# Find limit as n -> infinity

1. Jan 17, 2008

### dalarev

1. The problem statement, all variables and given/known data

Find limit as n -> infinity

[ (n+1)^2 ] / [ $$\sqrt{}3+5n^2+4n^4$$ ]

2. Relevant equations

L'Hopital won't do the job, I think.

3. The attempt at a solution

It's something really small I'm just completely missing.

2. Jan 17, 2008

### ircdan

divide the numerator and denominator by n^4

3. Jan 17, 2008

### dalarev

I found an example that said exactly that, but I guess I don't see clearly how, first of all,

the radical / n^4 comes out pretty.

4. Jan 17, 2008

### ircdan

you have, (n+1)^2/(sqrt(3) + 5n^2 + 4n^4), or

(n^2 + 2n + 1)/(sqrt(3) + 5n^2 + 4n^4), so dividing num and denom by n^4,

(1/n^2 + 2/n^3 + 1/n^4)/(sqrt(3)/n^4 + 5/n^2 + 4) and now the limit as n->inf is ...

5. Jan 17, 2008

### dalarev

Oh, I'm sorry. In the denominator, everything is actually under the radical. It doesn't matter, however, because the numerator comes out to zero. Thanks for the help.

6. Jan 17, 2008

### jdg812

$$\sqrt{3}+5n^2+4n^4$$ OR $$\sqrt{3+5n^2+4n^4}$$ ?

if the last one, then divide num and denom by n^2

Last edited: Jan 17, 2008
7. Jan 17, 2008

### dalarev

It's the last one, everything under the radical.

My problem is I'm not seeing how I would divide something like

$$\sqrt{3+5n^2+4n^4}$$ / n^4, or / any number, for that matter. I'm not seeing how to simplify that radical into individual terms.

8. Jan 17, 2008

### jdg812

You should not simplify the radical, just put n^2 INSIDE the radical... and remember that n^2 becomes n^4 when inside radical...

9. Jan 17, 2008

### ircdan

figured this would be your difficulty, here is an example
sqrt(n^2 + 2n)/n^2 = sqrt(n^2 + 2n)/sqrt(n^4) = sqrt((n^2 + 2n)/n^4) = sqrt(1/n^2 + 2/n^3)

try to see why it works, now mimic it for your problem.

10. Jan 17, 2008

### dalarev

Ahh, I see it. I forgot about that property where you're allowed to simply "take out" multiplying/dividing terms. I see it now, thanks a bunch.