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Question-217733

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Question Number 217733 by Samuel12 last updated on 19/Mar/25
Answered by vnm last updated on 19/Mar/25
x^(1/(ln (e^x −1))) =e^((ln x)/(ln (e^x −1))) =e^((ln x)/(ln x+ln ((e^x −1)/x))) = e^(1/(1+(1/(ln x))ln ((e^x −1)/x))) →_(x→0+) e^(1/(1+(1/(−∞))∙ln 1)) =e
$${x}^{\frac{\mathrm{1}}{\mathrm{ln}\:\left({e}^{{x}} −\mathrm{1}\right)}} ={e}^{\frac{\mathrm{ln}\:{x}}{\mathrm{ln}\:\left({e}^{{x}} −\mathrm{1}\right)}} ={e}^{\frac{\mathrm{ln}\:{x}}{\mathrm{ln}\:{x}+\mathrm{ln}\:\frac{{e}^{{x}} −\mathrm{1}}{{x}}}} = \\ $$$${e}^{\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{ln}\:{x}}\mathrm{ln}\:\frac{{e}^{{x}} −\mathrm{1}}{{x}}}} \:\:\:\underset{{x}\rightarrow\mathrm{0}+} {\rightarrow}\:\:\:{e}^{\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{−\infty}\centerdot\mathrm{ln}\:\mathrm{1}}} ={e} \\ $$$$ \\ $$
Answered by vnm last updated on 19/Mar/25
(ln (1+e^(−x) ))^(1/x) =((1/e^x )ln (1+e^(−x) )^e^x )^(1/x) = e^(−1) (ln (1+(1/e^x ))^e^x )^(1/x) →_(x→+∞) e^(−1) (ln e)^0 =e^(−1)
$$\left(\mathrm{ln}\:\left(\mathrm{1}+{e}^{−{x}} \right)\right)^{\frac{\mathrm{1}}{{x}}} =\left(\frac{\mathrm{1}}{{e}^{{x}} }\mathrm{ln}\:\left(\mathrm{1}+{e}^{−{x}} \right)^{{e}^{{x}} } \right)^{\frac{\mathrm{1}}{{x}}} = \\ $$$${e}^{−\mathrm{1}} \left(\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{e}^{{x}} }\right)^{{e}^{{x}} } \right)^{\frac{\mathrm{1}}{{x}}} \underset{{x}\rightarrow+\infty} {\rightarrow}{e}^{−\mathrm{1}} \left(\mathrm{ln}\:{e}\right)^{\mathrm{0}} ={e}^{−\mathrm{1}} \\ $$$$ \\ $$
Answered by vnm last updated on 20/Mar/25
((e^x −e^2 )/(x^2 +x−6)) =_(x=2+t) ((e^2 (e^t −1))/((2+t)^2 +2+t−6))=((e^2 (e^t −1))/(t(t+5)))= (e^2 /(t+5))∙((e^t −1)/t) →_(t→0) (e^2 /5)∙1=(e^2 /5) ((tan x−sin x)/(sin x (cos 2x−cos x)))=(((1/(cos x))−1)/(cos 2x−cos x))= ((1−cos x)/(cos x (2cos^2 x−1−cos x)))= (1/(cos x))∙((1−cos x)/((2cos x+1)(cos x−1)))= (1/(cos x))∙((−1)/(2cos x+1)) →_(x→0) −(1/3)
$$\frac{{e}^{{x}} −{e}^{\mathrm{2}} }{{x}^{\mathrm{2}} +{x}−\mathrm{6}}\:\:\underset{{x}=\mathrm{2}+{t}} {=}\:\:\frac{{e}^{\mathrm{2}} \left({e}^{{t}} −\mathrm{1}\right)}{\left(\mathrm{2}+{t}\right)^{\mathrm{2}} +\mathrm{2}+{t}−\mathrm{6}}=\frac{{e}^{\mathrm{2}} \left({e}^{{t}} −\mathrm{1}\right)}{{t}\left({t}+\mathrm{5}\right)}= \\ $$$$\frac{{e}^{\mathrm{2}} }{{t}+\mathrm{5}}\centerdot\frac{{e}^{{t}} −\mathrm{1}}{{t}}\:\underset{{t}\rightarrow\mathrm{0}} {\rightarrow}\:\frac{{e}^{\mathrm{2}} }{\mathrm{5}}\centerdot\mathrm{1}=\frac{{e}^{\mathrm{2}} }{\mathrm{5}} \\ $$$$ \\ $$$$\frac{\mathrm{tan}\:{x}−\mathrm{sin}\:{x}}{\mathrm{sin}\:{x}\:\left(\mathrm{cos}\:\mathrm{2}{x}−\mathrm{cos}\:{x}\right)}=\frac{\frac{\mathrm{1}}{\mathrm{cos}\:{x}}−\mathrm{1}}{\mathrm{cos}\:\mathrm{2}{x}−\mathrm{cos}\:{x}}= \\ $$$$\frac{\mathrm{1}−\mathrm{cos}\:{x}}{\mathrm{cos}\:{x}\:\left(\mathrm{2cos}^{\mathrm{2}} {x}−\mathrm{1}−\mathrm{cos}\:{x}\right)}= \\ $$$$\frac{\mathrm{1}}{\mathrm{cos}\:{x}}\centerdot\frac{\mathrm{1}−\mathrm{cos}\:{x}}{\left(\mathrm{2cos}\:{x}+\mathrm{1}\right)\left(\mathrm{cos}\:{x}−\mathrm{1}\right)}= \\ $$$$\frac{\mathrm{1}}{\mathrm{cos}\:{x}}\centerdot\frac{−\mathrm{1}}{\mathrm{2cos}\:{x}+\mathrm{1}}\:\underset{{x}\rightarrow\mathrm{0}} {\rightarrow}\:−\frac{\mathrm{1}}{\mathrm{3}} \\ $$

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