Why can we resolve indeterminate forms?
So you're looking at something of the form $$\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty}\frac{g(x)}{h(x)} $$ and if this limit exists, say the limit it $L$, then it doesn't matter how we rewrite $f(x)$. However, it's possible you can write $f(x)$ in different ways; e.g. as the quotient of different functions: $$f(x) = \frac{g_1(x)}{h_1(x)} = \frac{g_2(x)}{h_2(x)}$$ The limit of $f$ either exists or not, but it's possible that the individual limits in the numerator and denominator exist, or not. More specifically, it's possible that $$\lim_{x \to +\infty} g_1(x) \quad\mbox{and}\quad \lim_{x \to +\infty} h_1(x)$$ do not exist, while $$\lim_{x \to +\infty} g_2(x) \quad\mbox{and}\quad \lim_{x \to +\infty} h_2(x)$$do exist. What you did by dividing numerator and denominator by $x$, is writing $f(x)$ as another quotient of functions but in such a way that the individual limits in the numerator and denominator now do exist, which allows the use of the rule in blue ("limit of a quotient, is the quotient of the limits; if these two limits exist"): $$\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty}\frac{g_1(x)}{h_1(x)} = \color{blue}{ \lim_{x \to +\infty}\frac{g_2(x)}{h_2(x)} = \frac{\displaystyle \lim_{x \to +\infty} g_2(x)}{\displaystyle \lim_{x \to +\infty} h_2(x)}} = \cdots$$and in this way, also find $\lim_{x \to +\infty} f(x)$.
When you try to apply that rule but the individual limits do not exist, you "go back" and try something else, such as rewriting/simplifying $f(x)$; this is precisely what happens: $$\begin{align} \lim_{x \rightarrow +\infty} f(x) & = \lim_{x \rightarrow +\infty} \frac{x+1}{x-1} \color{red}{\ne} \frac{\displaystyle \lim_{x \rightarrow +\infty} (x+1)}{\displaystyle \lim_{x \rightarrow +\infty} (x-1)}= \frac{+\infty}{+\infty} = \; ? \\[7pt] & = \lim_{x \rightarrow +\infty} \frac{1+\tfrac{1}{x}}{1-\tfrac{1}{x}} \color{green}{=} \frac{\displaystyle \lim_{x \rightarrow +\infty} (1+\tfrac{1}{x})}{\displaystyle \lim_{x \rightarrow +\infty} (1-\tfrac{1}{x})} = \frac{1+0}{1+0} = 1 \\ \end{align}$$
Really, this has to do with the definition of continuity. The function $Q(x,y) = x/y$ is continuous except at $y = 0$. Thus, whenever $f(t) \to L_f$ and $g(t) \to L_g \neq 0$, we have $$ \lim_{t \to a} \frac{f(t)}{g(t)} = \lim_{t \to a} Q(f(t),g(t)) = \lim_{(x,y) \to (L_f,L_g)}Q(x,y) = Q(L_f,L_g) $$ However, $Q$ is not continuous at $(0,0)$. In particular, $\lim_{(x,y) \to (0,0)}Q(x,y)$ does not exist. We therefore state that $Q(0,0) = 0/0$ is an indeterminate form.
Similarly, $Q$ is "discontinuous at $\infty$", since $\lim_{(x,y) \to (\infty, \infty)}Q(x,y)$ does not exist. So, $\infty/\infty$ is an indeterminate form.
An inderterminate form just means that we have to take a closer look to understand what happens. Continuing with your example, we have $$ "\lim_{x \rightarrow +\infty} \frac{x+1}{2x-1}=\frac{+\infty}{+\infty}" $$ then seing things in more details: $$ \lim_{x \rightarrow +\infty} \frac{x+1}{2x-1}=\lim_{x \rightarrow +\infty} \frac{x(1+\frac{1}{x})}{x(2-\frac{1}{x})}=\lim_{x \rightarrow +\infty}\frac{(1+\frac{1}{x})}{(2-\frac{1}{x})}=\frac12. $$ The indeterminate form is the same, the result is different. An indeterminate form means the result is not automatic, many results are possible.