Limit as x approach to infinity of Log.

Question

I'm trying to solve this:

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

I tried L'Hôpital's rule but it didn't get me where I want to...

Any hints?

Answer 1

$x-\ln(1+2e^x)=$

$ \ln (e^x) -\ln(1+2e^x)=$

$\log (\dfrac{e^x}{1+2e^x}).$

Set $y:= e^x$, and consider $y \rightarrow \infty$ :

$f(y) := \dfrac{y}{1+2y} = \dfrac{1}{1/y +2}.$

$\lim_{y \rightarrow \infty} (\log f(y)) =$

$\log( \lim_{y \rightarrow \infty} f(y)) =$

$=\log(1/2)= -\log 2$.

Used:

1) $\lim_{y \rightarrow \infty} f(y) = 1/2;$

2) $\log$ is continuous.

Answer 2

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.

Answer 3

use that $$\ln(1+2e^x)=\ln\left(e^x\left(\frac{1}{e^x}+2\right)\right)=x+\ln\left(\frac{1}{e^x}+2\right)$$