To compute the infimum and supremum of $f(x)=\frac{x^2+x}{x-1}$. I have observed that is increasing for $$x\in\left[\frac1{\sqrt{3}},+\infty\right]\vee \left[-\infty, \frac{-1}{\sqrt{3}}\right]$$ Moreover $x_1=\frac1{\sqrt{3}}$ is local minimum and $x_2=\frac{-1}{\sqrt{3}}$ a local maximum.
But $\displaystyle \lim_{x\to \infty}f(x)=-\infty$ and $\displaystyle \lim_{x\to \infty}f(x)=-\infty$.
Can I conclude from the limit that the supremum of $f$ is $+\infty$ and $-\infty$ is the infimum?
On
this set has not supremum. $$A = \{\frac{x^2+x}{x-1} :x\in \mathbb{R}, x\neq1\}.$$
If it has supremun in particular it is bounded so let prove it isnt bounded.
Defining the function: $$f(x)=\frac{x^2+x}{x-1} .$$
and using the limit as $x$ goes to infinity of $f$
$$\lim_{x \to \infty}{\frac{x^2+x}{x-1} } = \infty.$$ Because of the definition of limit $\forall b\in \mathbb{R}$, exists $x_0\in \mathbb{R}-\{1\} $, such that $f(x_0) > b$ as $f(x_0) \in A$ for all real number we can choose a number from $A$ that is bigger than this real number so this set is not bounded, therefore it has not supremum.
We can prove this set does not have infimun neither in an analogous way.
I guess that you are trying to find inf and sup of the image of $f$. In this case yes, the sole information that $\lim_{x\to\pm\infty}f(x)=\pm\infty$ is sufficient. In the case the local minima and maxima are to be found, then one of course need to find the critical points of $f$ by posing $f'(x)=0$.