Why can we resolve indeterminate forms?

Question

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 1

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 $\lim_{x \to +\infty} f(x)$.

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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: $$\begin{align} \lim_{x \rightarrow +\infty} f(x) & = \lim_{x \rightarrow +\infty} \frac{x+1}{x-1} \color{red}{\ne} \frac{\displaystyle \lim_{x \rightarrow +\infty} (x+1)}{\displaystyle \lim_{x \rightarrow +\infty} (x-1)}= \frac{+\infty}{+\infty} = \; ? \\[7pt] & = \lim_{x \rightarrow +\infty} \frac{1+\tfrac{1}{x}}{1-\tfrac{1}{x}} \color{green}{=} \frac{\displaystyle \lim_{x \rightarrow +\infty} (1+\tfrac{1}{x})}{\displaystyle \lim_{x \rightarrow +\infty} (1-\tfrac{1}{x})} = \frac{1+0}{1+0} = 1 \\ \end{align}$$

Answer 2

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.

Answer 3

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.