# 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

$$\lim _{x\to \infty }\left(\ln\left(\frac{e^x-1}{x}\right)-x\right)$$

• for large values of $x$ one could say $e^x-1=e^x$ – imranfat Jan 19 '16 at 22:28

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 $-\infty$.

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 $-\infty$.

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 $- \infty$.

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 $-\infty$.

$$\frac{\mathrm e^x-1}x= \frac{\mathrm e^x(1-\mathrm e^{-x})}x,$$ hence $$\ln\frac{\mathrm e^x-1}x-x= -\ln x+\ln(1-\mathrm e^{-x})=-\ln x+o(1)\xrightarrow[x\to\infty]{}-\infty.$$