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Studying for exam, need help

  1. Nov 14, 2004 #1
    Studying for exam, need help!!

    Hi can someone help me out. My girlfriend has an exam tomorrow and she got stuck on this question. Her professor decided not to give out any solutions she's not too sure if she is heading in the right direction. Any help would be great, thank you in advance.

    Here is the question:
    Prove by using the definition of the limit of a sequence that:
    [tex]\lim_{n \to \infty} \frac{n + 1}{n^2} + 3 = 3[/tex]
     
    Last edited: Nov 14, 2004
  2. jcsd
  3. Nov 14, 2004 #2
    split up the fraction

    [tex]
    \frac{n}{n^2} + \frac{1}{n^2}
    [/tex]
     
  4. Nov 15, 2004 #3
    How do I get just one term of n?
     
  5. Nov 15, 2004 #4

    matt grime

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    Science Advisor
    Homework Helper

    You must show that given e>0 there is an N such that n>N implies

    [tex]\frac{n+1}{n^2}<e[/tex]

    agreed?

    Well,


    [tex]\frac{n+1}{n^2}<\frac{n+1}{(n+1)^2} = \frac{1}{n+1}[/tex]


    so pick N such that N+1>1/e
     
  6. Nov 15, 2004 #5
    Second question, if you could get back to me asap, we're at school cramming right now
    Let[tex]a_n = \frac{n^2-1}{2n^2+3}[/tex] Prove by using the definition of the limit of a sequence that [tex]\lim_{n \to \infty}a_n = \frac{1}{2}[/tex]
     
  7. Nov 15, 2004 #6

    matt grime

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    Science Advisor
    Homework Helper

    Well, have you simplified a_n -1/2?

    every question like this reduces to showing something tends to zero. that thing tends to zero for obvious reasons just like the previous example
     
  8. Nov 15, 2004 #7
    I just have [tex]|\frac{n^2-1}{2n^2+3} - \frac{1}{2}|[/tex] and I don't know how to get it to a single n term.
     
  9. Nov 15, 2004 #8
    Thanks for your help anyway. We gotta go now.
     
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