# 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?

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That looks good so far.... as n goes to infinity, what happens to 1/ln(n), and what happens to ln(1+1/n1/2+1/n1/6) and the similar term in the denominator?

Mentor
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?

Mentor
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.

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

1 person
Mentor
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.

1 person
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.