TITLE: How to compute $\lim _{x\to \infty }\left(\ln\left(\frac{e^x-1}{x}\right)-x\right)$?
QUESTION [5 upvotes]: 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
$$\lim _{x\to \infty }\left(\ln\left(\frac{e^x-1}{x}\right)-x\right)$$

REPLY [3 votes]: 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 $-\infty$.