Question: I'm trying to solve this:

$$\lim\_{x\rightarrow+\infty} (x-\ln(1+2e^x))= ?$$

I tried [L'Hôpital's rule](https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule) but it didn't get me where I want to...

Any hints?

Answer: 
Note that

$$x-\ln(1+2e^x)=x-\left[\ln e^x+\ln \left(\frac1{e^x}+2\right)\right]=x-x-\ln \left(\frac1{e^x}+2\right)\to -\ln 2$$

To figure out at first sight the result note that for $x\to +\infty$

* $\ln(1+2e^x)\sim \ln 2e^x=\ln2+\ln e^x=\ln 2 +x$

observe that the latter is not a proof but only a method to guess what the limit could be.
use that $$\ln(1+2e^x)=\ln\left(e^x\left(\frac{1}{e^x}+2\right)\right)=x+\ln\left(\frac{1}{e^x}+2\right)$$