,
,
,
,
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.
,
isn't even rigorously defined in this expression. What exactly do we mean by 
.
,
does not exist in the first place, although you will see why in a calculus course. So the point is that 
?
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 
,
or ![\[0/0, \infty/\infty, \infty-\infty, \infty \times 0\]](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 
.
is undefined because dividing by ![\[\overline{\mathbb{R}}=\mathbb{R}\cup \{-\infty,+\infty\}.\]](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 ![\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 ![\[\mathbb{C}^*=\mathbb{C}\cup\{\tilde{\infty}\}\]](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 ![\[\frac{z}{0}=\tilde{\infty},\frac{z}{\tilde{\infty}}=0.\]](http://latex.artofproblemsolving.com/1/6/0/160cb939707708ff4e0930724df2928af11d7fd1.png)
where 
and 
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 ![\[{\widehat {\mathbb {R} }}=\mathbb {R} \cup \{\infty \}.\]](http://latex.artofproblemsolving.com/c/0/4/c044ff37849acad3b8b280b9753539225a1276ca.png)
They behave in a similar way to the Riemann Sphere, with division by 
![$\underset{n\to \infty }{\mathop{\lim }}\,\sqrt[n]{lcm(1,2,3,...,n)}=e$](http://latex.artofproblemsolving.com/a/1/3/a139e5cb990f3386de109c701f8edc71b4054712.png)
be a function, two times derivable on 
for which there exist 
such that![\[ \frac { f(b)-f(a) }{b-a} \neq f'(c) ,\]](http://latex.artofproblemsolving.com/a/3/1/a31861ef909a6e17ed8b495b9b9b16d887ba573f.png)
for all 
.
.
be positive real numbers .
feet and area 
,![\[ \textbf{(A)}\ 24 \qquad
\textbf{(B)}\ 30 \qquad
\textbf{(C)}\ 46 \qquad
\textbf{(D)}\ 66 \qquad
\textbf{(E)}\ 74
\]](http://latex.artofproblemsolving.com/8/2/f/82ffc15072ca99f6cd1abc312781eb4827817356.png)
degrees about one side. The space through which the hexagon travels during the rotation forms a solid. Find the volume of this solid.
and height 
:
Volume of cone frustum with base radii 
,
has parallel bases 
and 
,
is the midpoint of side 
and 
.
cm and 
cm, what is 
?
,
,
Find
Y by KAME06
Could anybody possibly prove that the limit of 
is 
, maybe even with rather elementary means? As usual, 
denotes the 
-th prime number. The problem of that limit came up in my partial solution of this problem: https://artofproblemsolving.com/community/c7h3495516.
Thank you for your efforts.
is 
, maybe even with rather elementary means? As usual, 
denotes the 
-th prime number. The problem of that limit came up in my partial solution of this problem: https://artofproblemsolving.com/community/c7h3495516.Thank you for your efforts.
and this can be done using the Prime Number Theorem: 
is the "official" statement of the prime number theorem, with the prime-counting function 
.
Although I admit that the second statement expresses the same "idea" as the first, I cannot see that it follows immediately from the first. As I have never tried to understand the proof of the first statement, I never bothered to look for a proof of the second, but I assume that it still takes some real work to make the step from the first to the second statement. Do you agree?
,
Since 
we get 
i.e. 
Taking logarithm yields 
This implies 
so 
Since 
it follows that 
which combined with 
then 
and 
(recall that 
means 
)
and 
and 
does not exist, which illustrates that "additively" the sequences might be very different.
so
which is entirely new to me. And your other remarks were also very helpful!
