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Prove: $\lim\limits_{n \to\infty} a_n =\lim\limits_{n \to\infty} \dfrac{2n-1}{3n+2} = \dfrac{2}{3}$ using the definition of the limit.

This is what I have so far:

  1. Let $\epsilon > 0$ and take $N = \text{Max}\left(1, \dfrac{7}{9\epsilon}\right)$. This is my reasoning:

Solve: $\left|\dfrac{2n-1}{3n+2} - \dfrac{2}{3}\right| < \epsilon$

We get: $\left|\dfrac{-7}{9n+6}\right| < \epsilon$ $\iff$ $\left|\dfrac{-7}{9n+6}\right| < \left|\dfrac{7}{9n}\right| < \epsilon$

Take $n > 1$ so we can drop the absolute value sign and Solve for $n$:

$\dfrac{7}{9\epsilon} < n$

So where would I go from here? Also, does taking $n > 1$ mean we have to do $N = \text{Max}\left(1, \dfrac{7}{9\epsilon}\right)$?

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$\left|\dfrac{2n-3}{3n+2} - \dfrac{2}{3}\right| < \epsilon$

$\implies\left|\dfrac{-13}{9n+6}\right| < \epsilon$ $ \text{ and } $$\left|\dfrac{-13}{9n+6}\right| < \left|\dfrac{13}{9n}\right| < \epsilon$

Thus,

Given $\epsilon\gt 0,$ For $\forall n\gt \dfrac{13}{9\epsilon}, \left|\dfrac{2n-3}{3n+2} - \dfrac{2}{3}\right| < \epsilon $

Hence the limit of $\dfrac{2n-3}{3n+2} $ is $\dfrac{2}{3}$

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  • $\begingroup$ Do I need the Max() function? How did you take into account the absolute value sign? $\endgroup$ – CodeKingPlusPlus Sep 24 '12 at 5:07
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    $\begingroup$ @CodeKingPlusPlus: Since we take $n\in \Bbb N\implies n\geq 1$, thus there is no need of Max() function. $\endgroup$ – Aang Sep 24 '12 at 6:22

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