Limit of a Logarithm with Different Bases

Question

We are to compute

$$\lim_{n->\infty}{\frac{2^{\log_3 n}}{3^{\log_2 n}}}$$

Clearly the bases are reversed between the logarithm and exponents, so I can't seem to find any logarithm or exponential properties that can help simplify this problem.

Answer 1

Hint: We have $\log_2 n\ge \log_3 n$. So our fraction is $\le \left(\frac{2}{3}\right)^{\log_3 n}$.

Answer 2

Let us define the function $f(.)$ as follow:

$$f(n)=\dfrac{2^{\log_3n}}{3^{\log_2n}}.$$

Using the base change formula, we can easily transform $f(.)$ into the following function: (I used $\log$ and $\exp$ to denote the natural logarithm and the exponential function, respectively)

$$f(n)=\exp(\log2\log n\log_3e-\log3\log n\log_2e)$$ $$\Leftrightarrow$$ $$f(n)=\exp(C\log n)=n^C,$$

where $C$ is a constant. $C=\log2\log_3e-\log3\log_2e=\dfrac{\log2}{\log3}-\dfrac{\log3}{\log2}$.

Since $C$ is negative, i.e., $C<0$, then: $$\lim_{n \to +\infty}f(n)=0.$$