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 $\frac{3}{2}.$
 A: 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
$$\frac{\ln(1+\sqrt n+\sqrt[3]n)}{\ln(1+\sqrt[3] n+\sqrt[4]n)}\sim_\infty\frac{\frac12\ln(n)}{\frac13\ln(n)}=\frac32.$$
A: 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
$$\lim\frac{\frac1n+\frac2n}{\frac1n+\frac3n}=\lim\frac{0+\frac2n}{0+\frac3n}\ .$$
A: With all due respect to the answers already posted, I'm not in favor of using "equivalents" without enough explanation, or writing $\lim_{n\to \infty}$ before the limit has been shown to exist. Instead, I would do this: The expression for $n>1$ is less than
$$\frac{\ln (3n^{1/2})}{\ln n^{1/3}} = \frac{\ln 3 + (1/2)\ln n}{(1/3)\ln n} = \frac{\ln 3}{(1/3)\ln n} + \frac{3}{2} \to \frac{3}{2}.$$
Similarly, the expression is greater than 
$$\frac{\ln n^{1/2}}{\ln (3n^{1/3})} = \frac{(1/2)\ln n}{\ln 3 +(1/3)\ln n}  = \frac{(1/2)}{(\ln 3)/\ln n +(1/3)} \to \frac{3}{2}.$$
By the squeeze theorem, the limit is $3/2.$
A: 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.
