Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 213821 by hardmath last updated on 17/Nov/24

Find: lim_(x→0) (((sinx)/x))^((sinx)/(x − sinx)) = ?

$$\mathrm{Find}:\:\:\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{sinx}}{\mathrm{x}}\right)^{\frac{\mathrm{sinx}}{\mathrm{x}\:−\:\mathrm{sinx}}} \:\:=\:\:? \\ $$

Answered by mehdee7396 last updated on 17/Nov/24

lim_(x→0) (((sinx)/x)−1)((sinx)/(x−sinx)) =lim_(x→0) (((sinx−x)/x))((sinx)/(x−sinx))=−1 ⇒answer=e^(−1)

$${lim}_{{x}\rightarrow\mathrm{0}} \left(\frac{{sinx}}{{x}}−\mathrm{1}\right)\frac{{sinx}}{{x}−{sinx}} \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}} \left(\frac{{sinx}−{x}}{{x}}\right)\frac{{sinx}}{{x}−{sinx}}=−\mathrm{1} \\ $$$$\Rightarrow{answer}={e}^{−\mathrm{1}} \: \\ $$$$ \\ $$

Answered by lepuissantcedricjunior last updated on 20/Nov/24

lim_(x→0) (((sinx)/x))^((sinx)/(x−sinx)) =1^(FI) =>lim_(x→0) (((sinx)/x))^((sinx)/(x−sinx)) =lim_(x→0) e^((sinx ln(((sinx)/x)))/(x−sinx)) dl sinx a^� l′ordre 3 au vois de 0 sinx=Σ_(k=0) ^n (x^k /(k!))×f^k (0)=x−(x^3 /6)+x(0)^3 lim_(x→0) e^(((x−(x^3 /6))ln(1−(x^2 /6)))/(x−x+(x^3 /6))) or dl ln(1−t) a^� l′ordre 3 au vois de 0 donne ln(1−t)=−t+(t^2 /2)−(t^3 /3)+t(0)^3 =>lim_(x→0) (((sinx)/x))^((sinx)/(x−sinx)) =lim_(x→0) e^(((x−(x^3 /6))(−(x^2 /6)+(x^4 /(12))))/(x^3 /6)) =lim_(x→0) e^((−(x^3 /6))/(x^3 /6)) =e^(−1) .......................................................... prof cedric junior.....

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\boldsymbol{{sinx}}}{\boldsymbol{{x}}}\right)^{\frac{\boldsymbol{{sinx}}}{\boldsymbol{{x}}−\boldsymbol{{sinx}}}} =\mathrm{1}^{\boldsymbol{{FI}}} \\ $$$$=>\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\boldsymbol{{sinx}}}{\boldsymbol{{x}}}\right)^{\frac{\boldsymbol{{sinx}}}{\boldsymbol{{x}}−\boldsymbol{{sinx}}}} =\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\boldsymbol{{e}}^{\frac{\boldsymbol{{sinx}}\:\boldsymbol{{ln}}\left(\frac{\boldsymbol{{sinx}}}{\boldsymbol{{x}}}\right)}{\boldsymbol{{x}}−\boldsymbol{{sinx}}}} \\ $$$$\boldsymbol{{dl}}\:\boldsymbol{{sinx}}\:\grave {\boldsymbol{{a}}}\:\boldsymbol{{l}}'{o}\boldsymbol{{rdre}}\:\mathrm{3}\:\:\boldsymbol{{au}}\:\boldsymbol{{vois}}\:\boldsymbol{{de}}\:\mathrm{0} \\ $$$$\boldsymbol{{sinx}}=\underset{\boldsymbol{{k}}=\mathrm{0}} {\overset{\boldsymbol{{n}}} {\sum}}\frac{\boldsymbol{{x}}^{\boldsymbol{{k}}} }{\boldsymbol{{k}}!}×\boldsymbol{{f}}^{\boldsymbol{{k}}} \left(\mathrm{0}\right)=\boldsymbol{{x}}−\frac{\boldsymbol{{x}}^{\mathrm{3}} }{\mathrm{6}}+\boldsymbol{{x}}\left(\mathrm{0}\right)^{\mathrm{3}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\boldsymbol{{e}}^{\frac{\left(\boldsymbol{{x}}−\frac{\boldsymbol{{x}}^{\mathrm{3}} }{\mathrm{6}}\right)\boldsymbol{{ln}}\left(\mathrm{1}−\frac{\boldsymbol{{x}}^{\mathrm{2}} }{\mathrm{6}}\right)}{\boldsymbol{{x}}−\boldsymbol{{x}}+\frac{\boldsymbol{{x}}^{\mathrm{3}} }{\mathrm{6}}}} \boldsymbol{{or}}\:\boldsymbol{{dl}}\:\boldsymbol{{ln}}\left(\mathrm{1}−\boldsymbol{{t}}\right)\:\grave {\boldsymbol{{a}}}\:\boldsymbol{{l}}'\boldsymbol{{ordre}}\:\mathrm{3}\:\boldsymbol{{au}}\:\boldsymbol{{vois}}\:\boldsymbol{{de}}\:\mathrm{0}\:\boldsymbol{{donne}} \\ $$$$\boldsymbol{{ln}}\left(\mathrm{1}−\boldsymbol{{t}}\right)=−\boldsymbol{{t}}+\frac{\boldsymbol{{t}}^{\mathrm{2}} }{\mathrm{2}}−\frac{\boldsymbol{{t}}^{\mathrm{3}} }{\mathrm{3}}+\boldsymbol{{t}}\left(\mathrm{0}\right)^{\mathrm{3}} \\ $$$$=>\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\boldsymbol{{sinx}}}{\boldsymbol{{x}}}\right)^{\frac{\boldsymbol{{sinx}}}{\boldsymbol{{x}}−\boldsymbol{{sinx}}}} =\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\boldsymbol{{e}}^{\frac{\left(\boldsymbol{{x}}−\frac{\boldsymbol{{x}}^{\mathrm{3}} }{\mathrm{6}}\right)\left(−\frac{\boldsymbol{{x}}^{\mathrm{2}} }{\mathrm{6}}+\frac{\boldsymbol{{x}}^{\mathrm{4}} }{\mathrm{12}}\right)}{\frac{\boldsymbol{{x}}^{\mathrm{3}} }{\mathrm{6}}}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\boldsymbol{{e}}^{\frac{−\frac{\boldsymbol{{x}}^{\mathrm{3}} }{\mathrm{6}}}{\frac{\boldsymbol{{x}}^{\mathrm{3}} }{\mathrm{6}}}} =\boldsymbol{{e}}^{−\mathrm{1}} \\ $$$$.......................................................... \\ $$$${prof}\:{cedric}\:{junior}..... \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com