<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] Let's try the usual approach of standard limits. We have
\begin{align}
L &= \lim_{n \to \infty}\frac{\log(1 + \sqrt{n} + \sqrt[3]{n})}{\log(1 + \sqrt[3]{n} + \sqrt[4]{n})}\notag\\
&= \lim_{n \to \infty}\dfrac{\log\sqrt{n} + \log\left(1 + \dfrac{1 + \sqrt[3]{n}}{\sqrt{n}}\right)}{\log\sqrt[3]{n} + \log\left(1 + \dfrac{1 + \sqrt[4]{n}}{\sqrt[3]{n}}\right)}\tag{1}\\
&= \lim_{n \to \infty}\dfrac{\dfrac{\log n}{2} + \log\left(1 + \dfrac{1 + \sqrt[3]{n}}{\sqrt{n}}\right)}{\dfrac{\log n}{3} + \log\left(1 + \dfrac{1 + \sqrt[4]{n}}{\sqrt[3]{n}}\right)}\notag\\
&= \lim_{n \to \infty}\dfrac{\dfrac{1}{2} + \dfrac{1}{\log n}\cdot\log\left(1 + \dfrac{1 + \sqrt[3]{n}}{\sqrt{n}}\right)}{\dfrac{1}{3} + \dfrac{1}{\log n}\cdot\log\left(1 + \dfrac{1 + \sqrt[4]{n}}{\sqrt[3]{n}}\right)}\tag{2}\\
&= \dfrac{\dfrac{1}{2} + 0\cdot 0}{\dfrac{1}{3} + 0\cdot 0}\tag{3}\\
&= \frac{3}{2}\notag
\end{align}
Note that we can not go from step $(1)$ to step $(3)$ directly by just replacing the $\log$ expressions with their limits because replacing a sub-expression by its limit is justified only in two scenarios:


*

*When the sub-expression is connected to the rest of the expression in additive manner.

*When the sub-expression is connected to the rest of the expression in multiplicative manner and its limit is non-zero.


We have instead done a further simplification by dividing numerator and denominator by $\log n$. By doing this we ensure that each term in numerator and denominator has a well defined limit and we can apply algebra of limits to get the final limit.
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