I am trying to prove the following statement.

if $\displaystyle \lim_{n\rightarrow\infty}\frac{f(n)}{g(n)}= \infty$, then $f(n) = \Omega(g(n))$ but $f(n) \neq O(g(n))$

What I've done so far

Using the definition of limit: $\forall M>0 \quad \exists n_0 : \forall n \ge n_0 \quad \displaystyle\frac{f(n)}{g(n)} > M$
So I multiplied both halves by $g(n)$ obtaining $f(n) > M\cdot g(n)$
I therefore chose a value $c_1$ that respect the condition $0 \le M \le c_1$ in order to define the relation $0 \le c_1g(n)\le f(n)$ that proves that $f(n) = \Omega(g(n))$

So now my questions are:

  • Is the (first part of the) solution mathematically right?
  • How can I mathematically prove that $f(n) \neq O(g(n))$?

I'm struggling a bit to prove the second part of the demonstration.
The main idea I had is to do a reductio ad absurdum where I state that the value $c_2$ I get (following the definition of $O(g(n))$) leads me to a contradiction. But I'm not entirely sure it makes sense.

Thank you in advance for your time.


2 Answers 2


Your basic idea is good. Here is how you would approach it:

Let us assume towards contradiction that $\exists c_2,n_0:\forall n>n_0:f(n)\le c_2\cdot g(n)$

In particular, we can re-write this as:

$$\exists c_2,n_0:\forall n>n_0:\frac{f(n)}{g(n)}\le c_2$$

Now, since $\lim_{n\rightarrow \infty}\frac{f(n)}{g(n)}=\infty$, then we know by definition of the limit that $$\forall M>0:\exists n_0:\forall n>n_0:\frac{f(n)}{g(n)}>M$$

In particular, for any $c_2$ and $n_0$, there is some $n>n_0$ with $f(n)>c_2\cdot g(n)$. Hence, we get a contradiction to the assumption.

  • $\begingroup$ I wrote almost the same proof as you wrote above. I wasn't sure if that made sense, but now know for sure! Thank you very much! $\endgroup$ Jul 22, 2021 at 19:48
  • $\begingroup$ @LukeTheWolf this proof is directly following the definition of big-O and the limit definition that you see in calculus classes. It is very formal, but not always as understandable :o $\endgroup$
    – nir shahar
    Jul 22, 2021 at 20:00
  • $\begingroup$ Unfortunately, I know it is very formal, but I was asked to make a formal demonstration. Perhaps can you give me a few more tips on how to make that demonstration easier, but just as effective? $\endgroup$ Jul 22, 2021 at 20:04
  • 1
    $\begingroup$ @LukeTheWolf sadly, I do not know of another way, except for using the big-O definition using limits. In that definition, $f$ is said to be $O(g)$ if $\limsup_{n\rightarrow \infty} \frac{f(n)}{g(n)}<\infty$ and as you can see, it is exactly what you asked in the question (and hence not an interesting solution since it just assumes the answer) $\endgroup$
    – nir shahar
    Jul 22, 2021 at 20:27

Suppose $f(n)=n^4,g(n)=n$, so


That means $f(n)=\Omega(g(n))$ and $f(n)\neq\mathcal{O}(g(n)).$ As a result, if $f(n)=\omega(g(n))$, then we can conclude that, $f(n)=\Omega(g(n))$ and $f(n)\neq \mathcal{O}(g(n)).$

Note that in a such case that

$$\lim_{n\to\infty}\frac{f(n)}{g(n)}=\infty$$ easily we can conclude that $f(n)=\omega(g(n))$.

  • $\begingroup$ Thanks for the reply, but I would like to find the mathematical proof that determines that $f(n) \neq O(g(n))$, could you help me please? $\endgroup$ Jul 22, 2021 at 17:09
  • 1
    $\begingroup$ You can take $f(n)=\omega(g(n))$, as a result, $f(n)=\Omega(g(n))$ and $f(n)\neq \mathcal{O}(g(n))$ $\endgroup$
    – ErroR
    Jul 22, 2021 at 17:10
  • $\begingroup$ Ok, thanks for that! $\endgroup$ Jul 22, 2021 at 17:17

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