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.

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    $\begingroup$ The first "result" isn't a result. $\endgroup$ – Martín-Blas Pérez Pinilla Jan 12 '17 at 10:54
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    $\begingroup$ An indeterminate form is not a mathematical expression. It's a way of telling you: you can't do it using this method. So, what you get in the first scenario is really nothing. $\endgroup$ – user384138 Jan 12 '17 at 11:01

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

  • $\begingroup$ Well put :) ${}$ $\endgroup$ – user384138 Jan 12 '17 at 11:03
  • $\begingroup$ Thanks @OpenBall, but I'm sure it can be improved as it got a downvote as well. Comments are welcome. $\endgroup$ – StackTD Jan 13 '17 at 0:04
  • $\begingroup$ I upvoted, but I believe I understand the reason for the downvote; this is a good, rigorous answer, but it would be improved by also including an intuitive explanation. I would say it like so: Even outside of mathematics, you can make a true statement about some object that doesn't reveal its identity. Or, you can make a true statement which does reveal its identity. The first type of statement, in mathematics, results in an "indeterminate" answer; the second results in the correct answer of the "object's identity." They aren't two different answers; the first means, "We can't tell." $\endgroup$ – Wildcard Jan 13 '17 at 1:23

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.

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    $\begingroup$ While this explains why limits of ratios can be indeterminate, it does not address the question asked, which is why modifying the form of the expression can change the form into one that is determinate. $\endgroup$ – Paul Sinclair Jan 12 '17 at 17:37
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    $\begingroup$ @PaulSinclair I think it addresses why the situation is fundamentally different when (in OP's example) we've divided the top and bottom by $x$. $\endgroup$ – Omnomnomnom Jan 12 '17 at 17:44

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.


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