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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 321

Hi, I'm having trouble understanding the following fact about limits :

If f(x)<=g(x) for all x on (a,b) (except possibly at c) and a<c<b then,

lim f(x) <= lim g(x)

x -> c x->c

Here's how I interpret the definition : We have two functions f(x) and g(x), and the inequality f(x)<=g(x) hold true for all values that are not c. (That our interval (a,b)) If we were to evaluate the functions at c (considering that we can do it for our two functions.) then the inequality wouldn't hold anymore. (For example, f(x) would be superiro to g(x))

Please tell me if I have any errors.

THank you!

If you want to read more, go here : http://tutorial.math.lamar.edu/Classes/CalcI/ComputingLimits.aspx

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**Maburo****Member**- From: Alberta, Canada
- Registered: 2013-01-08
- Posts: 287

Hi Al-Allo,

If f(x) ≤ g(x) for all x ∈ (a,b), except maybe at x=c where a≤c≤b, that means that f(x) will never be greater than or equal to g(x), unless at x=c. For this to be true, at x=c, f(x)=g(x), or one of the functions is discontinuous at that single point (this will not change the limit). This should make sense, because if f(c)>g(c) and f(x) is continuous, it must have surpassed g(x) at some point other than x=c, so it must be true that either f(x)≤g(x), or f(x) has a disconinous point at x=c, making f(c)>g(c). The single disconinuous point will not change the limit of f(x) as x tends to c.

In other words, either f(c)<g(c), or f(c)=g(c), or f(c)>g(c) only occurs at a single point which will not alter the limit of f(x) as x tends to c.

Thus,

Does this help?

"Pure mathematics is, in its way, the poetry of logical ideas."

-Albert Einstein

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 321

...

*Last edited by Al-Allo (2014-07-09 06:57:54)*

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 321

Ok, my problem to why I was not understanding is because I thought that the limit inequality needed to hold ! So if the limit of f(x)= L and g(x)=M

I could have L>M

right ?

*Last edited by Al-Allo (2014-07-09 07:01:15)*

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**Maburo****Member**- From: Alberta, Canada
- Registered: 2013-01-08
- Posts: 287

No, that is not possible. We have f(x)≤g(x) in the interval (a,b), except at x=c. This means that it is possible to have f(x)≥g(x) ONLY at the point x=c. If f(x) and g(x) are both continuous, then you can have f(c)>g(c) only if at some point, d, with d<c, f(d)>g(d). But we said that f(x) can only be greater than or equal to g(x) at x=c, not at x=d. So if f(x) and g(x) are both continuous, then f(x) cannot be greater than g(x), and it should make sense that the limit would hold. But, it is still possible to have f(c)>g(c). Although this can only happen if either f(x) or g(x) are discontinuous at x=c. So imagine f(x) is smooth, but jumps up at the point x=c. In this case we have only f(c)>g(c). Now we must consider the limit of f(x) as x tends to c.

So, let

. Because f(x) is not continuous at x=c (it jumps), then f(c) ≠ L. In fact, in the case of this example, f(c)>L. So, the inequality still holds since the limit of f(x) as x tends to c remains less than or equal to g(x)."Pure mathematics is, in its way, the poetry of logical ideas."

-Albert Einstein

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 321

I was wondering, when we consider several functions at once in the same graph, is it ok if this whole is not a function itself ??? Do we care about whether this whole is function or not ?

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**Maburo****Member**- From: Alberta, Canada
- Registered: 2013-01-08
- Posts: 287

I am not sure I understand your question. What do you mean by the whole?

"Pure mathematics is, in its way, the poetry of logical ideas."

-Albert Einstein

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 321

I mean, when we define a function with the use of several sub functions, we call that a piecewise function, right ?

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**Maburo****Member**- From: Alberta, Canada
- Registered: 2013-01-08
- Posts: 287

Yes, that sounds right. A piecewise function could be a good example to explain that inequality.

For example:

If you graph these, you can see that f(x)≤g(x) everywhere except at x=0. So let c=0.

We still have f(x)≤g(x) on (a,b) and c=0, with a<c<b. But here, f(c)=3>g(c)=0. Yet,

I hope this clears everything up.

And yes, f(x), although piecewise, is still a function.

"Pure mathematics is, in its way, the poetry of logical ideas."

-Albert Einstein

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 321

Ok, but my question is, does this "sum" of function need to be a function itself ? We can have piecewise functions and non piecewise function. In other words, can we evaluate limits of non piece wise function (It doesn't satisfy the vertical line test.)

*Last edited by Al-Allo (2014-07-09 09:52:15)*

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 321

By the way, I created a sort of graph to help myself see the verity and that's how I see it :

http://i.imgur.com/emdBFyK.jpg

Of course, there are several other types of graphs that makes the "fact" true. Would my graph be correct? ( the limit is at the discontinuous part)

*Last edited by Al-Allo (2014-07-09 10:00:36)*

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**Maburo****Member**- From: Alberta, Canada
- Registered: 2013-01-08
- Posts: 287

A piecewise function isn't really a sum of functions. It is a function made up of sub-functions defined on subintervals its domain.

An example of a peicewise function is

I suppose you could create a graph with piecewise notation which does not satisfy the vertical line test. Just make the subintervals of 2 or more sub-functions overlap. An example:

You cannot evaluate the limit of y as x approaches any value between -2 and 0, because y takes on two values.

Your graph is correct in displaying the inequality. Let the red line be g(x), and the black one be f(x). f(x)≤g(x) at everywhere shown, except at x=5. So f(5)>g(5). Yet,

. I think you got it!*Last edited by Maburo (2014-07-09 10:10:37)*

"Pure mathematics is, in its way, the poetry of logical ideas."

-Albert Einstein

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 321

Ok, then I understod the fact. Now, concerning my function questiom, if we take my graph, we see that at the point x=-3 we have two y's. The black and red line are both functions. But with the vertical line test, we see that this isn't a function. So, my question is, this "fact" about the inequalities doesn't necessarily need the whole (non piece wise function) to be a function.

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**Maburo****Member**- From: Alberta, Canada
- Registered: 2013-01-08
- Posts: 287

The red and black curves are separate functions, so we should expect two values of y at each x where f(x) and g(x) are defined. One for each function.

I am not sure I understand the question.

If you are asking whether this fact is true wherever the f(x) and g(x) are defined, then no, that is not necessary. It works on any interval (a,b) where f(x) ≤ g(x). So maybe f(x)≤g(x) only on the interval (-5,2). This fact will still hold true.

"Pure mathematics is, in its way, the poetry of logical ideas."

-Albert Einstein

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 321

Ok, I understand now. My error was to think of both curves that I drew as forming one function, in the same manner as piece wise functions. But both these curves form distinct functions. Thank you!

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**Maburo****Member**- From: Alberta, Canada
- Registered: 2013-01-08
- Posts: 287

No problem! Glad I could help.

"Pure mathematics is, in its way, the poetry of logical ideas."

-Albert Einstein

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 321

Oh, I wouldn't want to annoy you, but I have another question.

If we don't consider my graph as two separate functions(So, we don't have a function anymore, only a relation.So it doesn't pass the vertical line test), and we still want to evaluate the limit at the same point, it would still be acceptable, even if its not a function?

*Last edited by Al-Allo (2014-07-09 10:53:09)*

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**Maburo****Member**- From: Alberta, Canada
- Registered: 2013-01-08
- Posts: 287

To find a limit for something that is not a function would probably require you to use parametric or polar coordinates. For example, you could find the limit of x as t tends to c, and the limit of y as t tends to c, for a parametric plot. Then you would have the point (x,y) for the limit as t tends to c. And for a polar plot, you would have to limit θ to avoid repetition.

In the case of your graph, there is no limit as x tends to -3 because it takes on two different values of y there.

"Pure mathematics is, in its way, the poetry of logical ideas."

-Albert Einstein

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 321

Maburo wrote:

To find a limit for something that is not a function would probably require you to use parametric or polar coordinates. For example, you could find the limit of x as t tends to c, and the limit of y as t tends to c, for a parametric plot. Then you would have the point (x,y) for the limit as t tends to c. And for a polar plot, you would have to limit θ to avoid repetition.

In the case of your graph, there is no limit as x tends to -3 because it takes on two different values of y there.

ok, but at -9, the limit would exist, right ?

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**Maburo****Member**- From: Alberta, Canada
- Registered: 2013-01-08
- Posts: 287

I suppose so, but the problem is that we cannot find the limit algebraically. This would be some sort of implicity defined plot, and cannot be expressed as y=... So we can't exactly write an expression describing the limit of y as x tends to any value.

If you are interested in finding limits with curves that are not functions, try doing it with some parametric plots. They can be very interesting looking graphs.

"Pure mathematics is, in its way, the poetry of logical ideas."

-Albert Einstein

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 321

Ok, thank you again for all. For some reason, my head wasn't here today... Do you want to know why I kept bugging ? I assumed, against my will, that limit and function value (the point c) were the same thing.... I don't why I did this.... Anyway...

*Last edited by Al-Allo (2014-07-09 11:17:41)*

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**Maburo****Member**- From: Alberta, Canada
- Registered: 2013-01-08
- Posts: 287

Ah yes, but not all functions are continuous or defined for all x. I'm glad we got all that cleared up, though!

"Pure mathematics is, in its way, the poetry of logical ideas."

-Albert Einstein

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 321

Maburo wrote:

Ah yes, but not all functions are continuous or defined for all x. I'm glad we got all that cleared up, though!

Yes, that's exactly the problem ! (I wasn't sure you would understand what I said.) Like you said, atleast I finally got it. I hate it when things like that happens In the best case, I should just forget everything that I assumed about a particular thing

*Last edited by Al-Allo (2014-07-09 11:23:40)*

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**Maburo****Member**- From: Alberta, Canada
- Registered: 2013-01-08
- Posts: 287

Don't worry, it happens to everybody! I think you are right, I don't think assumptions are necessary for understanding - sometimes it's easier to start learning something from scratch!

"Pure mathematics is, in its way, the poetry of logical ideas."

-Albert Einstein

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 321

Maburo wrote:

Don't worry, it happens to everybody! I think you are right, I don't think assumptions are necessary for understanding - sometimes it's easier to start learning something from scratch!

Exactly!Anyway, I need to go now. Thanks (for the millionth time) for the dedication!

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