Using $O(n)$ to determine limits of form $1^{\infty},\frac{0}{0},0\times\infty,{\infty}^0,0^0$?

Question

Is it sufficient to use $O(n)$ repeatedly on $1^{\infty},\frac{0}{0},0\times\infty,{\infty}^0,0^0$ to get determinate forms? For example if we look at $\frac{0}{0}$ then $$\frac{O(f(n))}{O(g(n))}$$ should simplify the matters, and if needed repeat again to get $$\frac{O(O(f(n)))}{O(O(g(n)))}$$

My question is: at most would it be sufficient to repeatedly apply $O(n)$ to find the limit? Using this approach what type of limits would still after an infinite application of above process remain indeterminate?

By using of $O(f(x))$ we mean what is given in this example , e.g. Just substitution of dominant terms in place of the function.

Update : separated the L'Hopitals aspect of the question into Examples of applying L'Hôpitals rule ( correctly ) leading back to the same state?, tried to clarify what I tried to mean by using of $O(f(x))$

Answer 1

Based on how you asked your question, I think you should take a look at the definition of big-O notation. It is not clear what you mean by $O(O(f))$ or $\frac{O(f)}{O(g)}$.

The answer to the title question is yes, we can use Big-O notation to simplify certain solutions for limits. Lets look at a few examples:

Example 1:$$\lim_{x\rightarrow 0}\frac{\sin x}{x}.$$ When $x\rightarrow 0$ we have $\sin x=x+O(x^3)$ so that $\frac{\sin x}{x}=1+O(x^2)$. Then taking the limit we conclude $$\lim_{x\rightarrow 0}\frac{\sin x}{x}=1.$$

Example 2:

Let $x$ be any positive real number, and define a sequence $\{a_n\}$ by $$ a_n = \frac{ [x] + [2x] + [3x] + \dotsb + [nx] }{n^2} $$ where $[x]$ is the largest integer less than or equal to $x$. Prove that $\displaystyle{\lim_{n \to \infty} a_n = x/2}$.

Solution: See this Math Stack Exchange answer.

Answer 2

It is fine to use $O$ notation, but only if you understand what you are doing. For example, $e^{n+3+O(1/n)}$ is not $e^{n+3}+O(1/n)$.

Answer 3

The only part of the question I think I understand is whether l'Hopital, applied repeatedly, can fail. Let $$f(x)={x\over\sqrt{x^2+1}}$$ and try to find $\lim_{x\to\infty}f(x)$, using l'Hopital. We get $$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}{1\over{x\over\sqrt{x^2+1}}}=\lim_{x\to\infty}{\sqrt{x^2+1}\over x}$$ Use l'Hopital again, and you get back to $\lim_{x\to\infty}f(x)$.

A different example: $$\lim_{x\to\infty}{e^{-x}\over1/x}=\lim_{x\to\infty}{e^{-x}\over1/x^2}=\lim_{x\to\infty}{e^{-x}\over1/2x^3}=\dots$$