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This may be a dumb question. I understand why $\frac{x}{x}$ when $x=0$ is undefined. This can causes errors if an equation is divided by $x$ without restrictions.

$\frac{\infty}{\infty}$ is undefined. So when I use $\frac{x}{x}=1$ to simplify an equation, can it also lead to errors because $x$ can equal infinity? Or is $x=\infty$ meaningless?

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Well, x cannot equal to infinity, so it is better to speak in terms of a limit. As a limit the answer is certainly 1. Yes, infinity ÷ infinity is "indetermine", but then again, as a limit, when x approaches infinity, the limit x/x exists and is 1. It's not a dumb question, lots of entry calc students need some time to grasp the limit constant and need to understand that infinity is not a number, but more a concept. –  imranfat Apr 28 '13 at 1:21
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Infinity is not a number. –  Andres Caicedo Apr 28 '13 at 1:45
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To say in short, "infinity" is not a noun; rather an adjective. –  Metin Y. Apr 28 '13 at 1:45
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@MetinY.: Your assertion doesn't make sense. While it can be used in some specific cases as an adjective (such as "infinity norm"), it is not an adjective. It is not the role of the word in language that matters, but the role in mathematics. The difference lies in quantifiability. –  Glen O Apr 28 '13 at 2:00
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I do not understand why there are 4 votes to close this question without any comments or reasons given. –  mixedmath Apr 28 '13 at 2:57

8 Answers 8

$$x = \infty \;\text{ is meaningless in standard contexts}.$$

Yes, when we have a fraction of the form $\dfrac xx$, where $x \in \mathbb R\setminus \{0\}$, then $\dfrac xx = 1, \quad \dfrac {3x}{x} = 3, \text{ etc.}.$ But $\infty \notin \mathbb R$; it is not itself a real number.

$$x \to -\infty ... \;\text{ or } \;\; x \to +\infty$$

is not meaningless. It denotes "x gets arbitrarily (very, very) small," or "x gets arbitrarily (very very) large, respectively,

Now, what we do have, is that for limits, e.g. $$\lim_{x\to +\infty} f(x) \;\;\text{ or}\;\; \lim_{x \to a} f(x), \;\; a \in \mathbb R$$ when evaluating a given limit, if we obtain the indeterminate form $\infty/\infty$, we cannot yet say anything about the existence of a limit. We certainly cannot reduce it to $1$ by "canceling" $\infty$ in both the numerator and denominator because it does not represent a number. It signifies that both the numerator and denominator are growing without bound. That doesn't necessarily mean that the limit does not exist: Obtaining an indeterminate form such as $\infty/\infty$ or $0/0$ is simply a signal that "more work needs to be done" to evaluate the limit, or to determine whether or not a limit exists.

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thumbs up from me! +1 –  Amzoti Apr 28 '13 at 2:48
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I know you know that, but $x=\infty$ is meaningful in various cases. Like in the extended real numbers, the projective line, the one-point compactification. Of course, the OP is obviously not there yet. But if he/she plans to get there later, maybe it would be good for him/her to be prepared for $x=\infty$ to make sense. –  1015 Apr 28 '13 at 3:02
    
Yes, @julien I suspect you are correct. I qualified my blanket statement. Thanks for addressing it. –  amWhy Apr 28 '13 at 3:11
    
"$x \over x$, where $x \in \mathbb{R}$". You may want to specify $x \in \mathbb{R} \setminus \{0\}$ –  stefan Apr 28 '13 at 11:03
    
@stefan yes, thanks. –  amWhy Apr 28 '13 at 14:31

$x/x = 1$ for any nonzero real $x$. However, $\infty$ is not a real number. Now, if you evaluate a limit and find a result that gives $\infty/\infty$, we cannot determine what the limit should be without further work. After all, "plugging in" infinity to any of $$ \lim_{x\to\infty}\frac{x^2}{x},\qquad \lim_{x\to\infty}\frac{x}{x},\qquad \lim_{x\to\infty}\frac{\log x}{x} $$ will give $\infty/\infty$. However, these limits evaluate respectively to $\infty$, $1$, and $0$ (showing that if you get $\infty/\infty$ when naively evaluating the limit, it could be any real number, or even infinity!). But if you are taking a limit $$\lim_{t\to\infty}\frac{X(t)}{X(t)}$$ for some expression $X(t)$ (where $X(t)\to\infty$ as $t\to\infty$), you can say that the limit is $1$ as long as $X(t)$ is defined for sufficiently large $t$, because $X(t)/X(t) = 1$ for $X(t)$ any real number besides $0$. In the limit, we only consider arbitrarily large real numbers (i.e. not actually infinity), so in taking the limit, we can reduce, and we see $$ \lim_{t\to\infty}\frac{X(t)}{X(t)} = \lim_{t\to\infty} 1 = 1. $$

Now, in some branches of mathematics, we can make sense of certain expressions with infinity. For example, we can make $\Bbb{R}$ or $\Bbb{C}$ compact, an important notion in topology and analysis, by adding a "point at infinity": meaning we create a new system given by $\tilde{\Bbb R} = \Bbb{R}\cup\{\infty\}$ or $\tilde{\Bbb C} = \Bbb{C}\cup\{\infty\}$, where $\infty$ is a formal symbol designed to be suggestive (notation not standard). However, in doing so, we remove nice properties of $\Bbb{R}$ and $\Bbb{C}$: namely, $\tilde{\Bbb R}$ and $\tilde{\Bbb C}$ are not fields. In these new systems, we can define many expressions involving $\infty$, such as $$\infty + \infty = \infty, \quad r/\infty = 0, \quad\textrm{and} \quad c/0 = \infty$$ (provided $r\neq\infty$, $c\neq 0$). There are expressions that we can't define, though. These include $$\infty/\infty, \quad\infty - \infty,\quad 0\cdot\infty,\quad \textrm{and}\quad 0/0:$$ which you might recognize as indeterminate forms. The reason they are indeterminate in this extended case of $\Bbb{R}$ and $\Bbb{C}$ is similar to the reason they are indeterminate in the case of normal calculus: defining them in one way would lead to contradictions. Say we defined $\infty/\infty = 1$. Then $$1 = \infty/\infty = (c/0)/\infty = c/(0\cdot\infty).$$ We would have to say that $0\cdot\infty = c$ for every nonzero $c$ in our original field, which simply does not make sense.

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$$\large{\bf{+}}\frac{\large{\bf{+1}}}{\large{\bf{+1}}}$$ –  Dylan Yott Dec 9 '13 at 17:37

In calculus, it is convenient to use the extended real numbers: these are the real numbers along with two extra numbers we call $+\infty$ and $-\infty$.

We can define arithmetic with extended real numbers. While I could just say that we've defined the arithmetic so that $(+\infty) / (+\infty)$ is undefined (e.g. just as $0/0$ and $1/0$ are undefined), it may help if I point out the motivation behind the arithmetic of the extended real numbers.

The extended real numbers are introduced for the purposes of making calculus and analysis more convenient, and to do so it is important that arithmetic be continuous: While most arithmetic operations with $+\infty$ and $-\infty$ can be defined -- e.g. $(+\infty) + 7 = +\infty$ and $(-\infty) \cdot (-47) = +\infty$ -- the following limits show that $(+\infty)/(+\infty)$ can't be defined by continuous extension:

$$ \lim_{x \to +\infty} x = +\infty \qquad \qquad \lim_{x \to +\infty} x^2 = +\infty $$ $$\lim_{x \to +\infty} \frac{x}{x} = 1 \qquad \qquad \lim_{x \to +\infty} \frac{x}{x^2} = 0 $$

If we could extend division continuously, then both of the limits on the second line should be the value of $(+\infty)/(+\infty)$. But they are different numbers. So we can't do it! Thus we opt to leave $(+\infty)/(+\infty)$ undefined.


Arithmetic with other systems of numbers you may know (integers, rationals, reals, complexes) were constructed so that they would obey the "familiar" laws of arithmetic. You're probably used to automatically assuming old laws of arithmetic continue to hold when you learn about new sorts of numbers.

The extended real numbers, however, were constructed for a different reason as I described above, and because of this, many familiar laws of arithmetic have been allowed to fail, so you should be wary about doing arithmetic with them until you've really gotten used to how they work. (it is impossible to fully satisfy both desires at once)

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There are branches of mathematics where infinite can be considered a number. For instance, in set theory you have a name for the infinity of the set of natural numbers. It is called $\omega$ and is called an ordinal number, which is a generalization of the concept of number. For such infinite number, if used in the context of the surreal numbers, you can write $\frac{\omega}{\omega}=1$ without using limits. Just my two cents

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In the set of real numbers $\bf R$ there are no infinite values, there are no numbers called "infinity" or given the label '$\infty$.' Thus "$x=\infty$" does not make sense in a context where $x$ is a real number. So in most standard analysis contexts, $x=\infty$ would never pose a problem because it is never a possibility to begin with. There are other contexts in which $x=\infty$ (or similar things) are allowed, though.

With cardinal numbers (which can be thought of as the possible sizes of sets), there are infinite cardinals. The smallest infinite cardinal $\aleph_0$ (aleph-null) is the size of $\bf N$, the natural numbers. It does not make sense to perform division of cardinals generally, because the results are not well-defined (neither is subtraction, which is the substance of the Hilbert's Hotel paradox). What this means is that if you partition a set of size $\alpha$ into pieces each of size $\beta$, the number $\gamma$ of pieces that result (which we may desire to think of as the ratio $\alpha/\beta$) is not uniquely determined merely by the numbers $\alpha$ and $\beta$ if they are infinite.

The so-called surreal numbers also involve infinities, in which we have quantities that are larger than any real number, as well as infinitessimals that are smaller than any real number. In this context it is possible to divide by anything that is not zero (we say that the surreals form a field).

With a topological mindset we may extend the number line or the complex plane by adjoining formal symbols like $\infty$ that function as an infinite quantity in some sense. In the case of the complex plane $\bf C$, the one-point compactification yields the Riemann sphere ${\bf C}\cup\{\infty\}$. Here it is actually possible and desirable to divide by $\infty$. If we wanted to (in the context of linear-fractional transformations), we could stipulate that $a/\infty=0$ for $a\ne\infty$ and $\infty/\infty=1$ otherwise. However then we lose the ability to speak about things like $a\cdot\infty$ without contradictions forming. Ultimately we just want $z\mapsto\frac{az+b}{cz+d}$ to be extended from a map from the plane to itself to a map on the sphere.

In conclusion, in the standard context of real numbers "$x=\infty$" does not literally mean anything, although it can be reinterpreted in many circumstances using limits. In more varied contexts, there may be multiple infinities so that "$x=\infty$" is ambiguous, and even then division by infinities is generally not well-defined - believing you can divide them may lead to errors. In other contexts division by any infinity is well-defined, and in another we can plug $x=\infty$ into certain fractional expressions but not all of the axioms of arithmetic are still applicable to infinity. Etc. etc.

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A simple way of looking at this is by examining the formal definition of the limit (which you can really find all over the place). Although, there is a rigorous way to prove the result using the definition, I think here we can get away with just the fact that in evaluating limits, it need not be the case that

$\lim_{x \to a} f(x) = f(a), \ (a \in \mathbb{R}, \text{ or } a = \infty) \tag{1}$

So, your limit, since $x \not = \infty$, simplifies to

$\lim_{x \to \infty} {x\over x} = \lim_{x \to \infty} 1 = 1. \tag{2}$

Assuming, of course, $x \not = 0.$ :)

I should also mention, that "$a = \infty$" here was used in, no means, a strict sense and there is a way to make $(1)$ rigorous as well.

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You cannot really say $x = \infty$ because $\infty \not \in \mathbb{R}$

What you do is, you take the limes. Limes means not, that $x=a$, but that $x$ is getting closeser and closer to $a$. For example: $$\lim_{x\mapsto 0}\frac{1}{x}=\infty$$ because the divisor gets smaller and smaller $$\frac{1}{2}=0.5 \\\frac{1}{1}=1 \\\frac{1}{0.5}=2\\..$$

So this is growing and growing, but $\frac{1}{0}$ is mathematically nonsense.

So it's quite simple.

$$\lim_{x\mapsto \infty}\frac{x}{x}=\lim_{x\mapsto\infty}\frac{1}{1}=1$$

In that case you simply cancel (is that the right word in english?) the $x$. That way the Limes and the $\infty$ vansishs.

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Are you seriously suggesting that l'Hospital should be used for this limit?? –  mrf Apr 28 '13 at 11:40
    
As you see, im suggesting that. Is that wrong? –  wegsehen Apr 28 '13 at 11:43
    
nothing wrong but over complicating stuff $\frac{x}{x}=1$ and that's it , no need for l'Hospital rule :) –  Albanian_EAGLE Apr 28 '13 at 13:44
    
Thanks, I simplified that. –  wegsehen Apr 28 '13 at 14:19

→ ∞/→ ∞ is not a defined value.

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Welcome to MSE! I realize you don't yet have enough reputation, but this would be better as a comment. Regards –  Amzoti May 19 '13 at 3:05

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