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math.stackexchange.com/questions/1812238
I'm having some trouble solving this limit:
$$\lim_{x\to\infty} x\left[\left(\cosh x\right)^ \frac1x - \left(1+\frac1x\right)^x\right]$$
I have tried several approaches, and the fastest seems the following [it does not require L'Hopital]:
$$
\lim_{x\to\infty} x[e^{\ln(\cosh x)\frac1x}-e^{\ln(1+\frac1x)x}]=\lim_{x\to\infty} e^{\ln x}[e^{\ln(\cosh x)\frac1x}-e^{\ln(1+\frac1x)x}]
$$
$$
\lim_{x\to\infty} e^{\phi_1(x)}-e^{\phi_2(x)}\ .
$$
Let us analyze the series expansions of the two functions in the exponent up to second next-to-leading order
$$
\phi_1(x)=\ln x+\frac{1}{x}\ln(\cosh(x))=\ln x+\frac{1}{x}\ln\left(\frac{1}{2}(e^x+e^{-x})\right)=\ln x-\frac{\ln 2}{x}+1+\frac{1}{x}\ln(1+e^{-2x})
$$
$$
\sim \ln x+1-\frac{\ln 2}{x}
$$
$$
\phi_2(x)=\ln x+x \ln(1+1/x)\sim\ln x +1-\frac{1}{2x}
$$
Therefore, Taylor-expanding the exponentials [of the type $e^{-c/x}$] around $x=\infty$
$$
e^{\phi_1(x)}\sim e x e^{-\ln 2/x}\sim e x -e \ln (2)+\frac{e \ln ^2(2)}{2 x}-\frac{e \ln ^3(2)}{6 x^2}+\ldots
$$
$$
e^{\phi_2(x)}\sim e x-\frac{e}{2}+\frac{e}{8 x}-\frac{e}{48 x^2}+\ldots
$$
we get the final result by subtracting one from the other
$$
\boxed{\lim_{x\to\infty} x[e^{\ln(\cosh x)\frac1x}-e^{\ln(1+\frac1x)x}]=\frac{e}{2}-e\ln 2\approx -0.525028...}
$$
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