Hello I am a beginner to this kind of notation and I would greatly appreciate an explanation which is easy to understand.
I need to prove
$$ \log_2(6 + \frac1x) = O(1) $$
and
$$ \log_2(6 + \frac1x) = \Omega(1) $$
My first thought would be to combine the equation inside, then separate the logs based on the division rule. But I tried, and I do not know what to do after that, or if that is the correct way.
Thank you for your help!
I assume that you are letting $x \to \infty$. From this, we can certainly assume that $x > 1$.
For the first equation, from the definition of "big-oh", we want to show that there is a postive $c$ such that $\log_2(6 + \frac1x) < c$ for all $x > \text{ some }b$. If $x > b$, then $\frac1{x} < 1/b$, so $\log_2(6 + \frac1x) < \log_2(6 + \frac1 b) $ and we can choose this vale for $c$.
Doing some cut, paste, and edit:
For the second equation, from the definition of "big-omega", we want to show that there is a postive $c$ such that $\log_2(6 + \frac1x) > c$ for all $x > \text{ some }b$. If $x > b$, then $\frac1{x} > 0$, so $\log_2(6 + \frac1x) > \log_2(6) $ and we can choose this value for $c$.