7
$\begingroup$

I know they can be $\frac{0}{0}$ or $\frac{\pm\infty}{\pm\infty}$, but what do they mean in English?

For example, $$\lim_{n\rightarrow \infty}A$$

In the context of limits, when the above example get a limit that is an indeterminate form. I assume it means that $A$ does not give enough information to "determine" a limit. Fine.

But why is it that when we cancel out some terms out in $A$, *poof* now we have enough information to get a definite limit even though no extra information (i.e. additional expressions) have been added to $A$.

Another example:

$\lim_{x\rightarrow3}\frac{(x-3)(x+3)}{x-3}=\frac{0}{0}$, not enough information to determine the limit.

$\lim_{x\rightarrow3}\frac{(x-3)(x+3)}{x-3}=\lim_{x\rightarrow3}(x+3)=6$, Hey! Now we have enough information to determine the limit! (I thought you said you didn't have enough information)

$\endgroup$
  • 2
    $\begingroup$ "Indeterminate form" is nonsense sanctified by tradition. In the next step, you determine the value, so it's misleading at best. $\endgroup$ – Professor Vector Mar 1 '18 at 7:45
17
$\begingroup$

$a\circ b$ is an indeterminate form (not expression!) if the knowledge of $a_n\to a$ and $b_n\to b$ alone tells us nothing about the behaviour (divergence/convergence/limit) of the sequence $a_n\circ b_n$. This may be because the form expresses an undefined operation (such as $\frac 00$) or that it is defined but not continuous there (such as $0^0$). Of course this does not prevent us from - "poof" - finding some better argument or transformation that does tell us all we need about $\lim(a_n\circ b_n)$.

$\endgroup$
4
$\begingroup$

It is not about individual cases, but rather the general case. If the limit exists, you can find it by some means. Contrast this with some other limits where we can say something general.

For example, if $\lim a_n = 0$ and $\lim b_n = 0$ then we can always say that $$\lim_n(a_n+b_n) = 0 $$ Hence "$0+0$" is not an "indeterminate form."

However, we cannot say anything in general about $$\lim_n \frac{a_n}{b_n} = \mathrm{?} $$ Therefore "$0/0$" is an "indeterminate form."

That does not mean the limit cannot be calculated by some means. It simply means that we have to approach each limit of this type individually.

$\endgroup$
3
$\begingroup$

This is a good question.

We like to evaluate limits using the “limit laws,” for instance: If two functions have a limit at a point, the limit of the sum of the functions is the sum of the limits of the functions. That is $$ \lim_{x\to a} (f(x) + g(x)) = \lim_{x\to a} f(x) + \lim_{x\to a}g(x) $$ The same is true for products of functions, and for quotients, with the caveat that we cannot divide by zero. $$ \lim_{x\to a}g(x)\neq 0\implies \lim_{x\to a} \frac{f(x)}{g(x)} = \frac{\lim_{x\to a}f(x)}{\lim_{x\to a}g(x)}, $$

Idiomatically, we say that a limit problem is in indeterminate form if the limit laws cannot be directly applied to the expression of the function. When you say

$$\lim_{x\rightarrow3}\frac{(x-3)(x+3)}{x-3} \color{red}{=\frac{0}{0}}$$

the problem is not that $\frac{0}{0}$ is undefined. The problem is that you broke the limit laws by applying it to a quotient where the denominator tends to zero. In this situation, I would avoid using the equals sign as well, because we're not asserting the limit is equal to anything, let alone an undefined thing. We're not making any “determination” about the limit at all.

We try to work around the situation by writing the function in such a way that we can legally apply the limit laws. As you did. Since $\frac{(x-3)(x+3)}{x-3} = x+3$ when $x\neq 3$, we have $$ \lim_{x\rightarrow3}\frac{(x-3)(x+3)}{x-3} = \lim_{x\rightarrow3}(x+3) = 6 $$ The cancellation happens before the limit is found.

We have shorthand for indeterminate forms, which I think can further muddy the waters. For instance, when we say $1^0$ is an indeterminate form, what we mean is that there is no limit law of the form: “If $\lim_{x\to a} f(x) = 1$ and $\lim_{x\to a} g(x) = 0$, then $\lim_{x\to a} f(x)^{g(x)} = (\text{something})$.” The form of the expression cannot be used to determine the limit.

$\endgroup$
2
$\begingroup$

One can define that a sequence $a_n$ has limit infinity: for all $\epsilon>0$ there exist only a finite number of elements of that sequence which are less than $\epsilon$, if any. In that sense we have some “determinate forms” as for example $\infty+\infty=\infty$, $\infty\cdot\infty=\infty$, meaning that the sum and the product of two such sequences also have limit $\infty$.

In that point of view no statement may be made in case of the “indeterminate forms” such as $\infty-\infty$ or $\infty/\infty$.

$\endgroup$
0
$\begingroup$

I think one key point is: the notation you use to write down an expression is not the expression. It is a shorthand for the expression. As in your example, an expression might have different representations. These are algebraically equivalent (you can pass from one to the other by means of addition or multiplication) but need not be equivalent from other points of view. Here by equivalent I mean equivalently suited to represent the expressions. Again, in your example, one of the two algebraically equivalent notation is ill equipped to describe the limiting behavior of the expression. For limiting operation we require a weaker consistency than algebraic consistency, namely that if the limit can be written across algebraically equivalent notation then they have to agree (and limits kindly oblige). When the limit can't be written out we just accept that that notation doesn't work to describe the limit.

Also, you can think this as the converse of imposing existence conditions when working algebraically. If you divide by an unknown you require it to be different than zero. Algebraic notation cannot handle "division by zero". But limiting notation can :)

$\endgroup$
-1
$\begingroup$

Let denote $A=\frac{a_n}{b_n}$ with $a_n\to L_1\in \mathbb{R}\,$ and $b_n=L_2\neq 0\in \mathbb{R}\,$ then in this case (by continuity) we can easily calculate the limit

$$\lim_{n\rightarrow \infty}\frac{a_n}{b_n}=\frac{L_1}{L_2}$$

Otherwise when both $a_n\to 0$ and $b_n\to0$ or $a_n\to \pm \infty$ and $b_n\to \pm\infty$ we write simbolically

$$\lim_{n\rightarrow \infty}\frac{a_n}{b_n}=\frac{0}{0}\quad \lim_{n\rightarrow \infty}\frac{a_n}{b_n}=\frac{\pm \infty}{\pm \infty}$$

to express the fact the the limit in these cases are in an indeterminate form in the sense that they can't be evaluated directly (by continuity) but we need some extra work by algebraic trick or theorems on limit.

In your example we write

$$\lim_{x\rightarrow3}\frac{(x-3)(x+3)}{x-3}=\frac{0}{0}$$

since the expression is in an indeterminate form, but since the limit implies that by definition $x\neq3$ we can cancel out the term $x-3$ and obtain

$$\lim_{x\rightarrow3}\frac{(x-3)(x+3)}{x-3}=\lim_{x\rightarrow3} \,(x+3)=6$$

since after this step (a trivial algebraic trick in this case) it is no more in an indeterminate form and can be evaluated directly plugging in the value $x=3$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.