True or False? (2 Viewers)

[brett86]

a) You can't use l'Hôpitals rule in this situation

proved u can use l'hopital's rule (look at my last posts)

you should never try to act like you know something when you don't or at least don't have a logical basis behind it

i did have a logical basis, all i did wrong was say x → 0 instead of x → 0⁺

this is just a notation error, my argument for 0⁰ was completely correct

maybe u should stop wasting our time with ur stupid, petty arguments

 

[acmilan]

hey acmilan, this is an interesting thread but should it be moved to the Appreciating the Beauty and Elegance (extracurricular topics) forum?

I agree, it may be given greater exposure to people who have more experience in it than me.

 

[Xayma]

yes u can!

to evaluate the limit of xlog_ex as x → 0 using l'hopitals rule: [EDIT: this should be x → 0⁺, just so Xayma doesnt have a cry lol]

yeah acmilan pointed that out to me. I prefer just to remember that xln x approaches 0 as x approaches 0.

Point still stands that from the left (using complex numbers) it approaches -∞+iπ

Whereas √x will approach 0 from both the left and right.

 

[samuelblayden]

true mate I think

 

[who_loves_maths]

Originally Posted by Slide_Rule
0/0 does not take the value of 1, is never agreed to be 1, in any cirumstance. It, unlike 0^0, is a violation of an axiom.

Do not say I am ignoring parts of your post where you explain how 0/0 is 1, because never have you explained how one can violate a fields axioms and remain mathematically correct.

You see I think perhaps you miss a fundamental distinction: Division by zero is undefined - it is a violation of an axiom. Division by a limit that goes to zero is indeterminate. In the case of two limits going to zero, one over the other, L'Hopitals rule quickly solves the problem. However, again, two limits which both approach zero, one over the other, are not said to equal one, as 0^0 is, but must evaluated on a case by case basis.

Examine:
Lim(ln(x+1))/Lim(x^3-sin(x)) as x->0
Which is Lim(1/(x+1))/Lim(3x^2-cos(x)) as x->0
Which is (1/2)/(-1)=-1/2. Not one.

Mathematicians often assume 0^0=1. Mathematicians never assume 0/0 is 1 (notably because it would say 0=1).

I'm in a cranky mood, so just ignore me if you want. I don't mean to upset you, but I do wish to be mathematical.

Slide_Rule, i maintain my stance on the fact that you have not read my initial posts carefully. once again, you pick a few sentences and words i have said informally, take them out of context, and criticise them unjustly.

i need to clarify to you that i have never acknowledged at all that 0/0 is an actual number! if you read my two initial posts in this thread carefully, then you'll see that i have always coupled 0/0 with the words "limits" or "limiting value" to make sure i do not cause confusion, and have always referred to them as indeterminate, not definite.
it was only in the latter posts that i started to loosely refer to "the limiting value of 0/0" simply as "0/0". but through these casual references i have alway thought that ppl, such as yourself, having read the previous posts, would understand that i was referring to its limiting value, not the undefined number itself. clearly i was wrong about this.

maybe i shouldn't have been so casual with the way i wrote, but be mindful that this thread, and this forum, is largely informal. the language of this forum is not stipulated to be noncolloquial by necessity. however, i also acknowledge that i should have been less lazy and properly referred to 0/0 in my latter posts as that of the limiting case, as this can very easily cause confusion like it did with you.

as for what you have said about division by zero, you are completely correct. and likewise, i am well aware of this restriction too. like i said, i have always implied the limiting value of 0/0 whenever i said 0/0 so this is a simple case of misunderstanding. however, you did clear this point up with whoever might be reading this thread, so thankyou for that.

the bottom line is, there is no real argument between us. we agree, and have always agreed, on the same things - that 0/0 is not a number in its own rights, but it has an accepted limiting value of 1 (in most cases).

 

[who_loves_maths]

Slide_Rule, here is my first two posts in this thread, i have not changed them in any way since. i've bolded the parts where i have specifically referred at all times to my definition of 0/0 = 1 as ONLY in the limiting value case:

1st post:

Originally Posted by who_loves_maths
we musn't forget that in the face of the Calculus branch of mathematics (even at the current HSC level), 0/0 is widely accepted to have the limiting value of 1.

it is incorrect and incomplete to generalise the statement that "0/0 is simply an undefined quantity" without the mention of possible limiting values. (of which 0 itself, other than 1, is also another)

Edit: in this situation, 0/0 is said to be indeterminate, rather than 'undefined'. the word 'indeterminate' in mathematics implies that there are more than one accepted or used value of, in this case, 0/0 depending on specific situations.

2nd post:

Originally Posted by who_loves_maths
as i mentioned in my last post, at the HSC level 0/0 is indeed defined. (as a limiting value)

the fact that Sin(x)/x = 1 as 'x' --> 0, depends on the definition that 0/0 =1

for the second post, you attacked me before on my line "depends on the definition that 0/0 =1" - when you can clearly see that had you taken that WHOLE post into account, then this debate would not have been necessary. i clearly specified in the first sentence of the post in brackets "as a limiting value".
this is what i mean when i said you take my words frequently "out of context".

i thought that most ppl who read that post would have taken that line into account in the second, but you did not - however, like i said in my last post, it's probably my fault since maybe i should have made it even clearer. but of course, i hope you can see now that the correct intentions were always there in the first place.

 

[ogmzergrush]

Slide_Rule, here is my first two posts in this thread, i have not changed them in any way since. i've bolded the parts where i have specifically referred at all times to my definition of 0/0 = 1 as ONLY in the limiting value case:

1st post:


2nd post:


for the second post, you attacked me before on my line "depends on the definition that 0/0 =1" - when you can clearly see that had you taken that WHOLE post into account, then this debate would not have been necessary. i clearly specified in the first sentence of the post in brackets "as a limiting value".
this is what i mean when i said you take my words frequently "out of context".

i thought that most ppl who read that post would have taken that line into account in the second, but you did not - however, like i said in my last post, it's probably my fault since maybe i should have made it even clearer. but of course, i hope you can see now that the correct intentions were always there in the first place.

That is all.

 

[who_loves_maths]

Originally Posted by acmilan
Sorry for this, but where does it state a function must be continuous for the delta-epsilon definition of a limit to apply? I am quite confident that you can find limits of functions that are not continuous.

i don't see how you could have missed it acmilan. it is near the middle of the web page i gave on Continuous Functions.

here's the quote:

" Many mathematicians prefer to define the continuity of a function via a so-called epsilon-delta definition of a limit. In this formalism, a limit... "

- check it out again if you don't believe me.

the point of that is that the epsilon-delta definition of a limit can and is used to define the continuity of functions about a particular point since it takes into account both sides within proximity of that point. {one of the three definitions of a continuous function, btw, is that such a limit does exist - this fact is also on that web page}
hence, since the curve y= x^x is NOT continuous about the point x= 0 {which is in fact a critical point} then you simply cannot apply the epsilon-delta definition of a limit to the point x= 0 on y= x^x , since this definition only works for limits on continuous intervals of a function/curve. in which case, l'Hopitals Rule is then applied to find the limit - and since its been acknowledged that the point is a critical point such that the epsilon-delta definition cannot be applied, then there is clearly no need to ask l'Hopitals Rule again to take into account both sides of the point x= 0 !

ie. put in another way, this is essentially what i am questioning:
before, when you applied the epsilon-delta definition to y= x^x at about x= 0, the definition failed. HOWEVER, my question is that how do you know it failed because (1) the limit really does not exist, or is it (2) because all its failure confirms is that the function is not continuous about x= 0 ?

ie. there is NO cause-effect relationship between a discontinuity on a curve and whether or not a limit there exists. just because a curve is discontinuous at a point does not mean a limit does not exist at that point!
in your previous posts you assumed that the result of applying the epsilon-delta definition, as you did, leads to implication number (1). however, you cannot be sure now that all it might indeed truly imply is in fact cause number (2), not (1).


i hope i've made my argument clearer now.


Edit: here's the webpage again:

Continuous Function and the Delta-Epsilon definition

 

[Xayma]

i don't see how you could have missed it acmilan. it is near the middle of the web page i gave on Continuous Functions.

here's the quote:

" Many mathematicians prefer to define the continuity of a function via a so-called epsilon-delta definition of a limit. In this formalism, a limit... "

- check it out again if you don't believe me.

the point of that is that the epsilon-delta definition of a limit can and is used to define the continuity of functions about a particular point since it takes into account both sides within proximity of that point. {one of the three definitions of a continuous function, btw, is that such a limit does exist - this fact is also on that web page}
hence, since the curve y= x^x is NOT continuous about the point x= 0 {which is in fact a critical point} then you simply cannot apply the epsilon-delta definition of a limit to the point x= 0 on y= x^x , since this definition only works for limits on continuous intervals of a function/curve. in which case, l'Hopitals Rule is then applied to find the limit - and since its been acknowledged that the point is a critical point such that the epsilon-delta definition cannot be applied, then there is clearly no need to ask l'Hopitals Rule again to take into account both sides of the point x= 0 !

Taking the limit from the left and right comes up with two different values, you can't just arbitarily take one to be the value of 0⁰.

 

[who_loves_maths]

Originally Posted by Xayma
b) you can extend logarithms into the complex field, you can see that lim sqrt (x) as x-->0 is equal to 0 both left and right however lim xx as x-->0- is equal to -∞+iπ which does not equal the same as from the right.

Originally Posted by Xayma
Taking the limit from the left and right comes up with two different values, you can't just arbitarily take one to be the value of 0^0.

a) we are not extending logarithms into the complex field here. we are taking into account the real limits, not complex ones which has not meaning in this thread. plz restrict yourself to the real number plane, which has no place for complex numbers. the limit, as seen on a graph of the function y= x^x, lies on a real number plane (esp. if you're going to apply techniques such as the epsilon-delta definition to the graph of the function). ie. complex numbers here is irrelevant.
{it's irrelevant because its impractical and unuseful by definition to assign a complex number as a limiting value for 0^0; you can also see this irrelevance in the arguments for the limiting value of 0^0 being = 1 in the webpages i've previously linked to. eg. 0 to 0 power }

b) simply because the point takes a real limit from one side and a complex one from the other does not mean the point does not have a limit at all or is undefined. two limiting values is NOT an argument of contradiction here - it simply means the limit can take different values depending on different situations. hence, you cannot use the reason that just because there's two limits (as calculated by you) then it must be undefined. it is not a contradiction in mathematics at all.

{btw, it is no joke that the fact is most professional mathematicians these days take the limiting value of 0^0 as = 1 . (some of these reasons are in the webpage i linked above) so if you are so sure about your argument, then maybe you can seek some expert opinion? (eg. your uni lecturers, etc...?)}

c) can you actually post up the method/technique/theory through which you arrived at the complex limit plz Xayma? i'm interested in having a look at it, thx in advance.

 

[who_loves_maths]

Originally Posted by Slide_Rule
I am not taking your posts out of context. I simply do not agree that the limiting value of two functions which tend to zero, each over the other, is commonly accepted to be 1 - it simply can't be (as I showed with my example). It is a different case to 0^0. Do not compare the two.

as i said already too, "(in most cases)" - ie. not all. and once again, i fail to realise what you are targeting with you last line of that quote (in bold). i did NOT compare 0/0 with 0^0 at all in my responses just before to your post?! nor have i since.

Originally Posted by Slide_Rule
The epsilon-delta definition of a limit is applicable so long as the limit is not being TAKEN at the discontinuity. It still applies to discontinuous functions. In fact, just to clarify, the epsilon-delta definition is the ONLY formal definition of a limit.

that's exactly what i said, and taking the limit AT the discontinuity is exactly what acmilan did. two posts ago, when i quoted from the mathworld site, i have already specified to acmilan that the reason why his application of the epsilon-delta definition of the limit does not work in this case is because he applied it to x= 0 for y= x^x, and at that point, the graph is discontinuous. hence, l'Hopitals Rule takes over, and his argument using the epsilon-delta definition falls apart.
and as i said, just because the epsilon-delta method fails to produce a limit does not mean a limit does not exist - just that at x = 0 the graph is discontinuous, which we already know is true.

you seem to say that i am implying that the epsilon-delta definition is not applicable in this case simply because the function y= x^x is not totally continuous - well this is not true. if you read my post to acmilan you'll see that in every instance i referred to the point x= 0 , at which acmilan used the technique, and not the generalisation of the entire curve.
clearly if you took a point at, say, x = 50, then the epsilon-delta definition would work since the graph is certainly continuous at that local region.
however, acmilan took it at the critical point.

 

[who_loves_maths]

Originally Posted by Slide_Rule
Yes it does. If you cannot use epsilon-delta, no limit exists.

you're very confusing to read. you agreed that you cannot apply the epsilon-delta definition to discontinuities... now you're saying that you can (since you've agreed that 0^0 DOES have a limiting value of 1)??? so if i can't apply the definition to y=x^x at x=0, then you're saying that there is no limiting value at x=0 on y=x^x??? where before you have already agreed that 0^0 has a limiting value of 1???

ie. by extension, you're saying that any discontinuities on any functions all do not have a limit because you can't apply the epsilon-delta definition to all such points. ?!

Originally Posted by Slide_Rule
The epsilon-delta definition of a limit is applicable so long as the limit is not being TAKEN at the discontinuity.

 

[brett86]

L'Hopital's rule has nothing to do with taking the limit of y=x^x as x->0

im sorry, but i dont understand what u mean:

i can use l'hopital's rule because:

also, many sources agree that 0⁰ = 1

Google:

Mathematica:

Maple:

 

[who_loves_maths]

Originally Posted by Slide_Rule
Correct. The point x=0 on y=x^x is a discontinuity. No limit exists at this point.

Didn't you wonder why there was no formal proof that 0^0=1?

ok, i see.

find lim(x --> 2) f(x) ; where f(x) = (x^2 -4)/(x -2)

so according to you, if this was in an exam, i should put "limit does not exist".

 

[acmilan]

ok, i see.

find lim(x --> 2) f(x) ; where f(x) = (x^2 -4)/(x -2)

so according to you, if this was in an exam, i should put "limit does not exist".

No that function is continuous at 2 and automatically you know the limit exists and equals 4. Remember you need to simplify before evaluating the limit, but you could also show it with l'Hopital's rule

 

[who_loves_maths]

Originally Posted by acmilan
No that function is continuous at 2, which already shows the limit exists and equals f(2).

lim x->2 (x^2 -4)/(x -2) = lim x->2 2x/1 = 4 = f(2)

We are guaranteed that lim x->2 exists because lim x->2- and lim x->2+ exist also and are equal to 4

no, that function is not continuous at x =2.

the function y= (x^2 -4)/(x -2) is not the same as y= x +2

i am surprised you don't know this, since you learn this fact as early as the start of year 11 i believe.

 

[brett86]

The point x=0 on y=x^x is a discontinuity. No limit exists at this point

but for the 0⁰ case the lhs limit cannot be looked at because the gradient for x^x is x^x(log_ex+1) which is not real if x <= 0 so no real gradient exists for the graph of x^x for x <= 0 and hence the graph does not exist for x <= 0

 

[brett86]

no, that function is not continuous at x =2.

the function y= (x^2 -4)/(x -2) is not the same as y= x +2

i am surprised you don't know this, since you learn this fact as early as the start of year 11 i believe.

good point! the limit exists by the ε - δ definition but the graph has a hole at x = 2 and is therefore not continuous

 

[who_loves_maths]

Originally Posted by acmilan
No that function is continuous at 2 and automatically you know the limit exists and equals 4. Remember you need to simplify before evaluating the limit...

no, the function is NOT continuous at x =2, it is a discontinuity.

and yes you do need to simplify before evaluating the limit. that is why i agree that the limit is indeed = 2. ie. i agree with you. but you didn't explain (Slide_Rule's) the use of epsilon-delta on this where x=2 is in fact discontinuous.

so my question was directed toward Slide_Rule's comments:

Originally Posted by Slide_Rule
The epsilon-delta definition of a limit is applicable so long as the limit is not being TAKEN at the discontinuity.

and,

Originally Posted by Slide_Rule
In fact, just to clarify, the epsilon-delta definition is the ONLY formal definition of a limit.

and,

Originally Posted by Slide_Rule
If you cannot use epsilon-delta, no limit exists.

so in actual fact, acmilan, you objections should be directed towards Slide_Rule, who is saying that since x=2 is a discontinuity on y=(x^2 -4)/(x -2), then you cannot apply the epsilon-delta definition (also being the "ONLY formal definition of a limit") to that point, in which case the limit according to him does not exist ?!

 

[Xayma]

Some limits dont exist because they are different from the left and right.

Such as f(x)=x, if x>0 and 7x+2, if x&le;0 doesn't have a limit at x=0.