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 $N = \text{Max}\left(1, \dfrac{7}{9\epsilon}\right)$?


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

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


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 $\dfrac{2}{3}$

  • $\begingroup$ Do I need the Max() function? How did you take into account the absolute value sign? $\endgroup$ – CodeKingPlusPlus Sep 24 '12 at 5:07
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.