Asymptotes

[rtz2k4]

$$f:R\rightarrow R,\:f\left(x\right)=x-ln\left(e^x+1\right)$$So I have this function and i need to find out the asymptotes. Vertical asymptotes don't exist.
For horizontal asymptotes I got:
$$\lim _{x\to \infty }f\left(x\right)=\lim _{x\to \infty }\left(x-ln\left(e^x+1\right)\right)$$So i get the indetermination : [inf-inf]
$$\lim _{x\to \infty \:}\left(ln\left(e^x\right)-ln\left(e^x+1\right)\right)=\lim _{x\to \infty }\left(\frac{e^x}{e^{x+1}}\right)=\lim _{x\to \infty }\left(\frac{e^x}{e^x\cdot e}\right)=\lim _{x\to \infty }\left(\frac{1}{e}\right)=-1$$But the answer is wrong. I don't know why. The correct answer it's actually 0, but I don't understand what I did wrong.
Also, for oblique asymptotes I get:
$$\lim _{x\to \infty }\left(\frac{f\left(x\right)}{x}\right)=\lim _{x\to \infty }\left(\frac{x-ln\left(e^x+1\right)}{x}\right)=\lim _{x\to \infty }\left(1-\frac{ln\left(e^x+1\right)}{x}\right)=\lim _{x\to \infty }\left(1-\frac{1}{e^x+1}\cdot e^x\right)=\lim _{x\to \infty }\left(\frac{1}{e^x+1}\right)=0$$Which is wrong, because I need to get 1.
What am I doing wrong?

 

[Steven G]

Why is $\displaystyle \lim_{x \to\infty}(1/e) = -1??$ Is it because ln(1/e) = -1?
What happened to the other limit? Why not calculate that one as well?

 

[rtz2k4]

$$\lim_{x\to\infty}ln(\frac{e^{x}} {e^{x}+1}) =\lim_{x\to\infty}ln\frac{e^{x} }{e^{x} (1+\frac{1}{e^{x}}) }=ln1=0$$I checked again and it seems that I've read $$e^{x} +1$$ as $$e^{x+1}$$, but now it's ok, I calculated again and the limit is correct. At oblique asymptoye I've just applied l'Hospital and calculated further and I got the wrong answer.

 

[Steven G]

$$\lim_{x\to\infty}ln(\frac{e^{x}} {e^{x}+1}) =\lim_{x\to\infty}ln\frac{e^{x} }{e^{x} (1+\frac{1}{e^{x}}) }=ln1=0$$I checked again and it seems that I've read $$e^{x} +1$$ as $$e^{x+1}$$, but now it's ok, I calculated again and the limit is correct. At oblique asymptoye I've just applied l'Hospital and calculated further and I got the wrong answer.

Again, you are leaving out one limit. Think about which one.

 

[Steven G]

$$\lim_{x\to\infty}ln(\frac{e^{x}} {e^{x}+1}) =\lim_{x\to\infty}ln\frac{e^{x} }{e^{x} (1+\frac{1}{e^{x}}) }=ln1=0$$

You really should put ln in front of the limit at some point.

 

[Steven G]

Again, you are leaving out one limit. Think about which one.

Comment??

 

[rtz2k4]

Again, you are leaving out one limit. Think about which one.

Alright, sorry for the late reply, I was really busy today.
So actually I forgot to calculate the other limit and I got.
$$m=\lim \:_{x\to \:-\infty \:}\left(\frac{f\left(x\right)}{x}\right)=\lim \:_{x\to \:-\infty \:}\left(\frac{x-ln\left(e^x+1\right)}{x}\right)=\lim \:_{x\to \:-\infty \:}\left(1-\frac{1}{e^x+1}\cdot \:e^x\right)=\lim \:\:_{x\to \:-\infty \:\:}\left(\frac{1}{e^x+1}\right)=1$$because $$e^x\rightarrow -\infty$$ is now 0 and the answer is correct.
$$n=\lim _{x\to -\infty }\left(f\left(x\right)-mx\right)=\lim _{x\to -\infty }\left(x-ln\left(e^x+1\right)-x\right)=-\lim _{x\to -\infty }\left(ln\left(e^x+1\right)\right)=-\lim _{x\to -\infty }\left(ln\left(0+1\right)\right)=0$$So m=1 and n=0 and my oblique asymptote at -inf is y=mx+n which is y=x
$$\lim _{x\to \infty }\left(f\left(x\right)\right)=\lim _{x\to \infty }\left(x-ln\left(e^x+1\right)\right)=\lim _{x\to \infty }\left(ln\left(\frac{e^x}{e^x+1}\right)\right)=ln\left(1\right)=0$$This is the horizontal asymptote to +inf.
$$\lim _{x\to -\infty }\left(f\left(x\right)\right)=\lim _{x\to -\infty }\left(x-ln\left(e^x+1\right)\right)=\lim _{x\to -\infty }\left(ln\left(\frac{e^x}{e^x+1}\right)\right)=\lim _{x\to -\infty }\left(ln\left(\frac{e^x}{e^x\left(\frac{1}{e^x}+1\right)}\right)\right)=\lim _{x\to -\infty }\left(ln\left(\frac{1}{\frac{1}{e^x}+1}\right)\right)=ln\left(0\right)=-\infty$$And this is the horizontal asymptote to -inf which doesn't exist because it's not a real number.
So, is this correct now?