Is $\infty + (\infty/\infty)$ indeterminate? I know $ (\infty/\infty)$ is indeterminate, but it can't be less than $0$. 
So can you assume $\infty + (\infty/\infty)$  is determinate because $\infty + n$ where $n\ge 0$ is still $\infty$ ?

The equation this question is based off of is $$\lim_{n \to \infty} \frac{n \log n + n}{\log n}.$$

This is in the context of big O notation. Would this be form be valid to use to determine the numerator's function is big Omega of the denominator? Or should l'hopitals rule be used to find a determinate and defined limit? 
 A: You can use the fact that if $\lim\limits_{x\to c}f(x)= \infty$ and $\liminf\limits_{x\to c}g(x)\ge 0$, then eventually $f(x)+g(x)\ge f(x)-1\to \infty$.
A: I think this is a great question! $\frac{\infty}{\infty} + \infty$ is not an indeterminate form, and usually it's considered undefined. However, it would be reasonable to consider it to be defined and equal to $\infty$; this is an alternate definition that is perfectly reasonable.

Question 1. Is $\frac{\infty}{\infty} + \infty$ an indeterminate form?
Answer: No. The expression $\frac{\infty}{\infty} + \infty$ is not an indeterminate form, because the words "indeterminate form" have a technical meaning, and that specifically means an expression which is one of the following:
$$\frac 00,~ \frac{\infty}{\infty},~ 0\times\infty,~ 1^\infty,~ \infty-\infty,~ 0^0 \text{ and } \infty^0.$$
Now, notice that $\frac{\infty}{\infty} + \infty$ is not one of the values on this list. Therefore, it's not an indeterminate form.

Question 2. Is $\frac{\infty}{\infty} + \infty$ defined?
Answer: Not according to the standard definition. Usually, we would say that an undefined value plus any other value (or times any other value, or multiplied by any other value, etc.) is just undefined. That is, if any part of an expression is undefined, the whole thing is undefined. Consider these examples:
$$
1 + \frac{0}{0} = \textbf{undefined} \quad \text{(subexpression } \frac{0}{0} \text{ is undefined so the expression is undefined})
$$
$$
0 \cdot \left(\lim_{x \to \infty} \sin x \right) = \textbf{undefined} \quad \text{(subexpression } \lim_{x \to \infty} \sin x \text{ is undefined so the expression is undefined})
$$
$$
\frac{\infty}{\infty} + \infty = \textbf{undefined} \quad \text{(subexpression } \frac{\infty}{\infty} \text{ is undefined so the expression is undefined})
$$

Question 3. OK, but is there some alternate definition that we can use so that $\frac{\infty}{\infty} + \infty = \infty$?
Answer: Yes!
There is something called a multivalued function that is often used in math, particularly in a branch of math called complex analysis. A "multivalued function" is where we assign the value of an expression to be all possible values that make sense instead of just one value. So in the multivalued-function world, we would say that
$$
\frac{0}{0} = \text{all real numbers and $\infty$ and $-\infty$}
$$
That is, we have said that the value of dividing $0$ by $0$ is not just one value, but all values because any value makes sense as the answer.
Using multi-valued functions, we would then say that
$$
\frac{\infty}{\infty} = \text{ all nonnegative real numbers and } \infty,
$$
because any nonnegative answer makes sense. Then, we would have that
$$
\frac{\infty}{\infty} + \infty = \infty,
$$
because if you add any nonnegative number to $\infty$, you get $\infty$.
This is an alternate, nonstandard definition, so it's important to know that your textbook or math teacher might not agree; but it's a perfectly valid alternate semantics, and it's useful in some areas of math.
A: So if we know that $\lim\limits_{n\to \infty} f(n) = +\infty$ and $\lim\limits_{n\to \infty} h(n) = +\infty$ and $\lim\limits_{n\to \infty} g(n) = +\infty$, can we assure that $\lim\limits_{n\to \infty}(f(n) + \frac {h(n)}{g(n)} = \infty$?
Yes.  
For any $M$ there is an $N_1$ so that $n > N$ implies $f(n) > M$.  And for $0$ there is an $N_2$ so that $n>N_2$ implies  that $h(n) > 0$  (a bit of overkill for $h(n) \to \infty$ ... but what the heck...) and an $N_2$ so that $n > N_3$ implies that $g(n) > 0$. 
So there is an $N \ge \max(N_1,N_2,N_3)$ so that if $n > N$ then $f(n) > M$ and $\frac {h(n)}{g(n)} > 0$ so $f(n) +\frac {h(n)}{g(n)} > f(n) > M$.
Thus $\lim\limits_{n\to \infty}(f(n) + \frac {h(n)}{g(n)} = \infty$.
For you example $\lim \frac {n\log n+n}{\log n} =\lim n(1+ \frac 1{\log n})$
We know $\lim\limits_{n\to \infty}\log n=\infty$ and $\lim\limits_{n\to \infty}\log (1+\frac 1{\log n}) = 1$.
Does that mean $\lim\limits_{n\to \infty}\log n(1+ \frac 1{\log n}) =\infty$?
Yes.  For every $M$  there is an $N_1$ so that if $n > N_1$ then $n > M$ and there is an $N_2$ so that if $n >N_2$ then $\frac 1{\log n} < \frac 1M$ so if $n > \max (N_1,N_2)$ we have $n(1+ \frac 1{\log n}) >M(1+\frac 1M)>M$.
A: In the context of limits of real numbers, $\infty+\frac{\infty}{\infty}$ is not an indeterminate form but can instead be said to be "equal to $\infty$".  To be precise, what this means is the following: 

Suppose $(a_n)$, $(b_n)$, and $(c_n)$ are sequences of real numbers such that $$\lim_{n\to\infty} a_n=\infty,\  \lim_{n\to\infty} b_n=\infty, \text{ and } \lim_{n\to\infty} c_n=\infty.$$ Then $$\lim_{n\to\infty}\left(a_n+\frac{b_n}{c_n}\right)=\infty.$$ 

The reasoning you suggest is the rough intuitive idea behind why this should be true, but it is not a rigorous proof since we are not actually literally adding $\infty+\frac{\infty}{\infty}$ but are instead taking a certain limit.  To prove it rigorously, fix $M\in\mathbb{R}$ and choose $N\in\mathbb{N}$ such that $a_n>M$ and $b_n,c_n>0$ for all $n>N$ (we can do this since the three sequences go to $\infty$ and so are eventually greater than any fixed real number).  Then for any $n>N$, $$a_n+\frac{b_n}{c_n}>a_n>M.$$  That is, $a_n+\frac{b_n}{c_n}$ is eventually greater than any fixed real number, so its limit is $\infty$.
A: This is a very interesting question... I believe you are correct. Below is a proof.


*

*Let $lim_{x \rightarrow a}[f(x)/g(x)]$ be some arbitrary indeterminate function such that $lim_{x \rightarrow a}[f(x)/g(x)]=\infty/\infty$ where $x\in \mathbb{R}$ and $a$ is a finite real number.

*Also let $lim_{x \rightarrow b}[h(x)]=\infty$ where $x\in \mathbb{R}$ and $b$ is a finite real number.

*As $x\rightarrow a$, we know that $f$ and $g$ either (i) grow at the same rate, (ii.) $f$ grows faster than $g$, or (iii.) $g$ grows faster than $f$.


Case i.


*

*If $f$ and $g$ grow at the same rate, then $lim_{x \rightarrow a}[f(x)/g(x)]=L$ where $L\in \mathbb{R}$.

*Then, $lim_{x \rightarrow b}[h(x)]+lim_{x \rightarrow a}[f(x)/g(x)]=\infty+L=\infty$.
Case ii.


*

*If $f$ grows faster than $g$, then $lim_{x \rightarrow a}[f(x)/g(x)]=\infty$.

*Then, $lim_{x \rightarrow b}[h(x)]+lim_{x \rightarrow a}[f(x)/g(x)]=\infty+\infty=\infty$.


Case iii.


*

*If $g$ grows faster than $f$, then $lim_{x \rightarrow a}[f(x)/g(x)]=0$.

*Then, $lim_{x \rightarrow b}[h(x)]+lim_{x \rightarrow a}[f(x)/g(x)]=\infty+0=\infty$.


Hence, in any case, $lim_{x \rightarrow b}[h(x)]+lim_{x \rightarrow a}[f(x)/g(x)]=\infty+(\infty/\infty)=\infty$
In this proof we are using limits to make sense of your statement, but I don't think you NEED limits to make sense of the statement. $\infty$ is simply a quantity increasing without limit... Some folks are saying $\infty/\infty$ is undefined; this is not true. It is indeterminate because the answer EXISTS, only it cannot be determined in the current form. By adding $\infty$ to the expression you obtain another expression that is suddenly in determinate form.
