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I want to evaluate $$\lim_{x \to \infty} \frac{\ln(1+e^x)}{x \cdot \arctan(x)}$$

Using L'Hospital, I get

$$\lim_{x \to \infty}\frac{\frac{e^x}{1+e^x}}{\arctan(x)+\frac{x}{1+x^2}}$$

Now I could use L'Hospital again, but it seems like it's not going to give me a result where I can conveniently take the limit.

Any hints?

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As said elsewhere, your second limit is not as difficult as you think (it is $\frac{1}{\frac\pi2+0}$).

To check your work, an intuitive approach is possible for the first limit: for large $x$, $e^x$ is huge in front of $1$, so that $\log(e^x+1)$ is virtually $x$.

Now after simplification all that remains is

$$\lim_{x\to\infty}\frac1{\arctan(x)}.$$

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Without L'Hospital

$$\frac{\ln(1+e^x)}{x \cdot \arctan(x)}=\frac{1}{ \arctan(x)}\frac{\ln(1+\frac1{e^x})+\ln (e^x)}{x}=$$

$$=\frac{1}{ \arctan(x)}\left(\frac{\ln(1+\frac1{e^x})}{x}+1\right)\to \frac 2 \pi\cdot1=\frac 2 \pi$$

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As $x \to \infty$ $\arctan x \to \pi/2$. So it is enough to find the limit of $\frac {ln (1+e^{x})} x$. You can apply L'Hopital's rule for this. The answer is $\frac 2 {\pi}$.

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We have $\frac{e^x}{1 + e^x}\to 1$, $\arctan(x)\to \frac\pi2$ and $\frac{x}{x^2 + 1}\to 0$. So the second limit is rather easy to calculate.

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Note that $1+\mathrm e^x\sim_{+\infty}\mathrm e^x$, so $\;\ln(1+\mathrm e^x)\sim_{+\infty}\ln(\mathrm e^x)=x$. On the other hand, $\;\arctan x\sim_{+\infty}\frac\pi2$, so that $$ \frac{\ln(1+e^x)}{x \cdot \arctan(x)}\sim_{+\infty}\frac x{x\,\frac\pi 2}=\frac2\pi.$$

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