# Getting the limit for a ratio of logarithms functions with different base

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:

$$f(x)/g(x)$$

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:

$$log_23$$

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.

• "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 $\lim \log_2 3 = \log_2 3 \approx 1.5849625.....$. Jul 26, 2019 at 1:14

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., $$\log_{2}3$$.