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Question Number 162925 by qaz last updated on 02/Jan/22

lim_(x→0) (1/x^4 )[(1+x+(x^2 /2)+(x^3 /3))^(1/(x+(x^2 /2)+(x^3 /3))) −(1+x+(x^2 /2)+(x^3 /3)+(x^4 /4))^(1/(x+(x^2 /2)+(x^3 /3)+(x^4 /4))) ]=?

$$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }\left[\left(\mathrm{1}+\mathrm{x}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}\right)^{\frac{\mathrm{1}}{\mathrm{x}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}}} −\left(\mathrm{1}+\mathrm{x}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{4}}\right)^{\frac{\mathrm{1}}{\mathrm{x}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{4}}}} \right]=? \\ $$

Answered by Ar Brandon last updated on 02/Jan/22

=lim_(x→0) {e^(((1/(x+(x^2 /2)+(x^3 /3))))ln(1+x+(x^2 /2)+(x^3 /3))) −e^(((1/(x+(x^2 /2)+(x^3 /3)+(x^4 /4))))ln(1+x+(x^2 /2)+(x^3 /3)+(x^4 /4))) } =lim_(x→0) {e^((1/(x+(x^2 /2)+(x^3 /3)))(x+(x^2 /2)+(x^3 /3))) −e^((1/(x+(x^2 /2)+(x^3 /3)+(x^4 /4)))(x+(x^2 /2)+(x^3 /3)+(x^4 /4))) } =e−e=0

$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{{e}^{\left(\frac{\mathrm{1}}{{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}}\right)\mathrm{ln}\left(\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\right)} −{e}^{\left(\frac{\mathrm{1}}{{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}}}\right)\mathrm{ln}\left(\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}}\right)} \right\} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{{e}^{\frac{\mathrm{1}}{{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}}\left({x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\right)} −{e}^{\frac{\mathrm{1}}{{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}}}\left({x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}}\right)} \right\} \\ $$$$={e}−{e}=\mathrm{0} \\ $$

Commented by qaz last updated on 02/Jan/22

what′s about Q162926 ?

$$\mathrm{what}'\mathrm{s}\:\mathrm{about}\:\mathrm{Q162926}\:? \\ $$

Commented by qaz last updated on 02/Jan/22

sorry .I typed wrong

$$\mathrm{sorry}\:.\mathrm{I}\:\mathrm{typed}\:\mathrm{wrong} \\ $$

Commented by Ar Brandon last updated on 02/Jan/22

(e/8)

$$\frac{{e}}{\mathrm{8}} \\ $$

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