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Kaylee
16th May 2007, 10:49 PM
This has bugged me for awhile.

2 ^0 = 1. (Any number raised to the zero power is equal to one, but I'll stick to 2s because the the only reason I know this is because the binary numeral system is used in programming.)

Intuitively this does not make sense to me. Exponents are just a shorthand way of saying how many times a number is multiplied by itself (http://www.math.com/school/subject1/lessons/S1U1L8GL.html).

So:
2^4 is the same as 2*2*2*2 = 16
2 ^3 is the same as 2*2*2 = 8
2 ^ 2 is the same as 2*2 = 4
2 ^ 1 is the same as 2

"Dr. Math" (http://mathforum.org/library/drmath/view/58233.html)(Yes, I know that web site is for kids, but I'm not proud. I'll take my information whereever I can get it. ) says another way of looking at this is that to get from 2^3 to 2^2 to 2^1 we can merely divide by 2 each step of the way. So since 8 divided by 2 = 4, 4 divided by 2 = 2, to get from 2^1 to 2^0 we would just continue to divide by 2 and thus get 1. So therefore 2 ^0 is 1.

But intuitively it seems to me that 2 raised to the zero power is the same thing as saying "we have no 2s", so therefore our answer should be zero.

So is the mathematical fact that 2^0 = 1 actually a necessary fiction to make the exponential system work? :confused:

17th May 2007, 12:10 AM
2^4 is the same as 2*2*2*2 = 16
2 ^3 is the same as 2*2*2 = 8
2 ^ 2 is the same as 2*2 = 4
2 ^ 1 is the same as 2

Try it this way:

2^4 is the same as 1*2*2*2*2 = 16
2 ^3 is the same as 1*2*2*2 = 8
2 ^ 2 is the same as 1*2*2 = 4
2 ^ 1 is the same as 1*2 = 2
2 ^ 0 is the same as 1

17th May 2007, 12:16 AM
Or look at it this way:

ab+c = abac

Hence ab = ab+0 = aba0

i.e. ab = aba0

So a0 = ab/ab = 1

---

Or this way:

ab-c = ab/ac

Hence

a0 = ab-b = ab/ab = 1

strathmeyer
17th May 2007, 12:20 AM
What happens when we graph f(x) = 2^x?

Ivor the Engineer
17th May 2007, 12:39 AM
Try it this way:

2^4 is the same as 1*2*2*2*2 = 16
2 ^3 is the same as 1*2*2*2 = 8
2 ^ 2 is the same as 1*2*2 = 4
2 ^ 1 is the same as 1*2 = 2
2 ^ 0 is the same as 1

A similar explanation can be used to show why 0! = 1.

mijopaalmc
17th May 2007, 12:44 AM
A similar explanation can be used to show why 0! = 1.

How so?

Ivor the Engineer
17th May 2007, 12:48 AM
How so?

2! = 1*1*2
1! = 1*1
0! = 1

Beausoleil
17th May 2007, 12:50 AM
But intuitively it seems to me that 2 raised to the zero power is the same thing as saying "we have no 2s", so therefore our answer should be zero.

So what do you make of 2^0.5

It seems to me your intuition would suggest you have half a two, or one.

(Posted to illustrate that your intuition breaks down in other places as well.)

mijopaalmc
17th May 2007, 12:51 AM
2! = 1*1*2
1! = 1*1
0! = 1

The other way I heard of how to interpret 0! is it is the number of ways to organize (or arrange) the empty set.

mijopaalmc
17th May 2007, 12:54 AM
So what do you make of 2^0.5

It seems to me your intuition would suggest you have half a two, or one.

(Posted to illustrate that your intuition breaks down in other places as well.)

It is the square root of 2. a^(1/b) is the bth root of a.

Cabbage
17th May 2007, 01:18 AM
Expanding on Dr Adequate's explanation somewhat:

What is the result if you add no numbers together?

I think most would agree the only reasonable answer would be 0. But why? I would say it is because 0 is the additive identity of the real numbers--adding 0 to any number does not change the number. Zero is the "nothing" of addition.

Similarly, what is the result if you multiply no numbers together?

Based on the previous idea, it seems to me that the result should be 1, because it is the multiplicative identity of the real numbers--multiplying any number by 1 does not change the number. One is the "nothing" of multiplication.

Additionally, if for any natural number n you define n! to be the product of all positive integers less than or equal to n, the above idea implies that 0!=1--Since we are taking a product over an empty set (no positive integers are less than or equal to 0), the result should be the "nothing" of multiplication, i.e., the multiplicative identity 1.

jsfisher
17th May 2007, 04:34 AM
But intuitively it seems to me that 2 raised to the zero power is the same thing as saying "we have no 2s", so therefore our answer should be zero.

Maybe we can come at this from the side:

There is a great tendency to equate "zero" and "no". "And at the half, there is no score." Not true; there is a score, and it is zero to zero.

What is 2 times nothing? It is not 0; that would be 2 times 0. Two times nothing is 2. (You had 2, and then you didn't multiply it by anything.)

For 20, we take the product over a set of zero 2's. We multiply nothing together. The result is the identity element for multiplication, namely 1.

ond_magiker
17th May 2007, 04:49 AM
"Exponents are just a shorthand way of saying how many times a number is multiplied by itself" isn't a formal mathematical definition. It's a way to make people understand exponents intuitively, and works well as long as you're not too curious. The moment you ask questions about k^0 or k^(1/2) and such, the definition is simply not good enough.

69dodge
17th May 2007, 05:18 AM
But intuitively it seems to me that 2 raised to the zero power is the same thing as saying "we have no 2s", so therefore our answer should be zero.

Indeed, we have no 2s. 1 isn't a 2.

Exponents are just a shorthand way of saying how many times a number is multiplied by itself (http://www.math.com/school/subject1/lessons/S1U1L8GL.html).

So:
2^4 is the same as 2*2*2*2 = 16
2 ^3 is the same as 2*2*2 = 8
2 ^ 2 is the same as 2*2 = 4
2 ^ 1 is the same as 2

Your definition talks about multiplication. But in the last line (2^1 = 2), there is no multiplication. Why doesn't this bother you too?

So is the mathematical fact that 2^0 = 1 actually a necessary fiction to make the exponential system work? :confused:

All mathematical definitions are fictions.

We are free to define the meaning of the symbol "^" however we want to.

It's more convenient to define it so as to satisfy simple and regular mathematical laws, such as "2^(x+y) = (2^x)*(2^y) for all x and y without exception", than to define it so as to admit of a simple formulation in English, such as "exponents are just a shorthand way of saying how many times a number is multiplied by itself", which is not entirely unambiguous anyway.

Fontwell
17th May 2007, 05:44 AM
Re OP: Yes

Soapy Sam
17th May 2007, 05:52 AM
All mathematical definitions are fictions.

I shall treasure this.:D

Beausoleil
17th May 2007, 06:12 AM
It is the square root of 2. a^(1/b) is the bth root of a.

No kidding. ;)

The point was to illustrate somewhere else where the intuition breaks down.

RenaissanceBiker
17th May 2007, 07:24 AM
Ah grasshopper, but what is 0^0? And what is the limit of 0^x as x approaches 0?

drkitten
17th May 2007, 07:32 AM
Ah grasshopper, but what is 0^0? And what is the limit of 0^x as x approaches 0?

"A suffusion of yellow."

RenaissanceBiker
17th May 2007, 07:42 AM
Whoa, dude. Yesterday I got a fortune cookie that said I should pay close attention to the color yellow. This little confirmation bias is totally freakin' me out, man.

JoeTheJuggler
17th May 2007, 09:48 AM
I learned addition and subtraction with the "ones column," the "tens column" etc. Another way of thinking about this is that each column is a power of ten. So 10^0 logically corresponds to one.

Or a number like 4,722 is the same as 4 * 10^3 + 7 * 10^2 + 2 * 10^1 + 2 * 10^0. This also works for numbers to the right of the decimal point and negative exponents.

Kaylee
17th May 2007, 12:32 PM
Thank you all! :) You're the best! This is extremely clear to me now, especially since I got the benefit of getting the answer from so many different points of view.

The last time I asked this question (from someone who claimed to have been a math major) I got the very unsatisfying and (now I know) incorrect answer that it was a kludge. I'm glad I asked again, here at this forum.

Kaylee
17th May 2007, 12:57 PM
"Exponents are just a shorthand way of saying how many times a number is multiplied by itself" isn't a formal mathematical definition. It's a way to make people understand exponents intuitively, and works well as long as you're not too curious. The moment you ask questions about k^0 or k^(1/2) and such, the definition is simply not good enough.

IMHO its worse than being not merely good enough, it actually prevents understanding what the exponential system is actually describing.

The intent behind this sort of "explanation" may be helpful, but the effect is truly disastrous. My sincere wish is that everyone could be spared this kind of "helpfulness."

Again, thank you all for helping me understand the mathematics behind a number being raised to the power of zero.

Indeed, we have no 2s. 1 isn't a 2.

Your definition talks about multiplication. But in the last line (2^1 = 2), there is no multiplication. Why doesn't this bother you too?

I wish it had. I would have been able to ask a better caliber of question, or perhaps not needed to ask the question at all.

All mathematical definitions are fictions.

"Do mathematical entities really exist?" (http://forums.randi.org/showthread.php?t=69660&highlight=math):)

We are free to define the meaning of the symbol "^" however we want to.

It's more convenient to define it so as to satisfy simple and regular mathematical laws, such as "2^(x+y) = (2^x)*(2^y) for all x and y without exception", than to define it so as to admit of a simple formulation in English, such as "exponents are just a shorthand way of saying how many times a number is multiplied by itself", which is not entirely unambiguous anyway.

It seems that some mathematical definitions are closer to reality than others. Definitions that work in all known cases vs. definitions that only work with a limited set of events or numbers shed more light as to what is actually occuring in reality -- as I think this thread just demonstrated in an extremely limited way. :blush:

Yllanes
17th May 2007, 01:30 PM
I wish it had. I would have been able to ask a better caliber of question, or perhaps not needed to ask the question at all.:)

It was a very good question. Many people never understand this.

I'll give you something to think about: It has been said that \footnotesize $x^{a/b} = \sqrt[b]{x^a}$, for a, b integers. Now what does something like \footnotesize $2^\pi$ mean?

mijopaalmc
17th May 2007, 01:32 PM
"Exponents are just a shorthand way of saying how many times a number is multiplied by itself" isn't a formal mathematical definition. It's a way to make people understand exponents intuitively, and works well as long as you're not too curious. The moment you ask questions about k^0 or k^(1/2) and such, the definition is simply not good enough.

What is the rigorous mathematical definition of an exponent then?

I am curious because I have only heard exponents described in the above way and have never hear even a hint that the definition was incomplete or incorrect.

JoeTheJuggler
17th May 2007, 02:13 PM
Kaylee, if you want to really have fun, try messing with exponents in other than the decimal system--like base four, or binary, or hexadecimal or whatever.

Again, each column in a number is a power of the base, and the first one to the left of the point is always the "ones" column (because that's the base to the 0th power).

T'ai Chi
17th May 2007, 02:47 PM
Just pick integer a and b such that a/b is very close to pi.

Thabiguy
17th May 2007, 02:50 PM
What is the rigorous mathematical definition of an exponent then?

In my math classes, we rigorously defined exponentiation (a^x) using the natural exponential function (exp(x)), and we defined that - well, differently in each school:
1. exp(x) = lim(n->inf) (1+x/n)^n
2. exp(x) = unique function that satisfies these two equations: a) f'(x) = x, b) f(0) = 1.

Curiously, we used definition 1 at the university and definition 2 in high school.

atese
17th May 2007, 03:00 PM
What is the rigorous mathematical definition of an exponent then?

I am curious because I have only heard exponents described in the above way and have never hear even a hint that the definition was incomplete or incorrect.

It's fairly simple.

an = a*a*a*...*a (n factors)
Also, a-n = (a-1)n So a-n is repeated multiplication of the multiplicative inverse of a.

You then begin a proof similar to Dr. Adequate's

1*ab = ab = a0+b = a0ab

so 1*ab = a0ab and a0 = 1 by cancellation

It is important to note here that a0 is giving the multiplicative identity because we defined the exponent to be repeated multiplication. Had we defined the exponent to be repeated addition(an= a+a+...+a), a0 would give the additive identity(in the case of real numbers, thats zero). By the same token, if the exponent is a repetition of permutation, a0 would give the identity permutation.

As a final note, this proof only holds because the real numbers are an integral domain. In some instances, like Z12 (Z mod 12), cancellation doesn't hold.

Kaylee
17th May 2007, 03:23 PM
You have all given me a lot to think about. Not having a strong math background and preferring things stated as simply as possible I was trying to think how would I explain this to a 5-year-old.

I think I would say something like exponents are an extremely fast way of counting. (To an older child I would add that its one of the ways we have of writing very large numbers in a manageable way.) I would say if you tell me how many marbles 3^4 marbles are -- I will give you that many marbles in exchange for one marble. An easy way to do this is to do this in pieces.

I will give you three marbles for each marble.
I'll keep doing that until we have done this 4 separate times.
(She would end up with 81 marbles.)

I would then explain that in math lingo 3 is the base and 4 is the power.

If we wanted to know what 5^3 is, we could play the same game in marbles by my giving her 5 marbles for each one, doing this 3 separate times.
(She would then end up with 125 marbles.)

The classic definition seems to leave out the concept that our starting point in exponents is always one -- we need to start with the multiplicative identity.

I believe 2 raised pi times is about 8.88. I would theoratically give the 5 year old 2 marbles pi amount of times. (Thanks for giving me something to think about Yllanes :wink:)

I have to go. I hope I got this right, and I'll read all your posts to learn how to phrase this in a more sophisticated way later.

Beleth
17th May 2007, 03:25 PM
But intuitively it seems to me that 2 raised to the zero power is the same thing as saying "we have no 2s", so therefore our answer should be zero.
Here's another way to look at it.

Remember that addition and multiplication have different identities -- numbers that answer the question "What number can I add/multiply any number by and get that same number back?"

For addition, the identity number is zero. X + 0 = X for any X.
But for multiplication, the identity number isn't zero; it's one. X * 1 = X for any X.

Exponents are tricksy. They look like they are all about addition (ax * ay = ax+y) but they are not. They are all about multiplication. They use addition in the calculation to achieve multiplication in the answer. So you can't use the addition identity (0) for the final answer. You have to use the multiplication identity, 1.

That's why X0 = 1 for all X. "I have no X's" in addition-speak means "I have 0", but in multiplication-speak, it means "I have 1."

JoeTheJuggler
17th May 2007, 04:45 PM
The classic definition seems to leave out the concept that our starting point in exponents is always one -- we need to start with the multiplicative identity.

Or that you're dealing with units until you reach the base.

Another way to think of it is the way an odometer works. Each column just has a wheel with the digits 0-9 on it. Their position gives them the value. When you've got a brand spanking new car, and you start rolling, you start at zero. The first number (to the left of the decimal point) is units. When a given wheel reaches its highest value (9), it goes back to zero and causes the next wheel to the left to increment.

To change to a different base, all you do is change the wheels to however many digits you're using. The system works the same, and the car still starts counting in units from zero.

For your marble thing, how about different colored marbles to represent the different powers of the base? A color for units, a color for x^1, a color for x^2, etc. In base 3 for example, a red might represent the base (x^1) and is worth 3 whites (units or x^0). Then a blue (x^2) would represent 3 reds or 9 whites.

If your 5 year old already understands that there are 10 pennies in a dime, and ten dimes in a dollar, you've really already got it.

69dodge
17th May 2007, 07:49 PM
I wish it had. I would have been able to ask a better caliber of question, or perhaps not needed to ask the question at all.

Oh. Sorry. I didn't mean to be so blunt. Nothing wrong with your question.

You have all given me a lot to think about. Not having a strong math background and preferring things stated as simply as possible I was trying to think how would I explain this to a 5-year-old.

I think I would say something like exponents are an extremely fast way of counting. (To an older child I would add that its one of the ways we have of writing very large numbers in a manageable way.) I would say if you tell me how many marbles 3^4 marbles are -- I will give you that many marbles in exchange for one marble. An easy way to do this is to do this in pieces.

I will give you three marbles for each marble.
I'll keep doing that until we have done this 4 separate times.
(She would end up with 81 marbles.)

I would then explain that in math lingo 3 is the base and 4 is the power.

If we wanted to know what 5^3 is, we could play the same game in marbles by my giving her 5 marbles for each one, doing this 3 separate times.
(She would then end up with 125 marbles.)

The classic definition seems to leave out the concept that our starting point in exponents is always one -- we need to start with the multiplicative identity.

That is very good!

Actually, my first reaction on reading it quickly was that it couldn't work, that some things are just too hard for 5-year-olds. But the more I think about it, the better I like it.

It makes the 1 nicely explicit: the kid has to start with one marble, in order to be able to exchange anything.

It even gives the "right" answer for 00, namely, 1: start as always with one marble, then exchange it zero times---i.e., don't exchange it---for zero marbles.

Why do I say that 00 rightfully equals 1?

Another definition of ez, besides the two that Thabiguy gave is:

\begin{align*}e^z& = \sum_{k=0}^\infty\frac{z^k}{k!}\\ & = \frac{z^0}{0!} + \frac{z^1}{1!} + \frac{z^2}{2!} + \frac{z^3}{3!} +\cdots\end{align*}

This formula works for complex z too, not just real z. It defines raising to any power in terms of raising to integral powers only. Raising to integral powers is easy; it's just repeated multiplication by the base.

We should have e0 = 1. If z = 0, the formula yields 00/0! + 01/1! + 02/2! + 03/3! + ... . All terms but the first certainly equal 0. So the first must equal 1.

The fact that 0! = 1 also sneaks itself in here, as you can see.

Everything fits together beautifully. Quite the opposite of a kludge.

I believe 2 raised pi times is about 8.88. I would theoratically give the 5 year old 2 marbles pi amount of times. (Thanks for giving me something to think about Yllanes :wink:)

8.88? I get (well, Windows Calculator gets) about 8.825. (Pi is 3.14..., not 3.15... .)

I have to go. I hope I got this right, and I'll read all your posts to learn how to phrase this in a more sophisticated way later.

thaiboxerken
17th May 2007, 07:51 PM
Any number raised to the power of 0 = 1. That sounds like inductive reasoning, thus a faith-based statement, according to Mijopaal.

Schneibster
17th May 2007, 08:00 PM
I actually thought of deriving the multiplicative identity from the Peano postulates, but I couldn't find a good derivation and I'm too lazy to do it. Also, once you show the multiplicative identity, at least two of the methods of deriving the theorem that x^0=1 given above are rigorous, so there was really no point (and I admit that I'm not entirely certain you can derive the multiplicative identity from the Peano postulates).

EHLO
17th May 2007, 11:02 PM
Ah, I'm reminded of the elegance and mental tourment of

$$e ^{ i\pi} = -1$$

(Actually I just wanted to try the laTeX features)

Kaylee
17th May 2007, 11:14 PM
Or that you're dealing with units until you reach the base.

That makes sense, thanks! :)

I like your odometer and currency examples, and using different colored marbles to show which exponential power we're at makes the marbles a much better teaching tool.

Oh. Sorry. I didn't mean to be so blunt. Nothing wrong with your question.
NP, I think I was just a little sensitive to not understanding this particular grade school concept.

That is very good!
Thanks! :)

It even gives the "right" answer for 00, namely, 1: start as always with one marble, then exchange it zero times---i.e., don't exchange it---for zero marbles.

Why do I say that 00 rightfully equals 1?

Hmm, I've been wrestling with this one on the net also. It seems to be controversial as to whether zero raised to the zero power is 1. Some web sites say yes (for consistency or because of set theory (?) ), others say no -- it's undefined.

Consistency for the sake of consistency isn't enough of a reason for me. Not every math problem is concerned with the number of possible permutations of an empty set. On the other hand I can follow the reasoning below as to why one would say zero raised to the power of zero is undefined:

Zero raised to the power of 1 or higher is always zero. Zero "raised" to a negative power is undefined because dividing by zero is undefined. (You can have four equal slices of pie, however you can't have zero equal slices of pie -- it's undefined.)

I don't see how the result of 1 would fit within within the other results of undefined or zero, so I agree with those that say that zero raised to the power of zero is undefined.
http://en.wikipedia.org/wiki/Wikipedia:Reference_desk_archive/Mathematics/2006_July_1 (http://en.wikipedia.org/wiki/Wikipedia:Reference_desk_archive/Mathematics/2006_July_1)

8.88? I get (well, Windows Calculator gets) about 8.825. (Pi is 3.14..., not 3.15... .)

Cuddles
18th May 2007, 06:34 AM
I was trying to think how would I explain this to a 5-year-old.

Wouldn't that count as cruel and unusual?

Mosquito
18th May 2007, 07:10 AM
What happens when we graph f(x) = 2^x?

Typically, the next day you'll be able to recognize all teachers at your school who understand some math on the way they point to you, laugh and make weird movements with their hands.

Mosquito - and you'd better believe it!
:D

mijopaalmc
18th May 2007, 11:58 AM
Any number raised to the power of 0 = 1. That sounds like inductive reasoning, thus a faith-based statement, according to Mijopaal.

Except there have been many deductive proofs posted here so that we needn't depend on the inductive one. I realize that mathematical induction is an important method of proof but it is different that simply saying "all the cases I have observed are true, therefore every case is true". You might want to check out computational approaches to the Golbach Conjecture or the Riemann Hypothesis and why neither is proven even though they have been tested for a huge number of values before you make asinine comments like the one above.

Ben Tilly
18th May 2007, 01:11 PM
I actually thought of deriving the multiplicative identity from the Peano postulates, but I couldn't find a good derivation and I'm too lazy to do it. Also, once you show the multiplicative identity, at least two of the methods of deriving the theorem that x^0=1 given above are rigorous, so there was really no point (and I admit that I'm not entirely certain you can derive the multiplicative identity from the Peano postulates).

It is more a question of defining than deriving.

The Peano axioms do not define multiplication. They do not even define addition. They just define the integers with enough detail that you can use proof by induction.

After that you define addition as the operation that satisfies the axioms:

1 + m = s(m)
s(n) + m = s(n + m)

(Where s(n) means successor of n, which means n + 1.) Then in order you prove the following:

1. + is well-defined.
2. + is commutative.
3. + is distributive.

After that you define multiplication as the operation that satisfies the axioms:

1 * m = m
s(n)*m = m + (n * m)

Then you are in a position to prove the rest of the common rules of algebra.

Cheers,
Ben

LostAngeles
18th May 2007, 01:56 PM
It was a very good question. Many people never understand this.

I'll give you something to think about: It has been said that \footnotesize $x^{a/b} = \sqrt[b]{x^a}$, for a, b integers. Now what does something like \footnotesize $2^\pi$ mean?

Ah, I'm reminded of the elegance and mental tourment of

$$e ^{ i\pi} = -1$$

(Actually I just wanted to try the laTeX features)

I was bothered by that and was screwing around with it one night, just because I don't like complex powers. I'm iffy on irrational powers, but I like irrationals better than complexes anyways.

Then I remembered $$e^{i\theta} = cos {\theta} + {i}sin {\theta}$$

For $${\pi}$$, it comes out to -1. I felt better, but it still hurts and I still hate Euler for it.

The Grave
12th June 2007, 04:04 PM
It was a very good question. Many people never understand this.

I'll give you something to think about: It has been said that http://www.randi.org/latexrender/latex.php?\footnotesize $x^{a/b} = \sqrt[b]{x^a}$, for a, b integers. Now what does something like http://www.randi.org/latexrender/latex.php?\footnotesize $2^\pi$ mean?

In comp. we used 22/7 for pi (to 2 dp). Our illustrious teacher gave us the task of finding a better approximation for pi... Here's some: 113/36~3.139. Or 135/43.

Or better still 157/50. Or 201/64. Or 223/71. Or 245/78.

As for 0^0...definitely =1. 1 times no zeros = 1. I think that's obvious.

Griff.

Yllanes
12th June 2007, 04:25 PM
In comp. we used 22/7 for pi (to 2 dp). Our illustrious teacher gave us the task of finding a better approximation for pi... Here's some: 113/36~3.139. Or 135/43.

Or better still 157/50. Or 201/64. Or 223/71. Or 245/78.

Valid to within one part in 10100:

\footnotesize
$\frac{39437283434272590306994370980763234507447310 2456264}{12553277201361201519554317372950508261618 6012726141}$

As for 0^0...definitely =1. 1 times no zeros = 1. I think that's obvious.

It doesn't work like that. Consider these functions

\footnotesize
\begin{align*}
f_1(x)&=\left(\mathrm{e}^{-1/x}\right)^x, &
f_2(x)&=x^{\exp(-1/x)}, &
f_3(x)&=0^x,&
f_4(x)&=x^0,&
f_5(x)&=\exp(-1/x^2)^{\log (x+1)}
\end{align*}

If 00 were a a well defined quantity, all of these functions would have the same limit as x->0. But actually

\footnotesize
\begin{align*}
f_1(x)&\xrightarrow{x\to0}\frac1{\mathrm{e}}, &
f_2(x)&\xrightarrow{x\to0}1, &
f_3(x)&\xrightarrow{x\to0} 0,&
f_4(x)&\xrightarrow{x\to0} 1, &
f_5 & \xrightarrow{x\to0} 0.
\end{align*}

Stir
13th June 2007, 07:53 AM
Ah grasshopper, but what is 0^0? And what is the limit of 0^x as x approaches 0?

Before you ask that question you must first confirm that such a limit exists. It is quite possible that more than one limit exists to a particular series or expression (perhaps depending on which direction you approach from) or that none at all exists [a simple example: limit as x->0 of 1/x sin(1/x) ]