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Math Help - Limit Proof

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    Limit Proof

    A sequence (sn) is said to converge to the real number s provided that for each ε > 0 there exists a real number N such that for all n that are natural numbers, n > N implies that lsn - sl < ε. If (sn) converges to s, then s is called the limit of the sequence (sn). If a sequence does not converge to a real number, it is said to diverge.

    Using only the above definition, prove that
    lim (3n + 1)/(n + 2) = 3.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by bearej50 View Post
    A sequence (sn) is said to converge to the real number s provided that for each ε > 0 there exists a real number N such that for all n that are natural numbers, n > N implies that lsn - sl < ε. If (sn) converges to s, then s is called the limit of the sequence (sn). If a sequence does not converge to a real number, it is said to diverge.

    Using only the above definition, prove that
    lim (3n + 1)/(n + 2) = 3.
    by the definition, you need to show that there exists an N such that \Bigg| \frac {3n + 1}{n + 2} - 3 \Bigg| < \epsilon for some \epsilon > 0 and n > N

    can you finish?
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    Not some epsilon. For any epsilon. No?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mr. frege View Post
    Not some epsilon. For any epsilon. No?
    yes
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    I am still having some difficulty constructing this proof...
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  6. #6
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    3n+1/n+2 - 3 is equal to 3n+1/n+2 - 3(n+2)/(n+2)=-5/n+2. So now since 1/n+2 converges, you're set.
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     \left| \frac{3n + 1}{n + 2} -  3 \right| = \left| \frac{-5}{n+2}\right| = \frac{5}{n+2} \leq \frac{5}{n}

    To make \frac{5}{n} < \varepsilon you need to choose n > \frac{5}{\varepsilon} so take N = \frac{5}{\varepsilon}.
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