# Limit as x Approaches Infinity

1. Sep 21, 2010

### MysticDude

I am starting this thread because when I was doing a homework problem I got interested in something:

$$\lim_{x \to +\infty} x^\frac{1}{x}$$

I already know that the answer is one because $$\frac{1}{\infty}$$ is the same as 0 and anything to the zeroth power is 1. The thing that got me interested is about the fact that I practically substituted infinity into the problem to get the answer (sorry if I confuse you here). So the limit is practically $$\infty^\frac{1}{\infty}$$. But I thought that that was indeterminate and what not. I also asked my very intelligent teacher about this and he said "Well, we are approaching infinity, not actually reaching it." But then again, infinity is a concept and not a number.

I want to know what you think about this, and help me clear this up. Does this show that $$\infty^\frac{1}{\infty}$$ is not indeterminate or is it because of the limit that it is equal to 1. The infinity is the part that got me interested.

Well thanks for your any input that you may have. Just don't call me an idiot :tongue:

Sorry If This Is Confusing X(

2. Sep 21, 2010

### Hurkyl

Staff Emeritus
The form $(+\infty)^0$ is indeterminate; for any positive y, $(+\infty)^y = +\infty$, however, for any positive finite x, $x^0 = 1$ -- therefore, exponentiation is not continuous at $(+\infty)^0$, and so there is not a limit law of this form.

(Note that the above can be interpreted either as a purely formal statement about kinds of limits, or it could be interpreted as an expression involving the arithmetic of extended real numbers. If you don't know what the latter means, then you have probably only been taught about the former)

3. Sep 21, 2010

### MysticDude

Okay so I understand that $$(\infty)^0$$ is indeterminate.
Then the only reason that $$\lim_{x \to +\infty} x^\frac{1}{x}$$ = 1 is because of the limit correct?

Oh and just so you know, I wasn't taught calculus by a teacher, I printed online worksheets and did everything on my own. The teachers only helped me when I really needed help.

4. Sep 22, 2010

### HallsofIvy

Typically, the simplest thing to do with exponentials is to take the logarithm.

If $y= x^{1/x}$, then $ln(y)= ln(x)/x$ and $x^{1/x}= e^y$. Taking the limit of $ln(x)/x$ as x goes to infinity give a form of "infinity over infinity" so we can use L'Hopital's rule.

The derivative of ln(x) is 1/x and the derivative of x is 1 so now we are looking at the limit, as x goes to infinity of 1/x which is 0. That is, $\lim_{x\to\infty} (1/x)ln(x)= 0$ so that $\lim_{x\to\infty} y= 0$ and then, since $e^x$ is continuous for all x, $\lim_{x\to\infy} x^{1/x}= e^0= 1$.

But certainly you could say, of any limit, "the only reason lim f(x)= L is because the limit is correct"! (Assuming, of course, that it is correct. Knowing that is the hard part!)

5. Sep 22, 2010

### Tac-Tics

In my opinion, "indeterminate form" is a vulgar term. It crops up out of nowhere so that calculus teachers have some justification about certain limits without having to break out the actual definitions of a limit.

That said, the definitions of limits are hard to for the uninitiated. Many calculus students never actually grasp them. (Many are not even taught them). On top of that, there are over a dozen kinds of limits (limits of a sequence, limits of a function at a point, limits towards infinity, limits from the right, etc).

When we talk about limits to infinity, the "infinity" is actually absorbed into the definition:

Contrast with this.

Notice how there is no infinity in the second definition other than the name. (We could have easily have called it a "left limit" or given it some other name, completely avoiding talk of infinity).

There are subtleties that are used without thought in practice. Both definitions identify what it means to be a limit -- but neither says a limit must exist. In fact, in either definition, a limit might fail to exist.

Often, we make a distinction about WHY it fails to exist.

For a function like f(x)=x^2, we say the limit is ＋∞. The limit doesn't exist, according to the above definition, but it's often handy to characterize functions.

If we wanted to include these "infinite" limits, we might extend the definition like this:

We call the infinite limits ＋∞ and －∞, but we could easily have called them "top" and "bottom". They never enter our arithmetic at all. You never divide by ∞, you never add it or exponentiate by it. Everything "inside" the $$\lim{x \to \infty}$$ is pure algebra. The outside is a single real number -OR- the symbols +∞ or －∞.

Hope that helps.

6. Sep 22, 2010

### planck42

$${\infty}^{0}$$ is certainly an indeterminate form. For those improper limits, I find it useful think about what would happen if I substituted in the limit in question plus an infinitesimal(large positive/negative finite numbers for infinite limits) so that it behaves the same as the original limit, but has a different value.

7. Sep 22, 2010

### MysticDude

Thanks everyone for your inputs on my question. Oh and Tac-Tics, I'm barely grasping epsilon-delta limits so your post was hard to understand, but I got through it. And planck42, I would also do that for some of my problems :D. Thanks HallsofIvy for your little proof! I also appreciate you guys for taking your time to post your thoughts and facts. Thanks again :D!

8. Sep 23, 2010

### Tac-Tics

Epsilon-delta definitions are really tough. Even after a few years of working with them, I still have to think hard to make sure I got it right.

It's basically an encoding of "if you want to approximate output of a function very well (with accuracy of ε), you have to be accurate enough with the input (with accuracy of δ)."

But my point doesn't really require you to understand the definitions :) It's enough to see that "infinity" can be kept away from the algebra. You don't ever divide by zero or infinity or any of that nonsense. When you see indeterminate forms, they are just shorthands that give the right answer very quickly most of the time (but leaves you unsure why).

9. Sep 23, 2010

### Hurkyl

Staff Emeritus
Indeterminate forms are the ones that don't give you the answer very quickly.

$5+7$ is a (determinate) form, equal to 12, and there is the corresponding limit law:
If $\lim_{x \rightarrow a} f(x) = 5$ and $\lim_{x \rightarrow a} g(x) = 7$, then $\lim_{x \rightarrow a} f(x) + g(x) = 12$​

$(+\infty) \cdot 5$ is a (determinate) form, equal to $+\infty$, and there is the corresponding limit law:
If $\lim_{x \rightarrow a} f(x) = +\infty$ and $\lim_{x \rightarrow a} g(x) = 5$, then $\lim_{x \rightarrow a} f(x) g(x) = +\infty$​

$(+\infty) \cdot 0$ is indeterminate form, because there isn't a corresponding limit law. Amongst limits of that form -- that is, limits that are $\lim_{x \rightarrow a} f(x) g(x)$ where $\lim_{x \rightarrow a} f(x) = +\infty$ and $\lim_{x \rightarrow a} g(x) = 0$ -- you can find examples where the limit can be any real number, or even $\pm \infty$, as well as limits that don't converge to anything at all.

We do arithmetic with forms because it is a quick and rigorous way to compute limits, if we can put the limit into a determinate form, in those cases where we cannot invoke continuity. (Of course, once we learn a little more, we use the extended real numbers and can invoke continuity in all of the cases we would otherwise use forms for. I really don't understand why those aren't taught in into to calculus classes, or even before)

Last edited: Sep 23, 2010
10. Sep 23, 2010

### Tac-Tics

Technicalities.

They are shorthands, and I have a distaste for the term :)