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Question-182441




Question Number 182441 by mathlove last updated on 09/Dec/22
Answered by JDamian last updated on 09/Dec/22
lim_(x→0) ((ln(Σ_(k=1) ^n e^(xk) )−ln n)/x)=9  lim_(x→0) (((Σ_(k=1) ^n ke^(xk) )/(Σ_(k=1) ^n e^(xk) )))=((Σ_(k=1) ^n k)/(Σ_(k=1) ^n 1))=((n(n+1))/(2n))=((n+1)/2)=9  n=17
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{ln}\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{e}^{{xk}} \right)−\mathrm{ln}\:{n}}{{x}}=\mathrm{9} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{ke}^{{xk}} }{\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{e}^{{xk}} }\right)=\frac{\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}}{\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{1}}=\frac{\cancel{{n}}\left({n}+\mathrm{1}\right)}{\mathrm{2}\cancel{{n}}}=\frac{{n}+\mathrm{1}}{\mathrm{2}}=\mathrm{9} \\ $$$${n}=\mathrm{17} \\ $$
Answered by qaz last updated on 09/Dec/22
lim_(x→0) (1/x)ln((e^x +e^(2x) +...+e^(nx) )/n)  =lim_(x→0) (1/x)ln((n+(1+2+...+n)x+o(x))/n)  =lim_(x→0) (1/x)ln(1+(1/2)(n+1)x+o(x))  =lim_(x→0) (1/x)((1/2)(n+1)x+o(x))=(1/2)(n+1)=9  ⇒n=17
$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\mathrm{1}}{{x}}{ln}\frac{{e}^{{x}} +{e}^{\mathrm{2}{x}} +…+{e}^{{nx}} }{{n}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\mathrm{1}}{{x}}{ln}\frac{{n}+\left(\mathrm{1}+\mathrm{2}+…+{n}\right){x}+{o}\left({x}\right)}{{n}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\mathrm{1}}{{x}}{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\left({n}+\mathrm{1}\right){x}+{o}\left({x}\right)\right) \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\mathrm{1}}{{x}}\left(\frac{\mathrm{1}}{\mathrm{2}}\left({n}+\mathrm{1}\right){x}+{o}\left({x}\right)\right)=\frac{\mathrm{1}}{\mathrm{2}}\left({n}+\mathrm{1}\right)=\mathrm{9} \\ $$$$\Rightarrow{n}=\mathrm{17} \\ $$
Commented by mathlove last updated on 09/Dec/22
what  o(x)???
$${what}\:\:{o}\left({x}\right)??? \\ $$
Commented by mathlove last updated on 12/Dec/22
????
$$???? \\ $$

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