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

Question

My only problem is the $\infty$ because otherwise it's a special limit. Do I use substitution? $$\lim_{x\to +\infty}\frac {\ln(e^x-1)}{x} $$

Answer 1

There are many ways that you could go about this; many people will suggest you use L'Hopital's rule. But, that ignores the intuition here, so let's try it a little differently.

Intuitively: when $x$ is large, $e^x-1$ is approximately $e^x$, so that $\ln(e^x-1)$ is approximately $\ln(e^x)=x$. So, we should expect this limit to approach $1$.

Let's try to codify this intuition. You could start by writing $$ e^x-1=e^x\cdot\frac{e^x-1}{e^x}=e^x\left(1-e^{-x}\right). $$ Then, using the properties of the logarithm, $$ \ln(e^x-1)=\ln(e^x)+\ln(1-e^{-x})=x+\ln(1-e^{-x}). $$ Substituting this into the expression, we get $$ \frac{\ln(e^x-1)}{x}=\frac{x+\ln(1-e^{-x})}{x}=1+\frac{\ln(1-e^{-x})}{x}. $$ The limit of this latter term as $x\to\infty$ is easily computed.

Answer 2

You estimate that $$ \ln(e^x-1) \approx \ln(e^x) = x $$ especially for large $x$ and thus this limit goes towards $1$.

More formally you could try L'Hôspital's rule, as this is a limit of type $\infty / \infty$.

Answer 3

HINT:

$$\ln(e^x-1)=\ln(e^x)+\ln(1-e^{-x})=x+\ln(1-e^{-x})$$

Now $\lim_{x\to\infty}=\ln(1-e^{-x})=\ln1=?$

Answer 4

If $x$ goes to $\infty$, then we can assume that $x>1$. Since $e^x -1 \le e^x$ and $\ln x$ increases, we can know that $$ \ln(e^x -1 ) \le \ln e^x =x. $$ Since $0<e^x - 2^x< e^x-1$, $$ \ln(e^x-2^x)\le \ln(e^x-1). $$ Thus this inequality holds: $$ 1+\frac{\ln\left(1-\frac{2}{e}\right)}{x}\le\frac{\ln(e^x-2^x)}{x} \le \frac{\ln(e^x-1)}{x}\le 1. $$ Since $\lim_{x\to \infty}\left(1+\frac{\ln\left(1-\frac{2}{e}\right)}{x}\right)=1$, $$ \lim_{x\to\infty}\frac{\ln(e^x-1)}{x}=1 $$ by sandwich theorem.