True or False? (1 Viewer)

[Xayma]

0⁰ can be looked at as x^x as x → 0

x^x = e^xlog_ex

using l'hopital's rule, xlog_ex → 0 as x → 0

:. x^x as x → 0 goes to e⁰ which equals 1

:. 0⁰ can be interpreted as 1

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

b) You stuffed up anyway.

x^x=e^{xln x}

lim (x-->0⁺) x^x=e^{(lim (x-->0⁺) xln x)}
=e⁰
=1

lim (x-->0^-) x^x=e^{(lim (x-->0^-) xln x)}
Which does not exist (negative logarithm).

∴ lim (x-->0) x^x does not exist.

∴ your argument falls to bit

∴ 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

 

[withoutaface]

log_a(1) = 0, a > 1

True or false?

log(1) = log(1*1) = log(1)+log(1) (log laws)
0 = log(1)

QED

 

[who_loves_maths]

Originally Posted by maths > english
point taken, i have sounded really stuck-up in this thread

hey maths > english, plz don't get me wrong... i was NOT being sarcastic when i said "is this maths > english demonstrating his mathematical prowess? "

i was quite serious ... and it was supposed to be a playful compliment
your 'proofs', or rather 'demonstrations', were informative, explanatory and helpful in my opinion :uhhuh:

Originally Posted by Xayma
[referring to maths > english]
...
a) You can't use l'Hôpitals rule in this situation.

b) You stuffed up anyway.

x^x=e^xln x

lim (x-->0+) x^x=e(lim (x-->0+) xln x)
=e0
=1

lim (x-->0-) x^x=e(lim (x-->0-) xln x)
Which does not exist (negative logarithm).

∴ lim (x-->0) x^x does not exist.

∴ your argument falls to bit

∴ 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 .

Xayma, your last statement was unnecessary and particularly hostile without basis. the demonstration of the limit by maths > english in relation to 0^0 is perfectly legitimate and accepted by the mathematics community - it is you who are out of touch.

unfortunately your argument fails even before it began. you see, the graph of y= x^x in ONLY defined or exists for x >= 0. that is to say, the graph is CONTINUOUS for x >= 0 . However, for x < 0 , the graph is discrete and discontinuous at irregular points - meaning that the function is NOT defined for x < 0 !

hence, you cannot not loosely apply l'Hopitals Rule or other limiting rules to y=x^x as 'x' --> 0 from the negative side - simply because the negative side does NOT exist.

your argument is its own downfall:

" ... Which does not exist (negative logarithm)... "

In addition, if would've been more wise of you if you'd taken the liberty and time to perhaps plot the graph y=x^x on a graphing program to confirm your own suspicions - in which case you would then not needed to have posted since the programs would have shown you that the limit y=x^x, for x--> 0, is indeed 1 as maths > english correctly pointed out.


let's take a very simple example of where your argument falls apart in school mathematics:

find the limit y= Sqrt(x) as x --> 0

of course, most ppl in this case would agree that the limit is indeed = 0 as x --> 0
as Sqrt(0) = 0 {even calculators would agree to this.}

However, based on your perculiar argument:
as x --> 0 from the negative side, y= Sqrt(x) is undefined since the square-root of a negative number does not exist since it's not real!

hence, y= Sqrt(x) as x--> 0 does not have a limit and is indeed of an indeterminate form meaning that it can take up to many or none different values, according to you!


... i hope you do see the farce in this argument.


P.S. plz don't take offense from this post. i am not denouncing what you did, only the hostility with which you did it just to unfairly discredit maths > english upon an unjust basis. i personally have confidence in the validity of his demonstration and i don't think it is fair to dismiss him, and his opinions, so nonchalantly.

 

[acmilan]

I think you may have mistaken Xayma's arguement, who_loves_maths.

He was saying that the limit as x->0 of x^x does not exist, since by definition, that implies both limit as x->0^- of x^x and limit as x->0⁺ of x^x exist AND are equal, which as he pointed out is not the case. If im not mistaken, the case is the same for sqrt(x).

 

[who_loves_maths]

^ just adding to my last post, for your convenience Xayma, and in support of maths > english, here's the graph:

'x' to the power 'x'

this is plotted on a graphing program. notice how, at x =0, it passes through the origin. also, this graph is not defined for x < 0 .

 

[who_loves_maths]

Originally Posted by acmilan
I think you may have mistaken Xayma's arguement, who_loves_maths.

He was saying that the limit as x->0 of x^x does not exist, since by definition, that implies both limit as x->0- of x^x and limit as x->0+ of x^x exist AND are equal, which as he pointed out is not the case.

hi acmilan, no actually i did not misunderstand. i know exactly what Xayma was trying to say, and if you read my post carefully then you'd see that i was trying to say that Xayma's reason is illegitimate because you can't say that the limit exists ONLY if it is true for both the positive and negative side since in this special case, the graph is not defined and does NOT exist on the negative side - hence, you only need to consider x --> infinity from the positive side.

i hope you can understand what i'm trying to say.

if you take a look at the example of y = Sqrt(x) as x--> 0 that i gave in that post of mine then you will understand why you cannot always consider the approach to 0 from both sides - esp. when one side simply does not exist.

if you agree with Xayma, then i am to assume that you consider Sqrt(0) as a non-determined and undefined value? and that you don't agree that Sqrt(0) = 0 ?

 

[acmilan]

Hmmm i see where you are coming from, but im not entirely convinced that is the case. The definition of limit as x->a of f(x) is:

This implies that in a small interval delta about a (in this case a is 0), the function lies withing an interval epsilon of L (in this case L is 1). However from -delta to 0 and, imo, this is not the case. The limit would exist if the following criteria are satisfied, which I dont believe to be the case:

Haha you've made me think of this area of maths for the first time in about a month.

 

[who_loves_maths]

^ i am not as well versed as you in the language of pure and technical mathematical definitions and restrictions on account of the lax nature of the HSC. as a result i do not understand in entirety the symbols you have posted.

however, i suspect your basic argument is on the fact that the negative side approach must give you the same result as the postive approach? if this is the case then i retain what i said before: " you cannot impose this quota on the limit, purely because of 'definition', as one of the sides simply do not exist."

that is to say, most negative value you substitute into the definition of the limit for f(x) = x^x will not produce a clear defining parameter or differential as some expressions, such as (-0.5)^(-0.5), simply does not exist.

i also believe that at times, esp. in sciences such as maths, common sense should be used rather than sticking to an inflexible formulaic 'definition'. lateral thinking and thinking outside the square is an important part of disciplines such as mathematics. clearly, logic would prevail in this case that a negative approach to x = 0 is not possible for this particular function. all i can do is urge you in this case to exercise some creativity and imagination - after all, definitions are based on common sense and perception too, who is to say this definition will not alter in the future as more new functions are discovered?
i personally do not appreciate a rigid approach to mathematics.

in addition, plz be mindful of what i originally said about the limit 0^0 - that it is indeed indeterminate by definition, however is currently generally accepted as holding true to the value of 1. this is not a unanimous consensus within the totality of the mathematics community, but is in favour by majority.
and as i already said before, there is very strong support and evidence that the most suitable value that should be assigned to it is 1.

however, i am glad that i have caused you to look into "this area of mathematics for the first time in about a month."

 

[acmilan]

Sorry about the symbols, ill translate the first one, and the rest would almost be the same with slight variations:

"Given an epsilon greater than 0, there exists a delta greater than zero such that for every x, 0 < |x - a| < delta implies |f(x) - L| < epsilon".

As for using common sense on the issue, I still believe that limit x->0⁺ x^x implies that the limiting value of 0⁰ is 1, and I am not questioning that, and I dont think that the limit not existing at 0 in any way degrades the result. Another function that I think is interesting to take a look at is f(x) = -x^-x, which is also in the form of 0⁰ as x->0

 

[who_loves_maths]

Originally Posted by acmilan
Sorry about the symbols, ill translate the first one, and the rest would almost be the same with slight variations:

"Given an epsilon greater than 0, there exists a delta greater than zero such that for every x, 0 < |x - a| < delta implies |f(x) - L| < epsilon".

As for using common sense on the issue, I still believe that limit x->0+ xx implies that the limiting value of 00 is 1, and I am not questioning that, and I dont think that the limit not existing at 0 in any way degrades the result. Another function that I think is interesting to take a look at is f(x) = -x-x, which is also in the form of 00 as x->0

can you please explain in greater detail for me? sorry, i still don't understand the connections between epsilon, delta, and L.
what are they and how are they connected?
ie. i don't understand how 0 < x -a < delta is related to f(x) - L < epsilon .

i apologise if this is a dumb question. but if you can then plz explain it to me, thanks in advance

P.S. on the issue of whether or not a limit exists, all i can say is that we can at least agree that we disagree on this issue. you belive that a limit does not exist. whereas i belive the limit is indeterminate, but most useful in the form = 1.

 

[Dumsum]

Math discussions/arguments amuse me it's nice to see people who are expanding their horizons though. I'm too focused on the HSC which on its own is rather useless. I admire y'all

 

[who_loves_maths]

Originally Posted by Slide_Rule
Quote:
Originally Posted by who_loves_maths
in relation to the original question of this thread:

the logarithmic equation is, as Slide_Rule correctly points out, defined for all real 'a' {positive or negative} EXCEPT for a=0.

the value 0^0 is, like 0/0, indeterminate in mathematics and may carry different values depending on its use.
HOWEVER, there is very strong (but unofficial) argument in the mathematics community at large currently in support of the proposition 0^0 = 1 ; since this is the most 'useful' value that can be assigned to 0^0 via the stipulations of Calculus atm.


No, not really. It is a violation of axioms to divide by zero. The same is not true of 0^0. 0/0 may be indeterminate, but it is not usefully so, as 0^0 is when it takes the value of 1. 0/0 is NOT 1. The log law does hold even for a=0. That does not mean that 0^0=1.

It is not true that division by zero is impossible, however since integral domains do not have zero divisors, to perform division by zero over a field is to contradict an axiom, which places you in all sorts of trouble and is nonsensical.

It is by no means 'unofficial' to note that 0^0 is 1. It is of course wrong to note that 0^0 is 1 in all circumstances.

Quote:
Originally Posted by Dreamerish*~
Except a = 1.


What? Absolutely not! log_a(1)=0, as 1=1^0.

Quote:
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


No. The fact that the LIMIT of sin(x)/x=1 as x->0 is due to epsilon-delta definitions of limits (the thing acmilan posted above). There is no such definition that 0/0=1. Reminds me of Maths in Focus once when it said 1/x=0. Bah humbug.

And people, when you say something doesn't exist or is an invalid operation, please state which set you are working with.

i am tired of typing the same thing up again and explaining myself. so i'm not going to respond in entirety to this. all i can say is that we disagree on whether the limit exists or not - which is entirely OK since there is no definite answer.
neither of us may be correct.

however, i want to remind you not to toss aside the mitigating circumstances, under which i said the limit is 1, and simply criticise the bits you see unfit.
for the last time, i said that BOTH 0/0 and 0^0 are INDETERMINATE by definition, i did NOT say that either 0/0 or 0^0 is DEFINITELY =1!
plz read what i type, not skim.

i continue to retain the statement, however, that the majority of the mathematical community agrees to take BOTH 0/0 and 0^0 as 1... and if you disagree with me on this fact, then you can go out and research and take surveys yourself if you wish.

 

[acmilan]

The best way i can think of to explain the formal epsilon-delta definition of a limit is graphically. Take a simple function like f(x) = x². The limit as x approaches 2 is 4. In this case, a is 2 and L is 4 in the definition i posted before.

Observe the graph below:

The definition of a limit is that given and epsilon > 0, there is a delta > 0 such that whenever the x-value is in that interval on the x-axis about 2, the function is guaranteed to be between the interval about 4 shown on the y-axis. This is the formal definition of a limit represented graphically

 

[who_loves_maths]

to save my energy and time, and for the convenience of both Slide_Rule and acmilan:

On 0^0:

0 to 0 power

0 to 0 power - arguments for = 1


On 0/0:

0 divided by 0


plz check them all out

 

[acmilan]

to save my energy and time, and for the convenience of both Slide_Rule and Acmilan:


0 to 0 power


check it out plz.

Im am not arguing that 0⁰ should not be taken as 1 (although i dont like saying it that way) what I am merely trying to state, and in a sense try to work out also, is if the limit of x^x as x->0 can be taken. I think if you could somehow get a graph which merged f(x) = x^x and f(x) = (-x)^(-x), or define one to be so then the limit as x->0 from both sides may exist and hence the actual limit as x->0 exists.

 

[who_loves_maths]

Okay, this is going to be my final attempt at trying to convince you acmilan & co. that maths > english approach to showing that lim 0^0 = 1 is legitimate:

plz take a CAREFUL read of the two following technical sites:

1) Continuous Function and the Delta-Epsilon definition

and,

2) the Limit


now, website 1) should immediately tell you that the delta-epsilon definition of the limit is ONLY applicable to CONTINUOUS functions, which y= x^x is NOT. so it is not satisfactory or even logical to apply the definition to y= x^x at around x=0 to test for a limit since x = 0 is a critical point. hence, stating that simply because the delta-epsilon definition is not compatible with x --> 0 for y= x^x is not a satisfactory argument for the non-existence of a limit since the graph is not a continuous one.

secondly, if, as you said in your last, we are too combine two functions together to form a continuous function such that the delta-epsilon definition can be applied, then i'm sure you'll agree that the outcome limit for 0^0 will be =1 ; in which case we both agree with maths > english, and so i don't know what we're arguing about.


now if this is not enough to convince you, then take a look at website 2). on it is clearly implied that IF the graph is discontinuous, or, exhibits points of indeterminate forms, then l'Hopitals Rule is to be applied to find the limit. and since y= x^x is not only discontinuous, but also has a critical point at x=0 where that point is indeterminate, then maths > english indeed correctly applied l'Hopitals Rule to evaluate the limit as = 1.
notice that l'Hopitals is applied in this situation ONLY because the delta-epsilon definition cannot be applied due to the discontinuity of the function - hence, as a substitute, it does not make sense for you (acmilan) or Xayma to require l'Hopitals Rule to take into account of the negative side of the graph, otherwise it'd be no different in trying to impose the delta-epsilon definition!
l'Hopitals Rule is only employed here because the approach from both sides cannot already be taken by the delta-epsilon definition - precluding the necessity of l'Hopitals Rule to take into account a twin-sided limiting approach.
therefore, reinforcing what i said before, it suffices that maths > english only took into account the positive side - since that's the only side from which any approach can be taken as the negative side does not even exist.


i hope i've at least made myself clear here.

 

[acmilan]

now, website 1) should immediately tell you that the delta-epsilon definition of the limit is ONLY applicable to CONTINUOUS functions, which y= x^x is NOT.

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.

Also I dont recall either myself or Xayma saying that you apply l'Hopital's rule on the left hand side of x = 0 for x^x?

I dont see any way to resolve this other than to agree to disagree for now and I must be off for tonight

 

[Xayma]

who_loves_maths: a) the last part was in response to the attitude present in some of his posts nothing more extreme then he has performed

edit: b) was wrong

 

[brett86]

0⁰ can be looked at as x^x as x → 0

x^x = e^xlog_ex

using l'hopital's rule, xlog_ex → 0 as x → 0

:. x^x as x → 0 goes to e⁰ which equals 1

:. 0⁰ can be interpreted as 1

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

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]

just for those interested, heres a proof of l'hopitals rule

the gradient of x^x is x^x(log_ex+1), so the graph of x^x is only defined for x > 0 when looking at real function values [gradient doesnt exist for x <= 0]

therefore the x^x justification for 0⁰ → 1 is only looked at from the rhs because an negative, infinitely small number to an equal negative, infinitely small number just makes no sense!

what i mean to say is the justification for 0⁰ → 1 is correct but the notation:

is wrong because the lhs limit does not exist, only the rhs one does

its a superficial error that is not relevant to the value of 0⁰

btw this is maths > english, i just got tired of my old username

 

[brett86]

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

thanks

also, i just want to take back my comment before when i said u werent an excellent mod. i read over more of ur posts and u were extremely helpful