\mathrm{x}}} \\ $$$$=\:\mathrm{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{x}}\right)} =\:\mathrm{e}^{\mathrm{1}} \:=\:\mathrm{e}\: \\ $$

Answered by Dwaipayan Shikari last updated on 10/Aug/20

Σ_(n=1) ^∞ (((−1)^(n+1) )/( (√n)))=0.609...

$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} }{\:\sqrt{\mathrm{n}}}=\mathrm{0}.\mathrm{609}… \\ $$$$ \\ $$

Commented by Dwaipayan Shikari last updated on 10/Aug/20

Answered by mathmax by abdo last updated on 10/Aug/20

let f(x) =(1+sinx)^(1/x) ⇒f(x) =e^((1/x)ln(1+sinx)) we have sinx ∼ ⇒1+sinx ∼1+x ⇒ln(1+sinx) ∼ln(1+x)∼x(x→0) ⇒(1/x)ln(1+sinx) ∼ 1 ⇒lim_(x→0) f(x) =e

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\left(\mathrm{1}+\mathrm{sinx}\right)^{\frac{\mathrm{1}}{\mathrm{x}}} \:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{x}}\mathrm{ln}\left(\mathrm{1}+\mathrm{sinx}\right)} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{sinx}\:\sim\:\Rightarrow\mathrm{1}+\mathrm{sinx}\:\sim\mathrm{1}+\mathrm{x}\:\Rightarrow\mathrm{ln}\left(\mathrm{1}+\mathrm{sinx}\right)\:\sim\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)\sim\mathrm{x}\left(\mathrm{x}\rightarrow\mathrm{0}\right) \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{x}}\mathrm{ln}\left(\mathrm{1}+\mathrm{sinx}\right)\:\sim\:\mathrm{1}\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{e} \\ $$

Answered by bemath last updated on 10/Aug/20

⊚BeMath⊚ (2) lim_(x→0) (1+sin x)^(1/x) = e^(lim_(x→0) ((ln (1+sin x))/x)) = e^(lim_(x→0) (((sin x−((sin^2 x)/2)+...)/x))) = e^1 = e

$$\:\:\:\:\:\circledcirc\mathcal{B}{e}\mathcal{M}{ath}\circledcirc \\ $$$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+\mathrm{sin}\:{x}\right)^{\frac{\mathrm{1}}{{x}}} =\:{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left(\mathrm{1}+\mathrm{sin}\:{x}\right)}{{x}}} \\ $$$$=\:{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{sin}\:{x}−\frac{\mathrm{sin}\:^{\mathrm{2}} {x}}{\mathrm{2}}+…}{{x}}\right)} =\:{e}^{\mathrm{1}} \:=\:{e} \\ $$

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