Homework Help: Simple limits question.

1. Oct 20, 2007

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 $$\lim_{x \to \infty}x^0 = 1$$?

2. Oct 20, 2007

arildno

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

Last edited: Oct 20, 2007
3. Oct 20, 2007

ace123

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

Edit: nvm just ignore me..

Last edited: Oct 20, 2007
4. Oct 20, 2007

atqamar

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 ;)

5. Oct 20, 2007

arildno

6. Oct 20, 2007

atqamar

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,
$$\lim_{x\to\infty}(2^{x})^{\frac{1}{x}} = \lim_{x\to\infty}2^{x^0} = 2^1 = 2$$.

7. Oct 20, 2007

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 $$\infty^{0}$$?

8. Oct 20, 2007

atqamar

9. Oct 20, 2007

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".

10. Oct 20, 2007

ZioX

Try this one!

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

And for the daring:

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

Last edited: Oct 20, 2007
11. Oct 20, 2007

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.

12. Oct 20, 2007

arildno

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

Not really.
The same procedure..

13. Oct 20, 2007

ZioX

You're assuming that constant functions are continuous.

14. Oct 20, 2007

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?

15. Oct 20, 2007

arildno

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

16. Oct 20, 2007

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.

Last edited: Oct 20, 2007
17. Oct 20, 2007

arildno

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.

Last edited: Oct 20, 2007
18. Oct 23, 2007

atqamar

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