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Math Help - Rigorous Proof on Limits

  1. #1
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    Rigorous Proof on Limits

    I'd someone to check my work:

    Using the definition of the convergence of a sequence, verify that the following converge to the limit that is proposed.

    NOTE, the definition of convergence for a sequence is as follows: A sequence (a_n) converges to a \in \mathbb{R} if, for every pos. number epsilon, there exists an N \in \mathbb{N} such that whenever n >= N, it follows that |a_n - a| < epsilon.

    1.) lim(1/(6n^2 + 1)) = 0

    2.) lim((3n+1)/(2n+5) = 3/2

    3.) lim(2/sqrt(n+3)) = 0

    MY WORK:

    1.) 1/(6n^2 + 1) < 1/(6n^2) < epsilon => n > sqrt(1/(6*epsilon))

    2.) |(3n+1)/(2n+5) - 3/2| = |[2*(3n+1) - 3*(2n+5)]/[2*(2n+5]| = 13/2 * 1/(2n + 5) < 13/4n < epsilon => n > 13/4*epsilon

    3.) 2/sqrt(n+3) < 2/sqrt(n) < epsilon => n > (2/epsilon)^2

    There we go.. my professor doesn't like that I didn't take it back and show |a_n - a| < epsilon or something like that.. not quite sure what he wants.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by seerTneerGevoLI View Post
    I'd someone to check my work:

    Using the definition of the convergence of a sequence, verify that the following converge to the limit that is proposed.

    NOTE, the definition of convergence for a sequence is as follows: A sequence (a_n) converges to a \in \mathbb{R} if, for every pos. number epsilon, there exists an N \in \mathbb{N} such that whenever n >= N, it follows that |a_n - a| < epsilon.

    1.) lim(1/(6n^2 + 1)) = 0

    2.) lim((3n+1)/(2n+5) = 3/2

    3.) lim(2/sqrt(n+3)) = 0

    MY WORK:

    1.) 1/(6n^2 + 1) < 1/(6n^2) < epsilon => n > sqrt(1/(6*epsilon))

    2.) |(3n+1)/(2n+5) - 3/2| = |[2*(3n+1) - 3*(2n+5)]/[2*(2n+5]| = 13/2 * 1/(2n + 5) < 13/4n < epsilon => n > 13/4*epsilon

    3.) 2/sqrt(n+3) < 2/sqrt(n) < epsilon => n > (2/epsilon)^2

    There we go.. my professor doesn't like that I didn't take it back and show |a_n - a| < epsilon or something like that.. not quite sure what he wants.
    here's what he wants. i'll just do the first one.

    Proof: \bold { \lim_{n \to \infty} \frac {1}{6n^2 + 1} = 0 }

    Let \epsilon > 0, choose N \in \mathbb {N} such that N > \sqrt {\frac 1{6 \epsilon}}.

    Then, n \ge N \implies \left| \frac 1{6n^2 + 1} - 0 \right| = \left|  \frac 1{6n^2 + 1}\right| < \left| \frac 1{6n^2} \right| < \left| \frac 1{6N^2}\right| < \left| \frac 1{6 \left( \sqrt { \frac 1{6 \epsilon}} \right)^2 } \right| = \epsilon

    Thus, \lim_{n \to \infty} \frac 1{6n^2 + 1} = 0

    QED


    Do you see the difference between what I did and what you did? Which one of our methods do you think conforms to the definition of the limit of a sequence that was given?
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  3. #3
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    Jhevon, can you look over these responses and tell me if these are fine?

    1-
    Let epsilon> 0 and let N be a natural number such that 1/(6N^2)< epsilon. Then if n>= N we will have:
    |1/(6n^2+1)|< |1/(6n^2)|<= |1/(6N^2)|< epsilon
    And that is what we were looking for.

    2-
    Let epsilon> 0 and let N be a natural number such that 2/sqrt(N)< epsilon. Then if n>= N we will have:
    |2/sqrt(n+3)|< |2/sqrt(n)|<= |2/sqrt(N)|< epsilon
    And that is what we were looking for.

    3-
    Let epsilon>0 and let N be a natural number such that 13/(4N)< epsilon. Then if n>= N we will have:
    |(3n+1)/(2n+5)-3/2|= |-13/(4n+10)|= |13/(4n+10)|< |13/(4n)|<= |13/(4N)|< epsilon
    And that is what we were looking for.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by seerTneerGevoLI View Post
    Jhevon, can you look over these responses and tell me if these are fine?

    1-
    Let epsilon> 0 and let N be a natural number such that 1/(6N^2)< epsilon. Then if n>= N we will have:
    |1/(6n^2+1)|< |1/(6n^2)|<= |1/(6N^2)|< epsilon
    And that is what we were looking for.

    2-
    Let epsilon> 0 and let N be a natural number such that 2/sqrt(N)< epsilon. Then if n>= N we will have:
    |2/sqrt(n+3)|< |2/sqrt(n)|<= |2/sqrt(N)|< epsilon
    And that is what we were looking for.

    3-
    Let epsilon>0 and let N be a natural number such that 13/(4N)< epsilon. Then if n>= N we will have:
    |(3n+1)/(2n+5)-3/2|= |-13/(4n+10)|= |13/(4n+10)|< |13/(4n)|<= |13/(4N)|< epsilon
    And that is what we were looking for.
    it is not really my taste to do these proofs like that, but those are fine. i don't know if i like the line "And that is what we were looking for," but hey, to each his own.

    why did you change the way you chose N?


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