AoPS Academy
, 
,
, 
, why 
or even expressions of those terms combined as if that would make them defined. I have made this post to answer these questions once and for all, and I politely ask everyone to link this post to threads that are talking about this issue.. It is usually regarded that 
, not because this works numerically but because it is convenient to define it this way. You will see the convenience of defining other undefined things later on in this post.? The issue here is that 
isn't even rigorously defined in this expression. What exactly do we mean by ? Unless the example in question is put in context in a formal manner, then we say that is meaningless.? Suppose that 
. Then we would have 
, absurd. A more rigorous treatment of the idea is that 
does not exist in the first place, although you will see why in a calculus course. So the point is that is undefined.
? An article from brilliant has a good explanation. Alternatively, you can just use a geometric series. Notice that
? Usually this is considered to be an indeterminate form, but I would also wager that this is also undefined.
indeterminate? In analysis (the theory behind calculus and beyond), limits involving an algebraic combination of functions in an independent variable may often be evaluated by replacing these functions by their limits. However, if the expression obtained after this substitution does not provide sufficient information to determine the original limit, then the expression is called an indeterminate form. For example, we could say that is an indeterminate form.
, 
or , obtained by applying the algebraic limit theorem (a theorem in analysis, look this up for details) in the process of attempting to determine a limit, which fails to restrict that limit to one specific value or infinity, and thus does not determine the limit being calculated. This is why it is called indeterminate. Some examples of indeterminate forms are![\[0/0, \infty/\infty, \infty-\infty, \infty \times 0\]](http://latex.artofproblemsolving.com/2/6/0/260f97a1cf1ea8722ff97a29b83bac7bfe83e577.png)
etc etc. So what makes something undefined? In the broader scope, something being undefined refers to an expression which is not assigned an interpretation or a value. A function is said to be undefined for points outside its domain. For example, the function 
given by the mapping 
is undefined for 
. On the other hand, 
is undefined because dividing by is not defined in arithmetic by definition. In other words, something is undefined when it is not defined in some mathematical context. is considered both indeterminate and undefined (but in the context of a limit then it is considered in indeterminate form). Additionally, this notion of something being undefined also means that we can define it in some way. To rephrase, this means that technically, we can make something that is undefined to something that is defined as long as we define it. I'll show you what I mean.![\[\overline{\mathbb{R}}=\mathbb{R}\cup \{-\infty,+\infty\}.\]](http://latex.artofproblemsolving.com/9/9/3/993c78b62ec2c84b5d77daa8e9a8cd6f70fcf904.png)
So instead of treating infinity as an idea, we define infinity (positively and negatively, mind you) as actual numbers in the reals. The advantage of doing this is for two reasons. The first is because we can turn this thing into a totally ordered set. Specifically, we can let 
for each 
which means that via this order topology each subset has an infimum and supremum and 
is therefore compact. While this is nice from an analytic standpoint, extending the reals in this way can allow for interesting arithmetic! In it is perfectly OK to say that,![\begin{align*}
a + \infty = \infty + a & = \infty, & a & \neq -\infty \\
a - \infty = -\infty + a & = -\infty, & a & \neq \infty \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \pm\infty, & a & \in (0, +\infty] \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \mp\infty, & a & \in [-\infty, 0) \\
\frac{a}{\pm\infty} & = 0, & a & \in \mathbb{R} \\
\frac{\pm\infty}{a} & = \pm\infty, & a & \in (0, +\infty) \\
\frac{\pm\infty}{a} & = \mp\infty, & a & \in (-\infty, 0).
\end{align*}](http://latex.artofproblemsolving.com/6/d/8/6d8fc1eca5c791d430a7168cd3b37a02104fa318.png)
So addition, multiplication, and division are all defined nicely. However, notice that we have some indeterminate forms here which are also undefined,![\[\infty-\infty,\frac{\pm\infty}{\pm\infty},\frac{\pm\infty}{0},0\cdot \pm\infty.\]](http://latex.artofproblemsolving.com/a/0/b/a0b67076efcc014c78b2023be1151c84b57c1503.png)
So while we define certain things, we also left others undefined/indeterminate in the process! However, in the context of measure theory it is common to define 
as greenturtle3141 noted below. I encourage to reread what he wrote, it's great stuff! As you may notice, though, dividing by is undefined still! Is there a place where it isn't? Kind of. To do this, we can extend the complex numbers! More formally, we can define this extension as![\[\mathbb{C}^*=\mathbb{C}\cup\{\tilde{\infty}\}\]](http://latex.artofproblemsolving.com/8/4/e/84ed3d6ab237df970333b87beaef5311caa46185.png)
which we call the Riemann Sphere (it actually forms a sphere, pretty cool right?). As a note, 
means complex infinity, since we are in the complex plane now. Here's the catch: division by is allowed here! In fact, we have![\[\frac{z}{0}=\tilde{\infty},\frac{z}{\tilde{\infty}}=0.\]](http://latex.artofproblemsolving.com/1/6/0/160cb939707708ff4e0930724df2928af11d7fd1.png)
where 
and are left undefined. We also have
Furthermore, we actually have some nice properties with multiplication that we didn't have before. In 
it holds that![\[\tilde{\infty}\times \tilde{\infty}=\tilde{\infty}\]](http://latex.artofproblemsolving.com/8/e/0/8e035b4841d20c5ad671c46355cc667ca01e533b.png)
but 
and 
are left as undefined (unless there is an explicit need to change that somehow). One could define the projectively extended reals as we did with , by defining them as![\[{\widehat {\mathbb {R} }}=\mathbb {R} \cup \{\infty \}.\]](http://latex.artofproblemsolving.com/c/0/4/c044ff37849acad3b8b280b9753539225a1276ca.png)
They behave in a similar way to the Riemann Sphere, with division by also being allowed with the same indeterminate forms (in addition to some other ones).
and ![$f:[a,b]\rightarrow\mathbb{R}$](http://latex.artofproblemsolving.com/6/7/c/67c320754a36aaec178e61f04bfcbac41c0bf675.png)
be a strictly monotonous function such that 
. Show that 
.
for which there exists a scrictly monotonous function 
such that 
+1 w
is a random variable that takes on only nonnegative integer values, with ![$E[X]=1,$](http://latex.artofproblemsolving.com/e/8/2/e826f5b79a2adeacdaccd3884658e94a3b1f8e22.png)
![$E[X^2]=2,$](http://latex.artofproblemsolving.com/5/9/1/5915ee4a36f8615ddc8075a197f3b44a902936c6.png)
and ![$E[X^3]=5.$](http://latex.artofproblemsolving.com/d/1/1/d11c925af9f188f0140a89a0fbd6c83e1dd4fc12.png)
(Here ![$E[Y]$](http://latex.artofproblemsolving.com/a/2/9/a29ea940a5d54ad2a5d668a596f9db0c88fa1123.png)
denotes the expectation of the random variable 
) Determine the smallest possible value of the probability of the event 
and 
. Define 
for 
as 
, 
. Prove that 
exists and find it's value .
be a function that satisfies the conditions:
has Darboux’s property
at 
exist.
. If 
is rounded to the nearest integer and then divided by 
, what is the remainder?
be a continuous function such that for all 
and for all 
, 
Show that 
is a constant function.
, then is constant on 
. Let 
to see that is constant on 
. By continuity is constant on 
. In the same way, with 
one shows that is constant on ![$(-\infty,0]$](http://latex.artofproblemsolving.com/5/0/d/50d25d3e486146edd225ea505aaa53016f41edb7.png)
. Hence is constant.
is differentiable.
be a continuous function such that for all and for all , Show that is a constant function.
by any varying continuous function of a specific class.
. We have 
for all . Fix 
. Then
The last one is true by continuity.
for all 
. Hence is a constant.
to get 
(say). So, 
. Put 
to get 
. So, ![$f(x)=c\forall x\in (0,1]$](http://latex.artofproblemsolving.com/6/0/6/606f6f63a21fe94edf149ecf26a60ea8bf2bbb90.png)
. Hence, 
.
to get 
(say). So, ![$f(x)=k \forall x \in (-\infty,-1]$](http://latex.artofproblemsolving.com/1/2/d/12de0cd9a936f2c3d2eb82fb694b95c3130e3182.png)
. Put 
to get 
. So, 
. Hence, 
.
is differentiable or not. 
, then is constant on . Let to see that is constant on . By continuity is constant on . In the same way, with one shows that is constant on . Hence is constant.
. Then for any fixed 
, it's clear that 
Take the limits of both sides as 
. Notice that 
as . By the continuity of 
, we have 
This shows that 
.
. We have for all . Fix . ThenThe last one is true by continuity. for all . Hence is a constant.
. We have for all . Fix . ThenThe last one is true by continuity. for all . Hence is a constant. as well.
be a continuous function such that for all and for all , Show that is a constant function. by any varying continuous function of a specific class. with any surjective function with domain either 
or best, 
.
, we assume for sake of simplicity
( any arbitrary) for which we define 
s.t. 
, then we have:
is continuous over we take limits both sides with 
and we know that for a continuous function over we have a sequence in say 
s.t. 
to be arbitrary we replace 
hence is constant function. 
. Consider the interval 
. Now consider 
. As 
, we have, 
. So, as 
, which satisfies the condition for 
, we have, 
. Therefore, if we consider any number 
s.t.
we see that, 
. Hence, it's meaningful to consider a sequence on either side of that converges to and simultaneously making use of the property that the function is continuous.
be a sequence converging to . Then 
converges to 
. And as, all the terms in the sequence are equal, the function is a constant function. 
be a continuous function such that for all and for all , Show that is a constant function.
and 
, then 
since, 
. Thus, we have 
Switch the roles of 
and 
we get 
and 
we get 
.
, since, 
is continuous, 
. Thus all in all, 
which proves is constant. 
is constant over 
. Consider . As 
is continuous and not bounded above, I can always find a suitable 
such that 
for any 
. Picking an 
arbitrarily close to assures that is a constant over , similarly we can argue that 
for all , which is the required. 
note that 
is defined on 
for any real 
Since we can adjust 
for all 
. Thus 
is constant for all 
greater than 
and plug 
in our original given function in the problem then 
thus 
must be constant function since we can adjust 
is positive then 
.
implies 
only
![$\ [f'(x)]'_{t} = e^t[f(e^tx)]_{t}'$](http://latex.artofproblemsolving.com/5/4/d/54dda66cd3df94224296f39630c047cce392ccfe.png)
![$\ \implies [f(e^tx)]'_t = lim_{\lambda \to 0} \frac{f(e^{t+\lambda}x) - f(x)}{\lambda}=0$](http://latex.artofproblemsolving.com/c/d/7/cd7d1b5a55a85b6e8c49fa0edcc420ead0615170.png)
where 
and as 
Q.E.D.
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