Question

Fast limit

Answer 1

\[\lim_{n \rightarrow \infty}n*\ln(\sqrt{n^2+2n+5}-n)\]

Answer 2

did you say fast limit just now?

Answer 3

I will try to work it on my paper.

Answer 4

Okay

Answer 5

I would use a numerical method, but I am fairly sure you aren't permitted to, right?

Answer 6

numerical method ? be more explicit!

Answer 7

Using a numerical method (and wolfram) I get 2.
n=10
http://www.wolframalpha.com/input/?i=%2810%29%C3%97ln%28%E2%88%9A%28%2810%29%5E2%2B2%2810%29%2B5%29-%2810%29%29
n=100
http://www.wolframalpha.com/input/?i=%28100%29%C3%97ln%28%E2%88%9A%28%28100%29%5E2%2B2%28100%29%2B5%29-%28100%29%29
n=1,000
http://www.wolframalpha.com/input/?i=%281000%29%C3%97ln%28%E2%88%9A%28%281000%29%5E2%2B2%281000%29%2B5%29-%281000%29%29
n=10,000
http://www.wolframalpha.com/input/?i=%2810000%29%C3%97ln%28%E2%88%9A%28%2810000%29%5E2%2B2%2810000%29%2B5%29-%2810000%29%29
n=100,000
http://www.wolframalpha.com/input/?i=%28100000%29%C3%97ln%28%E2%88%9A%28%28100000%29%5E2%2B2%28100000%29%2B5%29-%28100000%29%29

Answer 8

numerical approach, it is called.

Answer 9

SORRY BUT This is calculus not matLAB!

Answer 10

yeah-:(

Answer 11

@ganeshie8 @iGreen @eliassaab @Vincent-Lyon.Fr

Answer 12

I will be definitely back to see how has anyone solved this problem step by step, but I myself don't know how to do it-:(

Answer 13

I feel you

Answer 14

@.Sam. @uri @inkyvoyd

Answer 15

i think the inside limit is 2/2=1
after manupilating the expression inside
doing lim(n)*ln(ln(the radical expression -n))

Answer 16

i meant ln(lim()) not ln(ln)*

Answer 17

\[\lim_{n \rightarrow \infty}n*\ln(\sqrt{n^2+2n+5}-n)\]
Inside the logarithm as n becomes larger it approaches: \[\lim_{n \rightarrow \infty}n*\ln(\sqrt{n^2}-n)\] and the terms inside approach 0 causing the term itself to approach -infinity while the other term outisde the logarithm approaches +infinity by itself.

Answer 18

\(\sqrt{n^2+2n+5}=n(1+\dfrac2n + \dfrac{5}{n^2})^{1/2}\)

Developing at the second order yields:
\(n(1+\dfrac{1}{n}-\dfrac{1}{2n^2}+\dfrac{5}{2n^2})\)

simplify then develop \(n\ln(1+\dfrac{2}{n})\) yielding 2 as the limit.

Answer 19

hmm it is approaching -oo?

Answer 20

No, the limit is 2.00

Answer 21

I don't see how that works at all.

Answer 22

i don't how did he achieve the result lol
the second line is about linear approax?

Answer 23

Here's as far as I can get in as similar as I can see\[\lim_{n \rightarrow \infty}n*\ln(\sqrt{n^2+2n+5}-n)\] \[\lim_{n \rightarrow \infty}n*\ln(n\sqrt{1+\frac{2}{n}+\frac{5}{n^2}}-n)\] \[\lim_{n \rightarrow \infty}n*[ \ln(n)+ \ln(\sqrt{1+\frac{2}{n}+\frac{5}{n^2}}-1)]\]

Answer 24

That bottom term I've written is about a step away from looking like infinity - infinity, which is an indeterminant form, so I could see taking it further with L'Hopital's rule

Answer 25

@Kainui
first use second order expansion:

\((1+x)^a=1+ax+\dfrac{a(a-1)}{2!}x^2\;+\;...\)

then use the first order expansion:

\(\ln(1+x)=x\;+\;...\)

Answer 26

Thanks, that makes sense now, but it still leaves me feeling like there might be a better way.

Answer 27

numerical approach, it is, isn't that easier?

Answer 28

Numerical approach can tell you what the answer is and hint to what steps you should take, but it is not a mathematical proof.