Task that is related to the question
Can someone help me with a general way of solving a question of this type?
I know the rules for intervals and that the supremum is the smallest upper bound and the infimum is the biggest lower bound:
But how can I determine if it's accepted or not?
This kind of question is not really amenable to general methods. There are a number of things to look out for. What exactly is the domain? If it is an open set (typically an open interval), then the function may be well-behaved, but the inf or sup is realised at an endpoint of the interval rather than in the domain of the function. One also has to look out for the function being unbounded (above or below or both). Another classic is where the function is, at least technically, undefined at key point in the domain.
But I think it is best to work through the specific example you gave.
To save time, the plot shows the graph for the range $0<x<2$.
A quick look at the graph suggests that the function is strictly decreasing throughout the interval $(0,2)$ and so its infimum for the open interval $(0,1)$ is the value of the function at $x=1$, but the inf is not realised in the open interval. As you comment, it would be realised in the closed interval.
One also notes that the sup on the open interval is $+\infty$, or if one prefers, one can say the function is not bounded above on the open interval.
However, that is not the end of the story. Mathematica does sensible things, rather than the pedantic things people are expected to do when learning the subject. So it simply plots the value 1 for $x=1$.
But a professor will probably say that the function is undefined at $x=1$ because the denominator $x-1$ is 0.
Hmmm. But suppose we put $x=1+w$ and use the standard series $\ln(1+w)=w-w^2/2+\dots$. Then the expression becomes $$\frac{\ln(1+w)}{w}=1-w/2+\dots$$ so everyone except professors teaching math classes will automatically define the function to be 1 at $x=1$.
So - unless making a special effort for students - people like me will tend to say that the inf is realised on the closed interval $[0,1]$.
Can I leave it there?
Oh, I should add that there is less argument about the supremum. The function tends to $\infty$ as $x\to0$. So most people would say either that the function is not bounded above, or that the sup is $+\infty$. But in any case it is not realised on the open interval.
Added the following day.
The above was dashed off quickly in response to some of the comments by the OP. There are a few other points.
The function $\ln x$ is well-known to be well-defined, continuous and differentiable on the positive real line, but tends to $-\infty$ as $x\to 0^+$.
The function $\frac{1}{x-1}$ is well-defined on the entire real line, except for the point $x=1$. As $x\to1^+$ it tends to $+\infty$ and as $x\to1^-$ it tends to $-\infty$, so one cannot avoid a discontinuity at $x=1$ even if one allows the values $\pm\infty$.
However, the product $f(x)=\frac{\ln x}{x-1}$ is carefully chosen. $\ln x$ changes sign at $x=1$ and is continuous across $x=1$, meaning that $\lim_{x\to1^+}f(x)=\lim_{x\to1^-}f(x)$. Both limits are 0. So although the expression $\frac{\ln x}{x-1}$ is arguably not defined at $x=1$, one can easily make the function $f(x)$ continuous (and differentiable) and $x=1$ simply by defining $f(1)=0$. Practising mathematicians tend to make such extensions of the domain almost automatically, without thinking much about them. In complex variable theory, a course which usually follows the early calculus courses, one gets used to distinguishing a function from particular representations of it in particular subdomains.
Finally, on how much detail one is expected to go into in an exam or assignment, that is a matter of judgment. I would say that the log function is so well-known that one could deal with the question briefly. The only point that needs a little examination/explanation is why it is continuous (and still decreasing) near $x=1$.