Prove from definition of convergence that (-2n+5)/(3n+1) is convergent.

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Let $\epsilon>0$. We have

$$\left|\frac{-2n+5}{3n+1}+\frac23\right|=\frac{17}{9n+3}<\epsilon\iff n>\frac19\left(\frac{17}{\epsilon}-3\right)=:\alpha$$ so for $n_0=\max(0,\lfloor\alpha\rfloor+1)$ we have for $n\ge n_0$ $$\left|\frac{-2n+5}{3n+1}+\frac23\right|<\epsilon$$ hence we proved by definition that the limit is $-\frac23$.

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Skam
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Skam

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Updated on June 26, 2020

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  • Skam
    Skam over 3 years

    Prove directly from the definition of convergence that $\frac{-2n+5}{3n+1}$ is convergent.

    So I let $\epsilon > 0$ there exists $N \in \mathbb{N}$ such that $n \geq to N$ implies that the $\left | \frac{-2n+5}{3n+1}-0 \right | \leq \frac{-2n+5n}{3n} = 1 \leq \epsilon $

    But, I am looking to set $N$ equal to some number, so $1 \leq \epsilon$ isn't an ideal solution! Can anyone please help?

    • Admin
      Admin about 9 years
      Hint: the limit is $-\frac23$.
    • Admin
      Admin about 9 years
      You're right, I should've known that. Thank you.
    • Admin
      Admin about 9 years
      There is a trick: when $n$ is large, $$\frac{-2n+5}{3n+1}\approx\frac{-2n}{3n}=-\frac23.$$