I know similar questions have been asked in the past, but I need some clarification to know if what I'm doing is correct:

let $f(x) = log_2x$, and $g(x)= log_3x$

Now I need to take the limit as $x$ approaches $\infty$ from the ratio:


I have applied the base change formula for $g(x)$ which yielded $log_2x/log_23$

now, when I divide $log_2x/log_2x/log_23$ I get:


so, how would I go about the taking the limit of $log_23$ since I could not find a rule regarding logarithms so far. and I'm doing this exercises as a part of algorithms class I'm taking.

  • 1
    $\begingroup$ "how would I go about the taking the limit of $\log_2 3$?" $\log_2 3 $ is a constant. A constant doesn't vary. $\lim c = c$ always. So $\lim \log_2 3 = \log_2 3 \approx 1.5849625.....$. $\endgroup$
    – fleablood
    Jul 26, 2019 at 1:14

1 Answer 1


Taking the limit here doesn't involve any specific rules regarding logarithms. This is because since $\log_{2}3$ is a constant, i.e., doesn't involve $x$ at all, the limit of it as $x \to \infty$ would be the same value, i.e., $\log_{2}3$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .