I Can indeterminate limits always be evaluated?

1. Jan 23, 2017

Mr Davis 97

For example, if we have something of the form $\lim_{x -> 0} \frac{f(x)}{g(x)} = \frac{0}{0}$ or any other of the indeterminate forms involving 0 and infinity, is there always a procedure (such as l'Hopital's rule) by which we can find out the limit, whether it be a limit that converges to a value or one that diverges to infinity? In other words, does there exist a limit problem like this where we can only say that the limit is indeterminate and nothing else?

2. Jan 23, 2017

Stephen Tashi

We have to distinguish between two thoughts:
1) No human being can determine if the given limit is convergent or divergent
2) The given limit is neither convergent or divergent - thus explaining the failure of human beings to determine whether it is convergent or divergent.

Thought 2) is impossible if we accept that the definition of the limit produces a statement - i.e. that when applied to a specific limit, the requirements for a limit to be convergent are either a true statement or a false statement. Taking "divergent" to mean "not convergent", we have that each limit must be convergent or divergent. To argue against 2) , I think you would have to formulate an argument that there is some ambiguity in the definition of limit that would make the definition not actually a statement.

Thought 1) might be reformulated as a statement about "computability". There are technical definitions for what sorts of functions are "computable". If the limit involved a function that was not computable, that would suggest that no human being could evaluate the convergence of the limit. However, there might be a loophole in that argument. It may be possible to compute certain properties of an uncomputable function, even though we cannot compute its numerical value.

3. Jan 24, 2017

Mr Davis 97

Great response. For thought 1), what about just in terms of functions one might find in a standard calculus textbook? Does there always exist some method, whether its algebra l'Hopital's rule or both, by which the limit can be evaluated?

4. Jan 24, 2017

Staff: Mentor

The problems that appear in calculus textbooks are chosen so that the limits can be determined, at least that's been my experience.

5. Jan 24, 2017

Svein

One example that springs to mind is $\sin(\frac{1}{x})$ as x approaches 0.

The important thing here is the concept of cluster point. The difference between a limit and a cluster point is given as:
• Given a sequence {xn}, we say that y is a cluster point for the sequence if $(\forall \epsilon >0)(\forall N)(\exists n>N)(\vert y-x_{n}\vert <\epsilon)$
• Given a sequence {xn}, we say that y is a limit for the sequence if $(\forall \epsilon >0)(\exists N)(\forall n>N)(\vert y-x_{n} \vert<\epsilon)$
Thus a limit is a cluster point but not vice versa.

6. Jan 24, 2017

Stephen Tashi

A smart-alec answer would be "Yes, unless the problem has a misprint". Textbooks intend to pose problems that are supposed to be solvable. Instructors, on the other hand, can be eccentric. I've read that John Nash gave famous unsolved mathematical problems as homework or test problems to students in lower level college math classes. He defended this practice by saying that someone who didn't know about past attempts might come up with something completely new.

A more serious answer would be to formulate a precise definition for a "closed form expression" and then ask if any limit of a closed form expression can be determined to be convergent or divergent by some precisely defined set of mathematical techniques. I won't tackle that project because making the definitions would be elaborate and I have no idea how to reach a conclusion after the precise definitions are made.

7. Jan 24, 2017

Stephen Tashi

That's an example if we define "divergent" to mean $\lim_{x\rightarrow a} f(x) = \infty$ or $\lim_{x\rightarrow a} f(x) = -\infty$.

If we define "divergent" to mean "not convergent", the limit is divergent.

8. Jan 25, 2017

Svein

It does illustrate my take on "cluster points", though, since every point in [-1, 1] is a cluster point.

9. Jan 25, 2017

MAGNIBORO

so... the answer is that the limit $\sin(\frac{1}{x})$ as x approaches 0 is indeterminate?
(I don't know much about formal language)

10. Jan 25, 2017

FactChecker

If you don't put additional conditions on f and g (like continuity), you can probably make up examples that do anything you want.
Example: Suppose g(x) ≡x and f(x) = x if x is rational and -x otherwise. Then the values of f(x)/g(x) are ±1 and there is no limit.
You can always say something about the behavior, but it may not be what you are looking for.

Last edited: Jan 25, 2017
11. Jan 25, 2017

Staff: Mentor

I wouldn't call it "indeterminate." This limit just flat doesn't exist.

$\lim_{x \to 1}\frac{x^2 - 1}{x - 1}$ has the indeterminate form $[\frac 0 0]$, but it can be shown that this limit does exist and is equal to 2.