sequence limit with epsilon proof

Discussion in 'Calculus' started by Fifi, Nov 27, 2021.

  1. Fifi


    Oct 15, 2021
    Likes Received:
    Hi guys!
    I've got problem in solving this:
    - Prove by definition that:
    ln(2+5/n) --> ln2 as n--> infinity.
    Can you help me?
    Fifi, Nov 27, 2021
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  2. Fifi


    Nov 6, 2021
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    The definition of "limit as n goes to infinity f(n)= A| is
    "Given any epsilon> 0 there exist N such that if n>N then |f(n)- A|< epsilon.

    Here f(n)= ln(2+ 5/n) and A= ln(2) so |f(n)- A|= ln(2+ 5/n)- ln(2)= ln(1+ 5/2n). We want ln(1+ 5/2n)< epsilon which, since the logarithm is an increasing function, is the same as 1+ 5/2n< e^epsilon. 5/2n< e^epsilon- 1 and 5/2< n(e^epsilon- 1).

    epsilon is positive so e^epsilon is larger than 1 and so e^epsilon- 1 is positive. Dividing both sides by it n> 5/(2e^epsilon- 2).

    That is, given any epsilon> 0 taking n> N= 5/(2epsilon- 2) guarantees that |f(n)- A|> epsilon (since every step above is reversible we could go backwards from n> N to |f(n)- A|< epsilon).
    Last edited: Dec 5, 2021
    HallsofIvy, Dec 5, 2021
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  3. Fifi

    Country Boy

    Dec 15, 2021
    Likes Received:
    That "0 taking n> N= 5/(2epsilon- 2) guarantees that |f(n)- A|> epsilon" should, of course, have been "0 taking n> N= 5/(2epsilon- 2) guarantees that |f(n)- A|< epsilon".
    Country Boy, Dec 26, 2021
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