1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Proving a simple limit

  1. Oct 21, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove the limit by using the ε criteria
    [itex]\stackrel{lim}{n→∞}[/itex][itex]\frac{n^2+1}{n+3}[/itex] = +∞

    2. Relevant equations

    3. The attempt at a solution
    Hi guys.So i'm having trouble proving this limits(all limits that i need to prove that they equal +/- ∞) , all that i've learned in calculus so far at school is these e proprieties

    limit when x→∞ from xn=L ⇔ [itex]\forall[/itex]ε>0 [itex]\exists[/itex]nε so that for [itex]\forall[/itex] n≥nε we have |xn-L|<ε

    the second one is when limit whenx→∞ from xn= +∞ it's the same except the last part where you don't have |xn-L|<ε but we have xn≥ε

    fot the third one limit when x→∞ from xn= -∞ when xn≤-ε

    Last edited by a moderator: Oct 21, 2012
  2. jcsd
  3. Oct 21, 2012 #2


    Staff: Mentor

    You're not using the correct form for this type of limit. This is the one you want.
    $$ \forall M, \exists N \text{ such that } n \ge N \Rightarrow f(n) > M$$

    The idea here is that M and N are large numbers. You don't use ##\epsilon## in this type of limit proof.
  4. Oct 21, 2012 #3
    well what you said is the same with what i said "the second one is when limit whenx→∞ from xn= +∞ it's the same except the last part where you don't have |xn-L|<ε but we have xn≥ε
  5. Oct 21, 2012 #4


    Staff: Mentor

    Which means that it isn't the same. Also, ##\epsilon## is generally considered to be a small number (close to 0).
  6. Oct 22, 2012 #5
    Well yes but i tried solving it like that and i get lost in all that math , i have a polynominal equation in n....can't you show me?
  7. Oct 22, 2012 #6


    Staff: Mentor

    The way PF works is that we help you, but we don't do your work for you. The polynomial is a quadratic, and you should already know how to solve quadratic equations and inequalities. Show us where you got stuck.
  8. Oct 22, 2012 #7
    Ok here we go , so starting backwards from the theorem we have (n2+1)/(n+3)≥ε after solving this and by arranging it you have


    from here


    so it from here where i m not certain what to do i tried to put the condition that Δ≥0 but by solving that i get really ugly results.I don't know what to do from here!!
  9. Oct 22, 2012 #8


    User Avatar
    Homework Helper

    You want :

    [itex]\forall M > 0, \exists N > 0 \space | \space n>N \Rightarrow a_n > M[/itex]

    So start by massaging what you want into a suitable form :

    [itex]\frac{n^2 + 1}{n + 3} > M[/itex]
  10. Oct 22, 2012 #9


    Staff: Mentor

    You should not use ε for this type of limit, as it is almost universally used to represent numbers that are close to 0.

    From the inequality $$\frac{n^2+1}{n + 3} > M$$
    you get
    ##n^2 - Mn + 1 - 3M > 0##
    The values of n that make the expression above equal to 0 are
    $$n = \frac{M \pm \sqrt{M^2 + 12M - 4}}{2}$$

    It is not necessary to get an exact value for n. All that is needed is to find a value that is large enough. ##\sqrt{M^2 + 12M - 4}## ≈ M, but the square root expression is always a little larger. I think we can say that the radical is smaller than 2M as long as M is not too small. For example, if M = 10 (which isn't very large), ##\sqrt{M^2 + 12M - 4}=\sqrt{216}≈11 < 20 = 2M##. For any larger value of M the radical will be smaller than 2M.

    Can you carry on from here?
  11. Oct 22, 2012 #10
    Not really , the thing is i don't really have a solid understanding about what am I after to prove here.So we've proved that that radical will be smaller than 2M when M is a large number, but hor does that help me with my initial inequation ?I'm sorry i know i'm pretty much asking to solveit step by step to me , but it 's my first kind of this exercise from my first calculus homework ever , and i really want to have a solid foundation on calculus.
  12. Oct 22, 2012 #11


    User Avatar
    Science Advisor
    Homework Helper

    You can make life a lot easier by simplifying the problem first. [itex]\frac{n^2 + 1}{n + 3} > \frac{n^2}{n + 3} > \frac{n^2}{n + n}[/itex] (for n>3). Simplifying the last expression and showing its limit is infinity shows the first one has that limit as well.
  13. Oct 22, 2012 #12
    Oh!I get it now! And it's obvious that n2/2n's limit is +∞ so from that i can conclude that the limit of my equation is also +∞.This is very very useful , and i can use it to solve all my exercises of this type.Thanks a lot! But is it a valid method? Seems more of an observation than a demonstration to me :smile:
  14. Oct 22, 2012 #13
    For example here is how i could solve another exercise which looks similar

    Prove that limn→∞[itex]\frac{n3+2n}{n2+3n-1}[/itex]=+∞

    So [itex]\frac{n3+2n}{n2+3n-1}[/itex] < [itex]\frac{n3+2n}{n2+3n}[/itex]
    But [itex]\frac{n3+2n}{n2+3n}[/itex]=[itex]\frac{n(n2)+2}{n(n+3)}[/itex]=[itex]\frac{n2+2}{n+3}[/itex], now here 2 and 3 are pretty irrelevant when n→∞ so we coud say [itex]\frac{n2}{n}[/itex]=n
    limn→∞ n=+∞, that is obvious so from this i proved that limn→∞[itex]\frac{n3+2n}{n2+3n-1}[/itex]=+∞

    What do you say?Is it ok?

    P.S idk why the eq editor didn't work!
  15. Oct 22, 2012 #14


    User Avatar
    Science Advisor
    Homework Helper

    You can make it into a real proof. If n^2/(2n)>M then (n^2+1)/(n+3)>M. You can use the same bound for the simpler expession for the more complicated.
  16. Oct 22, 2012 #15
    Do not use [NOPARSE][/NOPARSE] and [itеx][/itеx] together, if you want to write supscripts, then please write

    Code (Text):

    Also, limits can be written nicely as

    Code (Text):

    [itex]\lim_{n\rightarrow +\infty} n^2[/itex]
    Here is a LaTeX guide:

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook