Why Do Similar Looking Limits Yield Different Results?

[HallsofIvy]

All I can say is that Mark44 is completely correct. The limit is NOT necessarily the value of the function and, in particular, some thing like $(x^2- 4)/(x- 2)$ is NOT defined at x= 2.

 

[belliott4488]

I assume you mean "at x=0", but in any case the rest is bunk.

Yes, I did mean that, sorry - but can you please tone down your language a bit? I'm just trying to understand this.

Have you read what L'Hopital's rule says? It does NOT require either continuity or differentiability at the limit point.

Yes, and the definition I read says that the limit of the ratio of the derivatives must exist as x approaches the limiting value, which I took to mean that the derivatives must exist at that point.

From, e.g., Apostol's "Calculus, Vol. 1":

Theorem 7.9. L'Hopital's rule for 0/0. Assume f and g have derivatives f'(x) and g'(x) at each point x of an open interval (a,b), and suppose that

$$\lim_{x\rightarrow a^+}f(x) = 0$$

and

$$\lim_{x\rightarrow a^+}g(x) = 0$$

Also assume that

$$g'(x) \neq 0$$

for each x in (a,b). If the limit

$$\lim_{x \rightarrow a^+} \frac{f'(x)}{g'(x)}$$

exists and has the value L, say, then the limit

$$\lim_{x \rightarrow a^+} \frac{f(x)}{g(x)}$$

also exists and has the value L. [end statement of theorem]

Neither f'(a) nor g'(a) (nor for that matter, f(a) or g(a)) is required to exist.

Okay, so f'(x) and g'(x) must exist on the open interval (a,b), but not necessarily at x=a? I don't understand how L'Hopital works on a discontinuous function such as the one you've suggested. The proof I have in front of me uses the Cauchy Mean Value Theorem to prove L'Hopital, but does that apply to discontinuous functions?

 

[belliott4488]

All I can say is that Mark44 is completely correct. The limit is NOT necessarily the value of the function and, in particular, some thing like $(x^2- 4)/(x- 2)$ is NOT defined at x= 2.

Okay, thanks, guys. As I said in my earlier posts, I suspected that I had a basic misunderstanding somewhere, and I guess this must be it.

I had thought that where a function is indeterminate, we simply can't say what its value is, but that we do not necessarily assert that it is discontinuous there. When you evaluate the limit at that point, I thought this was equivalent to evaluating the function at that point, but you're telling me that we must actually set the value of the function equal to the limit at that point, by hand, so to speak.

Okay, I think I've got it now.

 

[jbunniii]

Yes, I did mean that, sorry - but can you please tone down your language a bit? I'm just trying to understand this.

Sure, and I apologize for becoming a bit impatient. It sounded like you were telling us all that we didn't know what we were talking about and didn't understand basic calculus, but perhaps that was a misread on my part.

Yes, and the definition I read says that the limit of the ratio of the derivatives must exist as x approaches the limiting value, which I took to mean that the derivatives must exist at that point.

Okay, so f'(x) and g'(x) must exist on the open interval (a,b), but not necessarily at x=a? I don't understand how L'Hopital works on a discontinuous function such as the one you've suggested. The proof I have in front of me uses the Cauchy Mean Value Theorem to prove L'Hopital, but does that apply to discontinuous functions?

Right, you are free to change f(a) or g(a) or both, in any way you like, or even leave them completely unspecified, and the answer that L'Hopital gives will be the same.

I'm not sure of the specifics of your proof, but I bet it's stated in such a way that it never evaluates f or g at the point a, but only at points arbitrarily close to a. (i.e., on an interval of the form (a,b)). Under the hypotheses of L'Hopital's rule, the mean value theorem applies on (a,b), but not on any interval that INCLUDES the point a.

 

[Hurkyl]

In calculus, we usually don't actually care about the values of functions at individual points; it's the "bulk" behavior of the function that we're really interested in. So the thing we often do is to take the "continuous extension" of whatever partial function we have described. Unfortunately, this step doesn't often seem to be explicitly mentioned in introductory courses.


For example, consider the expression^* f(x) := x/x. f(x) is constructed by taking the "diagonal" function $\Delta(x) = (x, x)$ with the "division" partial function $q(x, y) = x/y$.

q is only a partial function, because it's defined only for those (x, y) with y nonzero.

Alas, this means f is also merely a partial function, because the image of $\Delta$ does not lie in the domain of q. In particular, f is undefined at 0.

However, the limit of f at 0 exists; the "continuous" extension of f plugs this hole, and is the constant function 1.



*: x is a variable whose domain is all real numbers

 

[jbunniii]

I don't understand how L'Hopital works on a discontinuous function such as the one you've suggested.

Well, it works the same as if the function were continuous, because it doesn't actually make any claim about the value of the ratio AT the point a, only the limit as you approach a.

It can be useful for example, if you WANT to define a continuous function but you need to know "what value should I assign at x=a that will make the function continuous?"

For example, if I want to define a function of the form sin(x)/x, this expression itself is undefined at x=0, but I may be able to continuously extend it to x=0 by defining

f(x) = sin(x)/x if x is nonzero
f(0) = c

Then L'Hopital's rule will tell me what value of c (if any) will make f continuous.

 

[belliott4488]

Sure, and I apologize for becoming a bit impatient. It sounded like you were telling us all that we didn't know what we were talking about and didn't understand basic calculus, but perhaps that was a misread on my part.

No, I probably got too snippy because it seemed like some of the earlier responses I got just kept repeating the same assertion without explanation, which gets frustrating. Sometimes that's due to people not actually having an explanation, but there was no reason to assume that here.

I'm not sure of the specifics of your proof, but I bet it's stated in such a way that it never evaluates f or g at the point a, but only at points arbitrarily close to a. (i.e., on an interval of the form (a,b)). Under the hypotheses of L'Hopital's rule, the mean value theorem applies on (a,b), but not on any interval that INCLUDES the point a.

Yes, that is what it does.

Well, it works the same as if the function were continuous, because it doesn't actually make any claim about the value of the ratio AT the point a, only the limit as you approach a.

Once again, my understanding was too naive. I had thought it was telling us exactly what value of the function is at a, in cases where we could not examine the function directly, but now I see that's really not correct.

It can be useful for example, if you WANT to define a continuous function but you need to know "what value should I assign at x=a that will make the function continuous?"

For example, if I want to define a function of the form sin(x)/x, this expression itself is undefined at x=0, but I may be able to continuously extend it to x=0 by defining

f(x) = sin(x)/x if x is nonzero
f(0) = c

Then L'Hopital's rule will tell me what value of c (if any) will make f continuous.

Thanks - I've got a much better picture in my head now. c is not the "true" value of the function, which we had simply been unable to find explicitly before application of L'Hopital; rather, it is a the value towards which the function tends, and therefore which makes the function continuous at that point, but which we must insert "by hand". Close enough?

I think I get it, but my intuition is going to take a little longer to catch up. I still want to say that the "hole" in the function reflects only our inability to state the "true" value of the function, but that the limit allows us to discover the "true" value that belongs in the hole .. but that's not very mathematical, is it? I'll work on fixing my intuition.

 

[belliott4488]

In calculus, we usually don't actually care about the values of functions at individual points; it's the "bulk" behavior of the function that we're really interested in. So the thing we often do is to take the "continuous extension" of whatever partial function we have described. Unfortunately, this step doesn't often seem to be explicitly mentioned in introductory courses.


For example, consider the expression^* f(x) := x/x. f(x) is constructed by taking the "diagonal" function $\Delta(x) = (x, x)$ with the "division" partial function $q(x, y) = x/y$.

q is only a partial function, because it's defined only for those (x, y) with y nonzero.

Alas, this means f is also merely a partial function, because the image of $\Delta$ does not lie in the domain of q. In particular, f is undefined at 0.

However, the limit of f at 0 exists; the "continuous" extension of f plugs this hole, and is the constant function 1.



*: x is a variable whose domain is all real numbers

Okay! You've now convinced me of what I suspected all along: a deeper understanding of this question requires more mathematical knowledge than I possess! I'm overwhelmed, but thanks again, guys.

 

[jbunniii]

Thanks - I've got a much better picture in my head now. c is not the "true" value of the function, which we had simply been unable to find explicitly before application of L'Hopital; rather, it is a the value towards which the function tends, and therefore which makes the function continuous at that point, but which we must insert "by hand". Close enough?

That's correct.

I think I get it, but my intuition is going to take a little longer to catch up. I still want to say that the "hole" in the function reflects only our inability to state the "true" value of the function, but that the limit allows us to discover the "true" value that belongs in the hole .. but that's not very mathematical, is it? I'll work on fixing my intuition.

I think it's because your intuition wants the function to be continuous, in which case the natural thing to do is to "plug in the hole" with the value indicated by L'Hopital. That's a completely reasonable way to think about it.

I think the main point of this thread is that there is indeed a hole, and it isn't automatically filled by the value L'Hopital gives you, and that you're free to completely ignore that value and "plug in the hole" with some other value if you like. You will get a discontinuous function in that case, but there's nothing invalid about that.

 

[HallsofIvy]

Unfortunately, many students in Calculus I or Precalculus get the impression that "limit" is just a fancy way of talking about the value of a function! It happens that for continuous functions, the limit is equal to the value of the function. It just happens that because continuous functions are easy to work our ways of writing functions have developed so that all of the functions that we can write as easy formulas are continuous. Actually "almost all" functions (in a very specific sense) are not continuous any where!

 

[belliott4488]

Unfortunately, many students in Calculus I or Precalculus get the impression that "limit" is just a fancy way of talking about the value of a function! It happens that for continuous functions, the limit is equal to the value of the function.

Well, now that you say it explicitly, it seems kind of obvious. I guess my intuition is catching up with my intellect.

just happens that because continuous functions are easy to work our ways of writing functions have developed so that all of the functions that we can write as easy formulas are continuous. Actually "almost all" functions (in a very specific sense) are not continuous any where!

Now, that's just crazy talk!

Any way to explain that to a non-mathematician? Are you using the word "function" in a more general and abstract way than the usual layman's way of using it?

 

[AUMathTutor]

A function associates elements in a domain with elements in a range. Specifically, every element of the domain is mapped to exactly one element of the range.

The reason there are more continuous nowhere functions than continuous somewhere functions? In the discrete case, I would guess a combinatorial answer would suffice. In the continuous case, I don't know but I still believe it.

 

[tiny-tim]

Hi belliott4488!

just happens that because continuous functions are easy to work our ways of writing functions have developed so that all of the functions that we can write as easy formulas are continuous. Actually "almost all" functions (in a very specific sense) are not continuous any where!

Now, that's just crazy talk!

Any way to explain that to a non-mathematician? Are you using the word "function" in a more general and abstract way than the usual layman's way of using it?

Mathematically, "function" has a very wide meaning.

But don't worry about it … all "sensible" functions that you're going to come across will be continuous and defined, except perhaps at a few "singular" points.