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

  • Thread starter Thread starter dalarev
  • Start date Start date
  • Tags Tags
    Infinity Limit
dalarev
Messages
94
Reaction score
0

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.
 
Physics news on Phys.org
divide the numerator and denominator by n^4
 
ircdan said:
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.
 
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 ...
 
ircdan said:
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.
 
\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:
jdg812 said:
\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.
 
dalarev said:
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...
 
dalarev said:
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.
 
  • #10
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.
 
Back
Top