How to solve $\lim_{x\to +\infty}\ ({\ln(e^x-1)}-{x})$

Question

I tried isolating the x but then it turns from a $\infty - \infty$ to $0 \times \infty$ indetermination.

$$\lim_{x\to +\infty}\ ({\ln(e^x-1)}-{x})$$ Our teacher says we need to write all the steps so we can't solve it just with intuition (because of finals).

Answer 1

$$x = \ln e^x $$ then $$\lim_{x\to +\infty}\ ({\ln(e^x-1)}-{x}) = \lim_{x\to +\infty}\ ({\ln(e^x-1)}-\ln e^x) = $$ $$ \lim_{x\to +\infty}\ {\ln\left(\frac{(e^x-1)}{e^x}\right)} $$ $$ \lim_{x\to +\infty}\ {\ln(1-e^{-x})} = \ln(1) = 0 $$

Answer 2

Hint Write $$e^x - 1 = e^x (1 - e^{-x})$$ and use the product identity for logarithms.

Answer 3

$$\lim_{x\to\infty}\left(\ln\left(e^x-1\right)-x\right)=\lim_{x\to\infty}\ln\left(\exp\left[\ln\left(e^x-1\right)-1\right]\right)=$$ $$\lim_{x\to\infty}\ln\left(1-e^{-x}\right)=\lim_{x\to\infty}\ln\left(1-\frac{1}{e^x}\right)=$$ $$\ln\left(1-0\right)=\ln(1)=0$$

Answer 4

Hint take $e^x$ common so we get $ln(e^x(1+e^{-x}))-x$ then $log(ab)=log(a)+log(b)$ so we get $x-x+ln(1+e^{-x})$ so as x approaches infinity $e^{-x}$ goes to $0$ thus we get $ln(1+0)=0$