<s>[INST] 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 $\frac{3}{2}.$
 [/INST] The answer is correct but I would say the working is not quite rigorous.  In effect you have said that
$$\lim a=0\ ,\quad \lim c=0$$
implies
$$\lim\frac{a+b}{c+d}=\lim\frac bd\ .$$
It's true in this case since $b,d\to\infty$; it would also be true if $b,d$ have any finite non-zero limits; but it is not always true if $b,d\to0$.
For example, it is not true that
$$\lim\frac{\frac1n+\frac2n}{\frac1n+\frac3n}=\lim\frac{0+\frac2n}{0+\frac3n}\ .$$
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