Computing $\lim\limits_{x\to\infty}\ln(x)\cdot \ln(1-e^{-x})$ Limit as $x$ approaches infinity of $\ln(x)\cdot  \ln(1-e^{-x})$:
$$
\lim_{x\to\infty}\ln(x)\cdot  \ln(1-e^{-x})
$$
The only thing I can think to do is rewrite the ln(x) on the bottom as $(lnx)^{-1}$ and use L’Hôpital’s rule, but I’ve done two iterations now and it keeps getting back to the 0/0 or infinity*0 indeterminate case. Any help on how to proceed will be much appreciated! Thanks
 A: We have 
$$\lim_{x\to\infty} e^{-x}  = 0 $$
We know that, $X=0$
$$\lim_{X\to 0^+}\frac{\ln(1-X)}{X} =-1 $$ 
By setting $X= e^{-x}$
$$ \lim_{x\to \infty}\frac{\ln(1-e^{-x})}{e^{-x}}=-1$$ 
Therfore,
$$\lim_{x\to \infty} \ln(x)\cdot  \ln(1-e^{-x})= \lim_{x\to \infty}\frac{\ln(1-e^{-x})}{e^{-x}}\frac{\ln(x)}{e^{x}} \\=-\lim_{x\to \infty} \frac{\ln(x)}{e^{x}}=-\lim_{x\to \infty} \frac{x}{e^{x}} \frac{\ln(x)}{x}\\=-\left(\lim_{x\to \infty} \frac{x}{e^{x}} \right)\left(\lim_{x\to \infty}\frac{\ln(x)}{x} \right)=0$$
A: We have $1-\frac{1}{t}\leq  \log t  \leq t-1$ for all $t>0$. If $ x>1$ then $t=1-e^{-x}>0$  and 
$$
1-\frac{1}{1-e^{-x}}\leq  \log (1-e^{-x})  \leq -e^{-x}.
$$
Multiplying the first, the second and third members of the above  inequality by the first, second and third members of the inequality $ 1-\frac{1}{x}< \log x< x-1$, for $x>1$, 
$$
\underbrace{\Big(1-\frac{1}{x}\Big)}_{(1-\frac{1}{x})\to 1 }\Big(1-\frac{1}{\underbrace{1-e^{-x}}_{(1-e^{-x})\to 0}}\Big)\leq  \log (x)\log (1-e^{-x})  \leq \underbrace{-e^{-x}(x-1)}_{-e^{-x}(x-1)\to 0}.
$$
Once 
$$
\lim_{x\to \infty}\Big(1-\frac{1}{x}\Big)\Big(1-\frac{1}{e^{x}}\Big)
 =
\lim_{x\to\infty}-e^{-x}(x-1)=0
$$
we have
$$
\lim_{x\to \infty} \log (x)\log (1-e^{-x}) =0
$$
A: $\displaystyle \lim_{x\to\infty}\ln(x)\cdot  \ln(1-e^{-x})$
$\ln(x)\cdot  \ln(1-e^{-x})=-\ln x\cdot e^{-x}\cdot\dfrac{\ln(1-e^{-x})}{-e^{-x}}=-e^{\ln e^{-x}+\ln(\ln x)}\cdot\dfrac{\ln(1-e^{-x})}{-e^{-x}}=...$
$-e^{-x+\ln(\ln x)}\cdot\dfrac{\ln(1-e^{-x})}{-e^{-x}} =-\exp\bigg(-x+\ln x\cdot\dfrac{\ln(\ln x)}{\ln x}\bigg)\cdot\dfrac{\ln(1-e^{-x})}{-e^{-x}}=...$
$=-\exp\bigg[-x\bigg(1-\underbrace{\dfrac{\ln x}{x}}_{0}\cdot\underbrace{\dfrac{\ln(\ln x)}{\ln x}}_{0}\bigg)\bigg]\cdot\underbrace{\dfrac{\ln(1-e^{-x})}{-e^{-x}}}_{1}\underset{x\to +\infty}{\longrightarrow}0$
