Solution.

$$
\begin{aligned}
& \lim _{x \rightarrow 1+0}(\ln x \cdot \ln (x-1))=(0 \cdot(-\infty))=\lim _{x \rightarrow 1+0} \frac{\ln (x-1)}{\frac{1}{\ln x}}=(-\infty /-\infty)= \\
& \quad=\lim _{x \rightarrow 1+0} \frac{(\ln (x-1))^{\prime}}{\left(\frac{1}{\ln x}\right)^{\prime}}=\lim _{x \rightarrow 1+0} \frac{\frac{1}{x-1}}{\frac{1}{x \ln ^2 x}}=\lim _{x \rightarrow 1+0} \frac{x \ln ^{2} x}{x-1}= \\
& \quad=\lim _{x \rightarrow 1+0} \frac{\ln ^{2} x}{1-\frac{1}{x}}=(0 / 0)=\lim _{x \rightarrow 1+0} \frac{\frac{\ln x}{x}}{\frac{1}{x^{2}}}=\lim _{x \rightarrow 1+0} 2 x \ln x=0 .
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_f5a9bb7960904462604bg-248.jpg?height=31&width=536&top_left_y=132&top_left_x=366)
6.55. $\lim _{x \rightarrow 1+0}(\ln x \cdot \ln (x-1))$.

6.55. $\lim _{x \rightarrow 1+0}(\ln x \cdot \ln (x-1))$. 

(No change needed as the text is already in English and contains mathematical notation which is universal.)