Natural log limits as n approaches infinity

[MSG100]

My question is:

Show the limit of
$$x_{n}=\frac{ln(1+\sqrt{n}+\sqrt[3]{n})}{ln(1+\sqrt[3]{n}+\sqrt[4]{n})}$$
as n approaches infinity


Solution:
$${x_n} = \frac{{\ln (1 + {n^{\frac{1}{2}}} + {n^{\frac{1}{3}}})}}{{\ln (1 + {n^{\frac{1}{3}}} + {n^{\frac{1}{4}}})}} = \frac{{\ln \left( {{n^{\frac{1}{2}}} \cdot (\frac{1}{{{n^{\frac{1}{2}}}}} + 1 + \frac{1}{{{n^{\frac{1}{6}}}}})} \right)}}{{\ln \left( {{n^{\frac{1}{3}}} \cdot (\frac{1}{{{n^{\frac{1}{3}}}}} + 1 + \frac{1}{{{n^{\frac{1}{{12}}}}}})} \right)}} =$$


$$= \frac{{\ln {n^{\frac{1}{2}}} + \ln (\frac{1}{{{n^{\frac{1}{2}}}}} + 1 + \frac{1}{{{n^{\frac{1}{6}}}}})}}{{\ln {n^{\frac{1}{3}}} + \ln (\frac{1}{{{n^{\frac{1}{3}}}}} + 1 + \frac{1}{{{n^{^{\frac{1}{{12}}}}}}})}} = \frac{{\frac{1}{2}\ln n + \ln (\frac{1}{{{n^{\frac{1}{2}}}}} + 1 + \frac{1}{{{n^{\frac{1}{6}}}}})}}{{\frac{1}{3}\ln n + \ln (\frac{1}{{{n^{\frac{1}{3}}}}} + 1 + \frac{1}{{{n^{^{\frac{1}{{12}}}}}}})}} =$$

$$= \frac{{\frac{1}{2} + \frac{1}{{\ln n}} \cdot \ln (1 + \frac{1}{{{n^{\frac{1}{2}}}}} + \frac{1}{{{n^{\frac{1}{6}}}}})}}{{\frac{1}{3} + \frac{1}{{\ln n}} \cdot \ln (1 + \frac{1}{{{n^{\frac{1}{3}}}}} + \frac{1}{{{n^{\frac{1}{{12}}}}}})}}$$


Now I'm stuck.
Is this the right way to do it?

 

[Office_Shredder]

That looks good so far... as n goes to infinity, what happens to 1/ln(n), and what happens to ln(1+1/n^1/2+1/n^1/6) and the similar term in the denominator?

 

[Mark44]

How did you (OP) go from (1/2)ln(n) to 1/2 + 1/ln(n) ? And same question for the similar part of the denominator?

 

[Mark44]

I would continue from here:
$$= \frac{{ \frac 1 2 \ln n + \ln (\frac{1}{{{n^{\frac{1}{2}}}}} + 1 + \frac{1}{{{n^{\frac{1}{6}}}}})}}{{ \frac 1 3 \ln n + \ln (\frac{1}{{{n^{\frac{1}{3}}}}} + 1 + \frac{1}{{{n^{^{\frac{1}{{12}}}}}}})}}$$
by factoring ln(n) from each term in the numerator and each term in the denominator, with the result looking like this:
$$ \frac{\frac 1 2 ln(n)(1 + \text{other stuff})}{\frac 1 3 ln(n)(1 + \text{some other stuff})}$$
As long as you can convince yourself the <other stuff> goes to zero as n gets large, you should be able to get the desired answer of 3/2.

 

[Office_Shredder]

Mark he just divided the numerator and denominator by ln(n)

 

[Mark44]

OK, I didn't see that. It looked to me like he was using an invalid log property. What he did is pretty close to what I'm suggesting in post #4.

 

[brmath]

I liked your first 3 steps until you got to $\frac{(1/2)logn + stuff}{ 1/3logn + stuff}$. What I'm calling "stuff" will all go to 0 as n $\rightarrow \infty$.

So from that point you can go directly to the answer of 3/2 -- I don't thing you need to fuss any further.

You should probably explain why the stuff goes to 0.

 

[MSG100]

Many thanks to all of you for the quick response! I can see it now!

These kind of problems are quite new for me so maybe I overdo it.

I need to do some more of these so I can get some grip of it.