Find $\lim_{x\to 0^+} \ln x\cdot \ln(1-x)$ Find $$\lim_{x\to 0^+} \ln x\cdot \ln(1-x)$$
I've been unable to use the arithmetic rules for infinite limits, as $\ln x$ approaches $-\infty$ as $x\to 0^+$, while $\ln(1-x)$approaches $0$ as $x\to 0^+$, and the arithmetic rules for the multiplication of infinite limits only applies when one of the limits is finite and nonzero.
Can anyone point me in the right direction for finding this limit? I've been unable to continue..
(Spoiler: I've checked WolframAlpha and the limit is equal to $0$, but this information hasn't helped me to proceed)
 A: Write
$$
\ln x\cdot\ln(1-x)=\frac{\ln(1-x)}{1/\ln x}.
$$
This is of the form $f/g$ with undetermined limit of the form $0/0$.
Let us compute $f'/g'$.
We find
$$
\frac{x(\ln x)^2}{1-x}.
$$
Now maybe you are allowed to use that $\lim_{x\rightarrow 0}x(\ln x)^2=0$.
In this case, you can conclude that your limit is $0$ by L'Hospital.
If you need to prove $\lim_{x\rightarrow 0}x(\ln x)^2=0$, do L'Hospital twice more.
A: Since $\lim_{x\to0} x \cdot \ln x=0$ and $\lim_{x\to0} \frac{\ln(1-x)}{x}=-1$
$$\lim_{x\to 0^+} \ln x\cdot \ln(1-x)=\lim_{x\to 0^+} (x\ln x)\times \lim_{x\to 0^+}\left(\frac{\ln(1-x)}{x}\right)=0\times-1=0.$$
A: $$\lim_{x \to 0^+}  \ln x\cdot \ln(1-x)= \lim_{x \to 0^+} x\ln x \cdot \frac{\ln(1-x)}{x} \,.$$
$$ \lim_{x \to 0^+}\frac{\ln(1-x)}{x}=-1$$
since this is just the definition of the derivative of $\ln(1-x)$ at $x=0$.
Also, with $y= \ln(x)$:
$$ \lim_{x \to 0^+} x\ln x =\lim_{y \to -\infty} e^yy =0$$ 
A: Hint: Another approach which is similar to @Mhenni's is:
When $x\to 0$ and we know that $\alpha(x)$ is very small then $$\ln(1+\alpha(x))\sim\alpha(x)$$
A: Hint:
$$ \ln(x)\ln(1-x)=-\ln(x)( x+\frac{x^2}{2}+\dots ) .$$
