Funky limit question

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I was hoping this thread would be about the limits of funk.

I am very disappointed.

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Funk has no limits...

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We all know

$\lim_{x\to 0}\; sin x \; = \; 0$

The real question is what is

$\lim_{x\to 0}\; x^{0} \;$ ?

So, what is $0^{ 0}$?

On one hand, 0 to any power equals 0...

On the other hand, any number to power of 0 equals 1...

If you can prove that its equal to 1, you can put it all together to get

$\lim_{x\to 0}\; x^{\sin x} \; = \; 1$

We all know it is equal to 1 in that little question, nobody knows why...

but the limit from each side is the same, there's no need for $0^{+}$ and $0^{-}$

Oh sorry, DrP has already explained pretty much all this in one line

Edited by coke
Consecutive posts merged.

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We all know

$\lim_{x\to 0}\; sin x \; = \; 0$

The real question is what is

$\lim_{x\to 0}\; x^{0} \;$ ?

So, what is $0^{ 0}$?

On one hand, 0 to any power equals 0...

On the other hand, any number to power of 0 equals 1...

If you can prove that its equal to 1, you can put it all together to get

$\lim_{x\to 0}\; x^{\sin x} \; = \; 1$

Your approach fails, you cannot break the exponential into separate parts and resolve them, a counter example of this approach would be:

$\lim_{x\to 0}\; x^{(1/x)} = 0$

$\lim_{x\to 0}\; x = 0$

$\lim_{x\to 0}\; [x^{(1/x)}]^x = 1$

This is incorrect, in this case it is actually 0.

We all know it is equal to 1 in that little question, nobody knows why...

but the limit from each side is the same, there's no need for $0^{+}$ and $0^{-}$

The limit does not exist for $0^{-}$. You cannot take a negative number to the an infitessimal power.

Edited by BigMoosie

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Your approach fails, you cannot break the exponential into separate parts and resolve them, a counter example of this approach would be:

$\lim_{x\to 0}\; x^{(1/x)} = 0$

$\lim_{x\to 0}\; x = 0$

$\lim_{x\to 0}\; [x^{(1/x)}]^x = 1$

This is incorrect, in this case it is actually 0.

The limit does not exist for $0^{-}$. You cannot take a negative number to the an infitessimal power.

yeah, didn't realize., you're right on that one

yes, my approach fails some questions, but not in others...

Same as

$\lim_{x\to 0}\; x+2 = 2$

because you can simply substitute

But the substituting approach fails in this one

$\lim_{x\to \infty}\; \frac{x+3}{x+2} = 1$

The limit does not exist for $0^{-}$. You cannot take a negative number to the an infitessimal power.
Edited by coke

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The real question is what is

$\lim_{x\to 0}\; x^{0} \;$ ?

So, what is $0^{ 0}$?

The answer is that $0^{ 0}$ is indeterminate.

In power series, mathematicians denote $x^{ 0}$ to be 1 for all x. This notation is not definitional. It is a very convenient abuse of notation. It is convenient because we can compactly write things like $f(x) = \sum_{n=0}^{\infty} a_n x^n$ and $f'(x) = \sum_{n=0}^{\infty} (n+1) a_{n+1} x^n$.

This notation is abuse of notation because defining $0^{ 0}$ to be 1 (or any other value) would lead to contradictions. Any mathematical system must, above all else, be consistent with itself. Lack of consistency (e.g. a contradiction) leads to all kinds of problems. For example, contradictory statements let one can prove any statement to be true or false. One single contradiction makes an entire mathematical system worthless.

Consider the expression $x^y$, and make x and y simultaneously approach zero. Given how you make x and y approach zero, you can make $x^y$ take on any value. For example, let $y=\ln(a)/\ln(x)$. Making x approach zero from above makes y approach zero, as $\lim_{x\to 0^+} y = 0$. With this definition of y, $x^y = a$ for all positive x, and thus $\lim_{x\to 0^+} x^y = a$. In short, you can make $0^0$ be any value whatsoever. To avoid problems, one must say that $0^0$ is indeterminate.

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yes, my approach fails some questions, but not in others...

Isn't that a bit like saying:

$\frac {65}{26} = \frac {5}{2}$

...because you can remove the 6 from the numerator and denominator? (rather than divide both by 13)

If you're going to use a technique that only works in some cases then you must justify why it works in this case.

Merged post follows:

Consecutive posts merged
In short, you can make $0^0$ be any value whatsoever.

It cannot be any value, only 1 or 0.

Try to find another value for a limit in the form of "0^0", it is not possible.

Edited by BigMoosie
Consecutive posts merged.

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If you're going to use a technique that only works in some cases then you must justify why it works in this case.

$\lim_{x\to 0}\; x^{\sin x} \; = \; 1$

it's obvious why it works in this case, because its a simple substitution (ok maybe I should have simply called it substitution from the beginning instead of making limits out of it)

The answer is that $0^{ 0}$ is indeterminate.

In power series, mathematicians denote $x^{ 0}$ to be 1 for all x.

Yeah but in many cases I think we can assume its 1. In school I put 0^0 in my calculator (nspire) it said its equal to 1. In my friend's calculator (ti-89) it said error...

Although, I can think of one problem where it's equal to 0:

$\lim_{x\to 0}\; 0^{x} \; = \; 0$

Because the 0 is fixed, whereas x is simply aprroaching 0, it has an advantage in the $0^0$ battle...

Graphing confirms this (even on the same calculator that said 0^0 = 1)

Of course, at this point this type of thinking is prob. above the calculator's programming

Edited by coke

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In school I put 0^0 in my calculator (nspire) it said its equal to 1. In my friend's calculator (ti-89) it said error...

Just because a calculator said one thing doesn't really mean anything. I can write you a program - a "calculator" - that adds two numbers together incorrectly. Does that mean that all of addition is wrong to? I mean, I have this calculator that says....

Just because the programmer who programmed that nspire calculator didn't handle $0^0$ correctly, doesn't mean that it is in fact equal to 1 or 0 or anything.

There is an attitude of implicit trust that a lot of people have about their calculators and computers that they really shouldn't have. The reason we still study math by hand is because calculators and computers can be programmed wrong and make mistakes. There is operator error a great deal of the time too. We do math problems by hand because when the answer gets spat out by the calculator or computer, we have to have some intuition that checks whether it was right or wrong. Otherwise, we are just going to be slaves to whatever the computer/calculator spits out.

The Simpsons had a great farcical examples of this. Mrs. Krabapple has her math text open and she asks the class "Whose calculator can tell me what 7 time 14 is?" and Milhouse is raising his hand high sayING "Oh! Oh! 'Low Battery!'". The joke is on people who just blindly accept what a calculator says.

I have a few more examples from when I was a TA for a fluid mechanics class that will always stick with me. In the first, I don't remember the specific problem, but at the end the students has to multiply a 2 digit number by a 3 digit number to get the final answer. When I was grading homeworks, the students were allowed and encouraged to work in groups, so I graded them together as a group (since all the homeworks from that group were the same anyway). Well, someone had mistyped what they put into their calculator, and got a 6 digit number, copied it onto their paper and circled it and thought they were done with that problem. And, then everyone else copied that same answer. Not a single one of the 10 of them thought "hey! a 2 digit number times a 3 digit number can at most be a 5 digit number... not 6. Someone probably entered the wrong numbers in the calculator." Not a single one! And it was really, really frustrating because these were juniors in an engineering program -- they should have known better. A lot better because they should have had done a lot of math by then.

The other was even more frustrating. They were supposed to calculate how much energy had to go into a pump in order to pump water from a reservoir up to the top of a water tower. And again, a group worked together, made a mistake or 3 along the way, punched some numbers into their calculator and slapped the answer on the paper. The problem? The amount of energy they put on their paper was negative! That's right, they were basically saying they would get energy out of pumping water up hill. When you think about it, everyone knows how ridiculous it was, but it is another example of implicitly trusting what the calculator spat out, and not thinking about the answer itself.

At the job I do now, while Finite Element Analysis and Computational Fluid Dynamics are invaluable tools for the modern design engineer, they are far, far, far from perfect. We very often brake things in stress tests in places that were not "hot" on the FEA analysis, and the CFD is often close to the real result, but rarely is it accurate enough to design control schemes and the like. Both should be used as "ballpark" tools, not tools that can resolve the details of most any real-world situation yet. Someday in the future, maybe, but not today. And yet, there are too many people that trust those simulations a great deal, despite our being burned by them time and time again.

Another example was that probe several years back that crashed into Mars because the subcontractor that built it used the English system of units instead of the metric that NASA was using. I'm sure the subcontractor's simulations seemed to be all in order, and no one said "hey! that just doesn't seem right to me... let's check that closer".

Long story short here, never ever blindly accept what your computer or calculator says. Mistakes go into programming those devices, so wrong answers will be generated once in a while. It can even be as simple as multiplication -- Excel 2007 had a pretty serious bug in it for a time: http://gcn.com/articles/2007/09/25/excel-multiplication-bug-unearthed.aspx

I hope I didn't come off too strong, but I really, really want to destroy that attitude of "my calculator said...", because it is a trust that is grossly misplaced. At least at this point in time. There will probably be a time when we can implicitly trust a computer's calculations, but it isn't today.

------------

Like D H wrote, depending on the way you approach the two zeros, you can make $0^0$ seem to approach any value you want. He gave a good example by having the exponent approach 0 along a logarithmic path, and getting the result of it being the constant a. There are other ways, too (an infinite number, of course).

And, like D H wrote, the only "right" answer is to leave $0^0$ indeterminate. Anything else and you introduce far, far more problems than you fix. There are several other indeterminate forms, too. Math doesn't advertise itself as being able to solve everything, so there are some things it just won't have answers for, like $0^0$.

Edited by Bignose

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It cannot be any value, only 1 or 0.

Try to find another value for a limit in the form of "0^0", it is not possible.

Let x and y be two complex numbers. One could define $0^0 = \lim_{x,y\to 0}x^y$ if the limit exists and is the same along all possible paths that can be taken by x and y as they approach zero. However, by choosing different paths one can come up with any complex value one desires (not just zero or one) as the value of the limit.

Attaining any positive real value

Let a be some positive real number. Define $y(x) = \frac{\ln a}{\ln x}$. Since $\lim_{x\to0^+} y(x) = 0$, this defines one family of paths for evaluating xy as x and y simultaneously approach zero. Note that for all positive x, $x^y=a$ with y defined as $y(x) = \frac{\ln a}{\ln x}$. This family of curves generates all positive reals as values for $\lim_{x,y\to 0}x^y$

Attaining any complex value

Use the same definition for y(x) as above, here using the principal value of the complex logarithm. Now one can chose a to be any complex value whatsoever, with $\lim_{x,y\to 0} x^y = a$.

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Just because a calculator said one thing doesn't really mean anything.....Just because the programmer who programmed that nspire calculator didn't handle $0^0$ correctly, doesn't mean that it is in fact equal to 1 or 0 or anything.

lol, yeah i realized that that's what I meant in the last line of the post:

Of course, at this point this type of thinking is prob. above the calculator's programming

ok, maybe i should have skipped out on the calculator reference altogether...

However even though $0^0$ is indeterminate, the limits to it don't have to be: namely

$\lim_{x\to 0}\; x^{0} \; = \; 1$

$\lim_{x\to 0}\; 0^{x} \; = \; 0$

And for some reason,

$\lim_{x\to 0}\; x^{x} \; = \; 1$

And in those cases I think it is better to put "0" or "1" rather than "indeterminate" or "no solution".

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And in those cases I think it is better to put "0" or "1" rather than "indeterminate" or "no solution".

Why do you keep skipping over D H's point that it is NOT just 0 or 1?!? It can be any value at all. Real or Complex. It could be 10.777. It could be $\frac{\pi}{142}$ it could be 7-6i. It could be 222 quadrillion. It can be anything.

THAT'S why it is called "indeterminate". Because ANY value is equally valid as any other value, and none of them are "right" or "wrong".

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name one example of a limit where its not 0 or 1? despite substitution granting $0^0$?

see, bigmoosie agrees...

It cannot be any value, only 1 or 0.

Try to find another value for a limit in the form of "0^0", it is not possible.

any counterexample?

(idk if DH already has one in there, in that case a simpler counterexample?)

Edited by coke

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coke, counterexample has already been given! Look at:

Attaining any positive real value

Let a be some positive real number. Define $y(x) = \frac{\ln a}{\ln x}$. Since $\lim_{x\to0^+} y(x) = 0$, this defines one family of paths for evaluating xy as x and y simultaneously approach zero. Note that for all positive x, $x^y=a$ with y defined as $y(x) = \frac{\ln a}{\ln x}$.

This family of curves generates all positive reals as values for $\lim_{x,y\to 0}x^y$

(I added the emphasis to the quote)

Do I have to put a number in it? Ok, I will. Do the limit of $x^\frac{\ln 5}{\ln x}$. This will go to 5. There. Counterexample.

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ah, i knew those natural logs would get me

alright point taken.... or is it?

in case anyone still feels like arguing on my behalf,

$\frac{\ln 5}{\ln x}$

is undefined at x = 0, right? since $\ln 0$ is undefined...

which is where the indeterminate comes from, not from the $0^0$... right?

it's like saying

$\lim_{x\to 0}\; x^{0/x} \; = \; a$

where a can be multiple constants apparently

because 0/0 can be any constant...its indeterminate (any number times 0 is 0)

but than you're only taking the case where 0/0 is 0 since you want it to be $0^0$

so you're attemting to take the indeterminacy and paste it onto $0^0$, no?

in that case you might as well say

$\lim_{x\to 0}\; \frac{0x^{x}}{0} \; = \; a$

because similarly first you're assuming that 0/0 is equal to 1, so that it equates to $0^0$

and then pretending like you didn't assume that, so that instead of eqauating to just $0^0$, it equates to $a0^0$ which can be 1,2,3,4,89,359436,etc.

but if you use this method, then every limit is indeterminate!

Edited by coke

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Let x and y be two complex numbers. One could define $0^0 = \lim_{x,y\to 0}x^y$ if the limit exists and is the same along all possible paths that can be taken by x and y as they approach zero. However, by choosing different paths one can come up with any complex value one desires (not just zero or one) as the value of the limit.

...

Ah right, I was so sure for some reason to the contrary.

Edited by BigMoosie

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Sounds a little like quantum foam, not that I know much of theory.

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Today's calculus warmup question has stumped everyone in the class, including our teacher. Here's the question:

$\lim_{x\to 0} x^{\sin x}$

we just need this little fact:

$\underset{x\to 0}{\mathop{\lim }}\,x\ln x=0,$ thus

$x^{\sin x}=\exp (\sin (x)\ln x)=\exp \left( \frac{\sin x}{x}\cdot x\ln x \right)\to 1$ as $x\to0.$

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Today's calculus warmup question has stumped everyone in the class, including our teacher. Here's the question:

$\lim_{x\to 0} x^{\sin x}$

We can't think of any safe way to evaluate the limit. We tried rewriting the problem to this:

$y = \lim_{x\to 0} x^{\sin x}$

and then changing to this:

$\ln y = \lim_{x\to 0}\sin x \ln x$

but in the end it didn't help.

Any ideas?

[hide]The answer turns out to be 1, but we have no idea to get it.[/hide]

Since the limit of $\frac{lnx}{\frac{1}{sinx}}$ is indeterminate as x goes to zero ,we use L'hospital"s rule and:

$\ lim_{x\to 0}\sin x \ln x$ = $\lim_{x\to 0}-\frac{\frac{1}{x}}{\frac{cosx}{sinx^2}}$ = $\lim_{x\to 0}-\frac{sinx^2}{xcosx}$ = $\lim_{x\to 0}-\frac{sinx}{x}\frac{sinx}{cosx}$ = -(1).$\frac{0}{1}$ = 0

Now since $lnx^{sinx}$ goes to 0 , $x^{sinx}$ goes to 1 as x goes to zero

Note that $\lim_{x\to 0} {xlnx}\neq 0$ ,but 0.$-\infty$ which is indeterminate

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