$Find $n$ such that $\lim _{x \rightarrow \infty} \frac{1}{x} \ln (\frac{e^{x}+e^{2x}+\dots e^{nx}}{n})=9$$

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$Find $n$ such that $\lim _{x \rightarrow \infty} \frac{1}{x} \ln (\frac{e^{x}+e^{2x}+\dots e^{nx}}{n})=9$$

Find $n$ such that
$$\lim _{x \rightarrow \infty} \frac{1}{x} \ln (\frac{e^{x}+e^{2x}+\dots e^{nx}}{n})=9$$.

Calculus Algebra Multivariable Calculus

User50

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Using the Hopital's rule we have
\[9=\lim _{x \rightarrow \infty} \frac{1}{x}\ln (\frac{e^x+e^{2x}+\dots +e^{nx}}{n})\]
\[=\lim_{x \rightarrow \infty} \frac{e^x+e^{2x}+\dots +e^{nx}- \ln n}{x}\]
\[=\lim_{x \rightarrow \infty} \frac{\frac{e^x+2e^{2x}+\dots +ne^{nx}}{e^x+e^{2x}+\dots +e^{nx}}}{1}\]
\[=\lim _{x \rightarrow \infty} \frac{e^{nx}(e^{-(n-1)x}+e^{-(n-2)x+ \dots +e^{-x}+n})}{e^{nx}(e^{-(n-1)x}+2e^{-(n-2)x+ \dots +e^{-x}+1})} =n.\]
So $n=9$.

Erdos

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$Find $n$ such that $\lim _{x \rightarrow \infty} \frac{1}{x} \ln (\frac{e^{x}+e^{2x}+\dots e^{nx}}{n})=9$$

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