I have been formulating clues about how a graph of $\frac{\ln x}{x-1}$ might look like. Here is what I found:
1) There are no stationary points
2) The graph at a first glance does not appear to be continuous, as $x \neq 1$
3) The domain is $x \in R^+$
4) The function is above the $x$-axis
Now, the struggle comes when I consider the limits. And the lack of this final clue, means I cannot sketch the graph:
$$\lim_{x \rightarrow 1^+} f(x) = +\infty ?$$ $$\lim_{x \rightarrow 1^-} f(x) = + \infty \ or \ 0?$$ $$\lim_{x \rightarrow +\infty} f(x) = 0$$ $$\lim_{x \rightarrow 0^+} f(x) = + \infty$$
I know for sure that the first two are wrong, but why, and what should the values be then? It also appears that the curve is continuos actually.. Can't understand how.
The clue that there are no stationary points and the fact that I have double asymptote around 1 and then at 0 are conflicting. Since, my derivatives are correct, the limits must be wrong.
By the L'Hôpital's rule, $$\lim_{x\to1}\frac{\ln x}{x-1}=\lim_{x\to1}\frac{(\ln x)^\prime}{(x-1)^\prime}=1$$