What is the Limit as n Approaches Infinity of a Rational Function?

[dalarev]

Homework Statement

Find limit as n -> infinity

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

Homework Equations

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

The Attempt at a Solution

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

 

[ircdan]

divide the numerator and denominator by n^4

 

[dalarev]

divide the numerator and denominator by n^4

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.

 

[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 ...

 

[dalarev]

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 ...

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.

 

[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

 

[dalarev]

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

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

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.

 

[jdg812]

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.

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

 

[ircdan]

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.

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.

 

[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.