# Thread: Wrong limit value, why?

1. ## Wrong limit value, why?

Hello, first of all i might excuse myself for my poor english, i'm Italian and my native language is far from being "english-friendly";

down to the question itself i had a test at university where i was called to solve this limit:

i might also want to specify that we could not use De L'Hopital or other "advanced tools" as we're studying "Sequences' limits" and not "Functions' limits", so we have to basically reduce this to a "collection of known limits" to be able to specify the value it assumes when n->+infinity.

Well, my solution said that the Sequence was convergent to 0, while Wolfram Alpha (as you can see from the image) says that it converges to 1...
How would you guys have approached the solution of said limit?

SkyWolf.

2. If you are allowed to use:

$\displaystyle\lim_{n \to{+}\infty}{\dfrac{\log (1+a_n)}{a_n}}=1\quad (a_n\to 0)$

then, you can substitute the denominator by:

$a_n=\dfrac{n}{n^3+1}$

and you can easily find the limit.

3. Originally Posted by FernandoRevilla
.. and you can easily find the limit.
Yeah... You are perfectly right; darn it ... One just needs the intuition to replace things with the right known limit to solve an exercise with the speed of light...

Muchas gracias,
SkyWolf.

4. Originally Posted by SkyWolf
Muchas gracias,

Benvenuti.

5. Originally Posted by SkyWolf
Hello, first of all i might excuse myself for my poor english, i'm Italian and my native language is far from being "english-friendly";

down to the question itself i had a test at university where i was called to solve this limit:

i might also want to specify that we could not use De L'Hopital or other "advanced tools" as we're studying "Sequences' limits" and not "Functions' limits", so we have to basically reduce this to a "collection of known limits" to be able to specify the value it assumes when n->+infinity.

Well, my solution said that the Sequence was convergent to 0, while Wolfram Alpha (as you can see from the image) says that it converges to 1...
How would you guys have approached the solution of said limit?

SkyWolf.
First You transform the expression as follows...

$\displaystyle \frac{\sqrt{n^{4}+1}-\sqrt{n^{4}-1}}{\ln (1+\frac{n}{1+n^{3}})} = \frac{2}{\ln (1+\frac{n}{1+n^{3}})\ \{\sqrt{n^{4}+1}+\sqrt{n^{4}-1}\}}$ (1)

... so that the problem is to find the limit of the denominator. Then You take into account that is ...

$\displaystyle \ln (1+\frac{n}{1+n^{3}})= \frac{n}{1+n^{3}} - \frac{n^{2}}{2\ (1+n^{3})^{2}} + ...$ (2)

... and...

$\displaystyle \sqrt{n^{4}+1}+\sqrt{n^{4}-1} = n^{2}\ (\sqrt{1+\frac{1}{n^{4}}} + \sqrt{1-\frac{1}{n^{4}}})$ (3)

... so that is...

$\displaystyle \lim_{n \rightarrow \infty} \ln (1+\frac{n}{1+n^{3}})\ \{\sqrt{n^{4}+1}+\sqrt{n^{4}-1}\}} = 2$ (4)

... and the 'goal' is met!...

Kind regards

$\chi$ $\sigma$

P.S. I'm Italian too and also my English is a little 'imperfect' ... never mind!... we have allways the opportunity to improve ourselves!...