How to compute $\lim _{x\to \infty }\left(\ln\left(\frac{e^x-1}{x}\right)-x\right)$? I have a problem with this limit, I don't know what method to use. I have no idea how to compute it. Can you explain the method and the steps used? Thanks
$$\lim _{x\to \infty }\left(\ln\left(\frac{e^x-1}{x}\right)-x\right)$$
 A: Notice that $x = \ln(e^x)$, therefore the original expression can be written as 
$$\ln\left(\frac{e^x - 1}{xe^x}\right) < \ln\left(\frac{e^x}{xe^x}\right) = -\ln x.$$
Since the $-\ln x \to - \infty$ as $x \to \infty$, the limit of the original expression is bound to be $-\infty$.
A: You may write, as $x \to \infty$,
$$
\begin{align}
\ln\left(\frac{e^x-1}{x}\right)-x&=\ln\left(e^x\times\frac{1-e^{-x}}{x}\right)-x\\\\&=\ln(e^x)+\ln\left(\frac{1-e^{-x}}{x}\right)-x\\\\&=\ln\left(1-e^{-x}\right)-\ln x\\\\&\sim -\ln x
\end{align}
$$ tending to $-\infty$.
A: Write $\ln\frac{e^x-1}{x}-x=\ln\frac{e^x-1}{x}-\ln e^x=\ln\frac{e^x-1}{xe^x}=\ln (\frac{e^x}{xe^x}-\frac{1}{xe^x})=\ln (\frac{1}{x}-\frac{1}{xe^x}).$
As $ x \to \infty, \frac{1}{x}-\frac{1}{xe^x} \to 0$. Since $\lim_ \limits{x \to 0} \ln x \to -\infty$, our original limit approaches $- \infty$.
A: I would try the following:
$$ \lim_{x \rightarrow \infty} \left( \ln{\left( \frac{e^x - 1}{x} \right)} - x \right) = \lim_{x \rightarrow \infty} \left( \ln{\left(e^x - 1 \right)} - x \right)- \lim_{x \rightarrow \infty}{\ln{x}}$$
You can show that the first term tends to zero and so the second term is what the limit evaluates to, which as $x \rightarrow \infty$ gives $-\infty$.
A: $$\frac{\mathrm e^x-1}x= \frac{\mathrm e^x(1-\mathrm e^{-x})}x,$$
hence 
$$\ln\frac{\mathrm e^x-1}x-x= -\ln x+\ln(1-\mathrm e^{-x})=-\ln x+o(1)\xrightarrow[x\to\infty]{}-\infty.$$
