Calculate the limit of: $x_n = \frac{\ln(1+\sqrt{n}+\sqrt[3]{n})}{\ln(1 + \sqrt[3]{n} + \sqrt[4]{n})}$, $n \rightarrow \infty$ Is it ok to solve the following problem this way? What I have done is to solve parts of the limit first (that converges to $0$), and then solve the remaining expression? Or is this flawed reasoning?
Question
Calculate the limit of:
$$x_n = \frac{\ln(1+\sqrt{n}+\sqrt[3]{n})}{\ln(1 + \sqrt[3]{n} + \sqrt[4]{n})}$$
when $n$ goes to infinity.
Answer
This can also be written as:
$$\lim_{n \to \infty} \frac{\ln(1+\sqrt{n}+\sqrt[3]{n})}{\ln(1 + \sqrt[3]{n} + \sqrt[4]{n})}$$
The denominator can be written as:
$$\ln(1 + \sqrt[3]{n} + \sqrt[4]{n}) = \ln(1 + \frac{1 + \sqrt[4]{n}}{\sqrt[3]{n}}) + \ln(\sqrt[3]{n})$$
From this we can see that:
$$\lim_{n \to \infty} \ln(1 + \frac{1 + \sqrt[4]{n}}{\sqrt[3]{n}}) \rightarrow 0$$
The numerator can be written as:
$$\ln(1 + \sqrt{n} + \sqrt[3]{n}) = \ln(1 + \frac{1 + \sqrt[3]{n}}{\sqrt{n}}) + \ln(\sqrt{n})$$
From this we can see that:
$$\lim_{n \to \infty} \ln(1 + \frac{1 + \sqrt[3]{n}}{\sqrt{n}}) \rightarrow 0$$
This means that we have the following limit:
$$\lim_{n \to \infty} \frac{\ln(\sqrt{n})}{\ln(\sqrt[3]{n})} = \lim_{n \to \infty} \frac{\ln(n^{\frac{1}{2}})}{\ln(n^{\frac{1}{3}})} = \lim_{n \to \infty} \frac{\frac{1}{2}\ln(n)}{\frac{1}{3}\ln(n)} = \lim_{n \to \infty} \frac{3 \ln(n)}{2\ln(n)} \rightarrow \frac{3}{2}$$
The limit converges towards $\frac{3}{2}.$

It is much shorter using equivalents:
$1+\sqrt n+\sqrt[3]n\sim_\infty\sqrt n$, hence $\;\ln(1+\sqrt n+\sqrt[3]n)\sim_\infty\frac12\ln(n)$.
Similarly, $\ln(1+\sqrt[3] n+\sqrt[4]n)\sim_\infty\frac13\ln(n)$, whence
$$\frac{\ln(1+\sqrt n+\sqrt[3]n)}{\ln(1+\sqrt[3] n+\sqrt[4]n)}\sim_\infty\frac{\frac12\ln(n)}{\frac13\ln(n)}=\frac32.$$
