limits of sequences with ln(n)

**Lisa91** · November 11th 2012, 05:07 PM

I have a problem with these sequences: $\lim_{x\to\infty} \frac{ln(3n^2+20n+5)}{ln(n^9-3n+12)}$
$\lim_{x\to\infty} n ln(n^{2}+1)-2n(lnn)(lnn)^{1/n}$
$\lim_{x\to\infty}\frac{1*4*7*...*(3n+1)}{2*5*8*... *(3n+2)}$

The first one - I don't know how to get rid of the ln(something) function or maybe I'm just doing it in a wrong way. The second one - everything would be great if there wasn't this $ln(n)[ln(n)]^{1/n}$ which I don't know what to do with. The third one - I don't have any idea about this one... I would appreciate your hints or solutions. I've been trying to solve them for so long but I constantly get wrong answers (I used wolfram alpha).

**Plato** · November 11th 2012, 05:19 PM

Originally Posted by Lisa91

I have a problem with these sequences: lim n-> infinity [ln(3n^2+20n+5)]/[ln(n^9-3n+12)] lim n-> ininity (n ln(n^2+1)-2n(lnn)(lnn)^(1/n)) lim n> infinity (1*4*7*...*(3n+1))/(2*5*8*...*(3n+2)) The first one - I don't know how to get rid of the ln(something) function or maybe I'm just doing it in a wrong way. The second one - everything would be great if there wasn't this lnn)(lnn)^(1/n) which I don't know what to do with. The third one - I don't have any idea about this one... I would appreciate your hints or solutions. I've been trying to solve them for so long but I constantly get wrong answers (I used wolfram alpha).

I wonder why you do not learn to post is a readable format?

LaTex Code is easy to learn.

**Lisa91** · November 11th 2012, 05:33 PM

Sorry, I've just done it

**hollywood** · November 12th 2012, 10:43 PM

That is a lot more readable.

For the first one, I think you can use L'Hopital's rule.

- Hollywood

**Prove It** · November 12th 2012, 10:46 PM

Are you sure you're making x approach infinity?

**hollywood** · November 13th 2012, 12:53 AM

Wolframalpha says that the second one is $-\infty$ and the third one is zero. On the third one, I think it might help to take the log so you have the sum of $\log{\frac{3k+1}{3k+2}}$ .

- Hollywood

**Lisa91** · November 13th 2012, 12:57 AM

Oh, yes, obviously, it's n which approaches infinity. I'm done with the third one but I can't do the first and the second one.

**darthjavier** · November 13th 2012, 03:53 AM

Hello

1. Using l'Hôpital:

$\\\lim_{n\to\infty}\dfrac{\ln(3n^2+20n+5)}{\ln(n^9-3n+12)}=\dfrac{\lim_{n\to\infty}\dfrac{6n+20}{3n^2 +20n+5}}{\lim_{n\to\infty}\dfrac{9n^8+3}{n^9-3n+12}}\\\lim_{n\to\infty}\dfrac{\ln(3n^2+20n+5)}{ \ln(n^9-3n+12)}=\lim_{n\to\infty}\dfrac{(6n+20)(n^9-3n+12)}{(9n^8+3)(3n^2+20n+5)}$

You can apply l'Hôpital few more times, but you know that:

$\\\lim_{n\to\infty}\dfrac{\ln(3n^2+20n+5)}{\ln(n^9-3n+12)}=\lim_{n\to\infty}\dfrac{6n^{10}+\ldots}{27 n^{10}+\ldots}\\\lim_{n\to\infty}\dfrac{\ln(3n^2+2 0n+5)}{\ln(n^9-3n+12)}=\dfrac{6}{27}=\dfrac{2}{9}$

2. Making this change of variable:

$\\n=\dfrac{1}{t}\\\lim_{x\to\infty}n\ln(n^2+1)-2n(\ln(n))(\ln(n))^{\frac{1}{n}}=\lim_{t\to0} \dfrac{1}{t}\ln{}\left({}\frac{1}{t^2}{}+1\right)-2\dfrac{1}{t}\ln\left(\frac{1}{t}\right){}\left({} \ln{}\left({}\frac{1}{t}\right)\right)^t\\=\lim_{t {}\to{}0}\dfrac{\ln\left(\frac{1}{n^2}+1\right)-2\ln\left(\frac{1}{n}\right)^{t+1}}{t}$

You can continue, just applies L'hopital (because is $\dfrac{0}{0}$ , before that it was $\infty-\infty$ ) many times.... The answer is $-\infty$

I can't write you the full solution because I don't know how to use $\mathrm{\LaTeX}$ well....

Greetings

**darthjavier** · November 13th 2012, 04:14 AM

Originally Posted by Plato

I wonder why you do not learn to post is a readable format?

LaTex Code is easy to learn.

Are you serious?

Originally Posted by Prove It

Are you sure you're making x approach infinity?

lol I'm sure he wanted to type $n$ .

**Plato** · November 13th 2012, 04:43 AM

Originally Posted by darthjavier

Are you serious?

Yes! I was quite serious.
If you look at the OP before it was edited, you would why.
As you can see it was made readable.

And LaTeX is easy to learn.
If I had my way, all regular posters should be required to us LaTeX code.

**darthjavier** · November 13th 2012, 05:17 AM

Originally Posted by Plato

Yes! I was quite serious.
If you look at the OP before it was edited, you would why.
As you can see it was made readable.

And LaTeX is easy to learn.
If I had my way, all regular posters should be required to us LaTeX code.

Yes, I'm agree with you. I discovered recently $\mathrm{\LaTeX}$ , and I try to practice every day because is very useful.

But I mean that $\mathrm{\LaTeX}$ is difficult to learn, especially on this page. When I post it's too annoying see the equations combined with the HTML code. I can't see the HTML when I'm editing my $\mathrm{\LaTeX}$ code, I don't know why. In other pages I use an editor (in my case use TextMaker) and I only had to copy-paste. But here I have to write everything on one line, and it seems that LaTeX doesn't work well here, sometimes there are mistakes, I have to write "{}" all the time. I don't know if I'm the only one with this problems.

**Lisa91** · November 13th 2012, 08:08 AM

Thank you very much, guys. I am wondering whether I am allowed to apply the de L'Hospital rule because we have a sequence, not a function here. I think I need to proof that let's say $ln(n^2+20n+5)$ and $ln(n^9-3n+12)$ are differentiable. Is there anyone who's studying math?

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