# Use L'Hospital's rule to evaluate $\lim_{x\to\infty} \frac{\ln(1+e^x)}{x \cdot \arctan(x)}$

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?

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

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

Without L'Hospital

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

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

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 $$\frac 2 {\pi}$$.

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.

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 $$\frac{\ln(1+e^x)}{x \cdot \arctan(x)}\sim_{+\infty}\frac x{x\,\frac\pi 2}=\frac2\pi.$$