Aren't indeterminant forms misleading?

[tahayassen]

So my calculus professor last semester said that 1^∞ is just 1 if 1 is exactly 1. He said that 1^∞ is an indeterminant form because the rate of change of x as x approaches 1 competes with the rate of change of ∞ as it gets larger in x^∞. He also said that 0/0 is an indeterminant form because the rate of change of x as x approaches 0 competes with the rate of change of y as y approaches 0 in x/y.

I'm confused now. So does that mean 1^∞ is exactly 1 if we mean 1 is just 1?

 

[HallsofIvy]

So my calculus professor last semester said that 1^∞ is just 1 if 1 is exactly 1.

I don't believe for a moment that your professor said a silly thing like that! What in the world could "if 1 is exactly 1" possibly mean? You must have misunderstood.

He said that 1^∞ is an indeterminant form because the rate of change of x as x approaches 1 competes with the rate of change of ∞ as it gets larger in x^∞. He also said that 0/0 is an indeterminant form because the rate of change of x as x approaches 0 competes with the rate of change of y as y approaches 0 in x/y.

I'm confused now. So does that mean 1^∞ is exactly 1 if we mean 1 is just 1?

"1^∞" does NOT have any meaning! It is just a shorthand way of writing $\lim_{x\to a}f(x)^{g(x)}$ in the special case where f(x) goes to 1 and g(x) goes to infinity. One can show that there are cases, where f(x) and g(x) are specific functions, that will have different limits. For example, if f(x)= 1 for all x (that might be what was meant by that strange phrase "if 1 is exactly 1"!) and g(x)= x, then we can see that, taking x just along the integers, 1¹= 1, 1²= 1, 1[/sup]3[/sup]= 1, etc. so that the limit is 1. But if we have f(x)= a^1/x for a any positive number, so that $\lim_{x\to \infty} a^{1/x}= a^{0}= 1$, and g(x)= x again, we have $f(x)^{g(x)}= (a^{1/x})^x= a^((1/x)x)= a$ and the limit is a.

Similarly, "0/0" is just a shorthand way of writing $\lim_{x\to a} f(x)/g(x)$ in the special case where f(x) and g(x) both go to 0. Again one can show that there are cases, where f(x) and g(x) are specific functions, that will have different limits. If, for example, f(x)= g(x)= x, then both f(x) and g(x) go to 0 as x goes to 0 but f(x)/g(x)= x/x= 1 for all positive x and so the limit is 1. Or, if we take f(x)= ax and g(x)= x, we still have f(x) and g(x) both going to 0 as x goes to 0 but now f(x)/g(x)= ax/x= a for all positive x and so the limit is a.

 

[Tac-Tics]

I'm confused now. So does that mean 1^∞ is exactly 1 if we mean 1 is just 1?

Like Halls said above, this is nonsense.

Then again, the value 1^∞ is too. It's like asking "what is 2 times a walrus". A walrus isn't a real number, and neither is ∞.

When people SAY 1^∞, they MEAN some kind of limit towards infinity.

 

[lurflurf]

Indeterminate forms are not misleading. Also who cares about real numbers, there are other numbers. There are limits that can be determined by a simple analysis, extended real numbers can be used to formalize this. Indeterminate forms are simply cases that cannot be so handled. As for "1^∞ is just 1 if 1 is exactly 1" that is not an ideal explanation.
Let [*] be a limit
consider
[x^y]
under favorable conditions
[x^y]=[x]^[y]

suppose [x]=1, [y]=∞
we cannot determine [x^y] from this information
if in addition x=1 we can conclude [x^y]=[x]^[y]=x^[y]=1^∞=1

It is important to distinguish between functions that are equal and functions that merely have the same limit.

consider some examples

$$\lim_{x \rightarrow \infty} \left(1+\frac{1}{x}\right)^x=e$$

$$\lim_{x \rightarrow \infty} 1^x=1$$

$$\lim_{x \rightarrow \infty} \left(1-\frac{2}{x}\right)^x=\frac{1}{e^2}$$

 

[lurflurf]

Saying "1^∞" does NOT have any meaning! It is just a shorthand way of writing.." is like saying "15 does NOT have any meaning! it is just a shorthand way of writing fifteen." If something is a shorthand way of writing something it has meaning, the trouble is shorter ways of writing things tend to be less explicit and at times can lead to confusion or mistake. If we say a limit is of indeterminate form 1^∞ we cannot determine the limit without more information. If we say a limit is of determinate form 0^+∞ we can conclude the limit is 0.

 

[That Neuron]

Perhaps I am over simplifying this a bit too much, but I believe that he was simply saying lim x $\rightarrow$ ∞ 1^x = 1.

Which is simply an intuitive proof, 1 $\bullet$ 1 $\bullet$ 1 $\bullet$ 1 = 1.

If your professor was speaking with experimentally determined constants taken to an exponent with an attached limit, if we have an approximated 1, like 1.0000001 = k, then lim x $\rightarrow$ ∞ k^x = ∞. or 0.9999999 = k then lim x $\rightarrow$ ∞ k^x = 0.

That's what I got out of it anyway.

 

[tahayassen]

So if I understood correctly...

$$\frac { 0 }{ 0 } = \lim_{x \to a}\frac { f(x) }{ g(x) }$$ where f(a) and g(a) approach 0

I understand that 1^∞ doesn't mean anything, but why don't we define it as:

$$\\ \\ { 1 }^{ \infty }=\lim_{x \to \infty} { 1 }^{ \infty }=1*1*1...$$

The problem I have with indeterminant forms is that they don't actually represent what you would intuitively think they would represent. 0/0 represents two functions that approach 0 but are not equal to 0. 1^∞ seems to represent two functions where one function approaches 1 and the other function approaches ∞.

 

[jgens]

So if I understood correctly...
$$\frac{0}{0} = \lim_{x \to a} \frac{f(x)}{g(x)}$$
where f(a) and g(a) approach 0.

You really should not write "=" in this case, since the expression 0/0 is meaningless with 0 and / interpreted in terms of real numbers, while the limit on the right-hand side may be well-defined. But it is true that the limit above is an example of an indeterminate form of the type 0/0.

I understand that 1^∞ doesn't mean anything, but why don't we define it as:
$$1^{\infty} = \lim_{x \to \infty} 1^x$$

In the context of the real numbers, the symbol 1^∞ can at best be interpreted as shorthand for something like lim_n→∞1ⁿ since ∞ is not actually part of the number system. In the extended and projective real number systems we certainly could extend exponentiation to the infinite case by defining 1^∞ = 1, but this turns out not to be very useful. One of the reasons is that we have limits like the following:
$$\lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n = e$$
In fact, for each positive real number, we can actually find an indeterminate form of the type 1^∞ which converges to that number. All you have to do is make a slight modification of the indeterminate form I gave above.

The problem I have with indeterminant forms is that they don't actually represent what you would intuitively think they would represent. 0/0 represents two functions that approach 0 but are not equal to 0. 1^∞ seems to represent two functions where one function approaches 1 and the other function approaches ∞.

This is not quite right. The expression 0/0 does not represent lim_x→0f(x)/g(x) where lim_x→0f(x) = lim_x→0g(x) = 0 and g(x) ≠ 0 in some nbhd of 0. But that limit is an indeterminate form of the type 0/0. Same thing goes for indeterminate forms of the type 1^∞.

P.S. I tried to fix your tex and this required a little guessing about what you meant at times. I am sorry if I misunderstood something and fixed it incorrectly.

 

[tahayassen]

Ah, thanks for answering. I wonder why my tex broke. This new update for daum equation editor made it buggy. :(

 

[micromass]

Ah, thanks for answering. I wonder why my tex broke. This new update for daum equation editor made it buggy. :(

If you want to write limits, then use this code

\lim_{x\rightarrow +\infty} f(x)

There is no need to use matrices.

 

[Mark44]

If you want to write limits, then use this code

\lim_{x\rightarrow +\infty} f(x)

There is no need to use matrices.

Or more simply:

\lim_{x \to +\infty} f(x)

 

[That Neuron]

lim x->0 x/x, -> 0/0 x'/x' = 1/1 =1

 

[Mark44]

lim x->0 x/x, -> 0/0

No, we don't say that x/x "approaches" 0/0 or that the limit is 0/0.

$$ \lim_{x \to 0} \frac{x}{x} = 1$$
The reason is that for any nonzero value of x, x/x = 1, and this is true as we pass to the limit. You can get this using L'Hopital's Rule, but it's not needed.

x'/x' = 1/1 =1

 

[HallsofIvy]

The reason is that for any nonzero value of x, x/x = 1, and this is true as we pass to the limit. You can get this using L'Hopital's Rule, but it's not needed.

What we do need is the theorem "If f(x)= g(x) for all x except a, then $\lim_{x\to a} f(x)= \lim_{x\to a} g(x)$.