limits - 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 $\infty$ ?

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

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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?

Answer

This is a very interesting question... I believe you are correct. Below is a proof.

1. 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.


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


3. 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.

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




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

Case ii.

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


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

Case iii.

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


2. 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.