Question-145491

Question Number 145491 by Mrsof last updated on 05/Jul/21

Commented by Mrsof last updated on 05/Jul/21

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Answered by puissant last updated on 05/Jul/21

lim_(x→0) e^((cos(x)−1)ln(tan(x))) =lim_(x→0) e^(−(x^2 /2)×x) =lim_(x→0) e^(−(x^3 /2)) =1 True

$$\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\mathrm{e}^{\left(\mathrm{cos}\left(\mathrm{x}\right)−\mathrm{1}\right)\mathrm{ln}\left(\mathrm{tan}\left(\mathrm{x}\right)\right)} \\ $$$$=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{e}^{−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}×\mathrm{x}} =\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\mathrm{e}^{−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{2}}} =\mathrm{1} \\ $$$$\mathrm{True} \\ $$

Answered by mathmax by abdo last updated on 05/Jul/21

f(x)=(tanx)^(cosx−1) ⇒f(x)=e^((cosx−1)log(tanx)) cosx∼1−(x^2 /2) ⇒cosx−1∼−(x^2 /2) ⇒ (cosx−1)log(tanx)∼−(x^2 /2)log(tanx)∼−(x^2 /2)logx →0 ⇒ lim_(x→+∞) f(x)=e^0 =1

$$\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{tanx}\right)^{\mathrm{cosx}−\mathrm{1}} \:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\mathrm{e}^{\left(\mathrm{cosx}−\mathrm{1}\right)\mathrm{log}\left(\mathrm{tanx}\right)} \\ $$$$\mathrm{cosx}\sim\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\mathrm{cosx}−\mathrm{1}\sim−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow \\ $$$$\left(\mathrm{cosx}−\mathrm{1}\right)\mathrm{log}\left(\mathrm{tanx}\right)\sim−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\mathrm{log}\left(\mathrm{tanx}\right)\sim−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\mathrm{logx}\:\rightarrow\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \mathrm{f}\left(\mathrm{x}\right)=\mathrm{e}^{\mathrm{0}} \:=\mathrm{1} \\ $$

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