What Are the Limits of 0^0 and ∞^0?

[atqamar]

I just have a question of principles.
I know 0^0 and \infty^0 are both undefined.
But isn't this true:
\lim_{x \to 0}x^0 = 1 and \lim_{x \to \infty}x^0 = 1?

 

[arildno]

Certainly.
Now, compute:
\lim_{x\to{0^{+}}}0^{x} and \lim_{x\to\infty}(2^{x})^{\frac{1}{x}}

 

[ace123]

I just argued the same thing in of your posts. I thought it should be. But i guess that's only if we take the x to be neglegible..

Edit: nvm just ignore me..

 

[atqamar]

Thanks, this question did arise from this thread: https://www.physicsforums.com/showthread.php?t=192598

Since x^0 = \frac{x}{x}, I would go as far as saying \infty^0=\frac{\infty}{\infty}=1 without the limits... as long as the infinities are the same, and they should be since that is what the equation requires. But again, messing with infinities is not my thing ;)

 

[arildno]

Read again my comment:

Thanks, this question did arise from this thread: https://www.physicsforums.com/showthread.php?t=192598

Since x^0 = \frac{x}{x}, I would go as far as saying \infty^0=\frac{\infty}{\infty}=1 without the limits... as long as the infinities are the same, and they should be since that is what the equation requires. But again, messing with infinities is not my thing ;)

Certainly.
Now, compute:
\lim_{x\to{0^{+}}}0^{x} and \lim_{x\to\infty}(2^{x})^{\frac{1}{x}}

 

[atqamar]

Certainly.
Now, compute:
\lim_{x\to{0^{+}}}0^{x} and \lim_{x\to\infty}(2^{x})^{\frac{1}{x}}

For the first one, \lim_{x\to{0^{+}}}0^{x} = 0. However, I can do this only intuitively, by mentally graphing the equation. Is there an algebraic way?

And for the second one,
\lim_{x\to\infty}(2^{x})^{\frac{1}{x}} = \lim_{x\to\infty}2^{x^0} = 2^1 = 2.

 

[arildno]

Indeed correct!

But, wouldn't you agree to that the first limit is of the shape 0^{0}, and the other of the shape \infty^{0}?

 

[atqamar]

Thank you very much arildno, your point is well-made!

 

[arildno]

So, as you can see, for symbol groups like 0^{0},\infty^{0}, and for that matter, 1^{\infty},\frac{\infty}{\infty},0*\infty,\infty-\infty,\frac{0}{0}, we cannot relate to such a group a unique number so that all limiting processes "tending" towards that symbol group will converge to that number.


That is why we prefer to say these expressions are "indeterminate", or "undefined".

 

[ZioX]

Try this one!

\lim_{t\to 0} (\cos (x))^{-\frac{1}{x^2}}

The answer may surprise you.

And for the daring:

\lim_{t\to 0} (\cos (x))^{-\frac{1}{x^n}}

 

[coomast]

Found this forum yesterday. Just registered and here's my first post.
The solution to the first one is $\sqrt{e}$. Write the function as an exponential
function and take the limit inside. Using 'l Hopital gives 1/2.

The second one uses the same principal, however the solution is depending on
the value of n. Assuming it can only take on integer numbers (pos and neg) we have
for the negative ones always 1. For n=0 and n=1 it is also 1. The equation with n=2
is allready given. For values of n higher than 2 the limit is infinity.

Hopefully the latex thing works.

 

[arildno]

\lim_{t\to 0} (\cos (x))^{-\frac{1}{x^2}}=(\cos (x))^{-\frac{1}{x^2}},x\neq{0}

The answer may surprise you.

Not really.

And for the daring:

\lim_{t\to 0} (\cos (x))^{-\frac{1}{x^n}}

The same procedure..

 

[ZioX]

You're assuming that constant functions are continuous.

 

[coomast]

It seems there is something not right.
*) The limit for t? It isn't in the function.
*) How do you display the formulas the latex way?

 

[arildno]

You're assuming that constant functions are continuous.

No, I'm assuming that t is a different and independent variable from the x variable.

 

[ZioX]

Independence implies constancy, as otherwise x would be a function of t. That is, f(x) is constant in t if x and t are independent.

But whatever. You know what I meant and my example was another illustration of your point.

 

[arildno]

Independence implies constancy, as otherwise x would be a function of t. That is, f(x) is constant in t if x and t are independent.

But whatever. You know what I meant and my example was another illustration of your point.

No, what you said was that I specifically assumed that constant functions are CONTINUOUS.

That is:

a) A trivial fact, and not an assumption.

b) Irrelevant to my post, or indeed, to your example.



As limit exercises taking x to zero, rather than t, your examples are nice, though.

 

[atqamar]

Try this one!

\lim_{t\to 0} (\cos (x))^{-\frac{1}{x^2}}

The answer may surprise you.

And for the daring:

\lim_{t\to 0} (\cos (x))^{-\frac{1}{x^n}}

Hmm... do you mean \lim_{x\to0}?