Find: $\lim_{x\to -\infty} \frac{\ln (1+e^x)}{x}$ (no L'Hospital) 
Find: $\displaystyle \lim_{x\to -\infty} \dfrac{\ln (1+e^x)}{x}$ (no L'Hospital)

I'm getting a hard time solving this limit. The book shows 1 as the answer, Wolfram Alpha shows 0. 
I could solve easily another problem when the denominator was $e^x$ but got stuck on this one. 
No L'Hospital rule can be used.
Hints and full answers are appreciated. Sorry if this is a duplicate.
 A: Recall that $\lim\limits_{x \rightarrow x_0} = f(x)g(x) = \lim\limits_{x \rightarrow x_0} f(x) \lim\limits_{x \rightarrow x_0} g(x)$ whenever $\lim\limits_{x \rightarrow x_0} f(x)$ and $\lim\limits_{x \rightarrow x_0} g(x)$ exist.  
Since $$\lim\limits_{x\rightarrow -\infty}\ln(1 + e^x) = \ln(1) = 0$$ and 
$$\lim\limits_{x\rightarrow -\infty}\frac{1}{x} = 0$$
We have $$\lim\limits_{x\rightarrow -\infty}\ln(1 + e^x) \frac{1}{x} = \lim\limits_{x\rightarrow -\infty}\ln(1 + e^x)\lim\limits_{x\rightarrow -\infty}\frac{1}{x} = 0$$
As Gimusi's answer shows, this can be used to rigorously verify that $\lim\limits_{x \rightarrow \infty} \frac{\ln(1+e^x)}{x} = 1$.  Note the identity  $$\ln(1 + e^x) = \ln(e^{-x}\frac{1 + e^x}{e^{-x}}) = \ln(1+e^{-x}) - \ln (e^{-x})$$
Thus 
$$\lim\limits_{x\rightarrow \infty}\ln(1 + e^x) \frac{1}{x} = \lim\limits_{x\rightarrow \infty} \left(\ln(1+e^{-x}) - \ln (e^{-x})\right) \frac{1}{x} = \lim\limits_{x\rightarrow -\infty} \left(\ln(1+e^{x}) - \ln (e^{x})\right) \frac{1}{-x} = \lim\limits_{x \rightarrow -\infty}\frac{\ln(e^x)}{x} = 1$$
where the last equality follows because limits distribute over addition (so we apply our previous work).
A: For $x\to +\infty$
$$\dfrac{\ln (1+e^x)}{x}=\dfrac{\ln (1+e^{-x})+\ln e^x}{x}=\dfrac{\ln (1+e^{-x})+x\ln e}{x}\to\ln e=1$$
For $x\to -\infty$
$$\dfrac{\ln (1+e^x)}{x}\to\frac{\ln 1}{-\infty}=0$$
