limits - Why can we resolve indeterminate forms?

I'm questioning myselfas to why indeterminate forms arise, and why limits that apparently give us indeterminate forms can be resolved with some arithmetic tricks. Why $$\begin{equation*}
\lim_{x \rightarrow +\infty}
\frac{x+1}{x-1}=\frac{+\infty}{+\infty}
\end{equation*} $$

and if I do a simple operation,

$$\begin{equation*}
\lim_{x \rightarrow +\infty}
\frac{x(1+\frac{1}{x})}{x(1-\frac{1}{x})}=\lim_{x \rightarrow +\infty}\frac{(1+\frac{1}{x})}{(1-\frac{1}{x})}=1
\end{equation*} $$

I understand the logic of the process, but I can't understand why we get different results by "not" changing anything.

Answer

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}$$