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I need to prove by definition that the limit of sequence :

$$a_n = \frac{5n^3-3n^2+1}{4n^3+n+2}$$

is $\dfrac54$ which means I need to show that :

$$\left|\frac{5n^3-3n^2+1}{4n^3+n+2} - \frac54\right| < \varepsilon$$

I tried for hours to solve it but could not, Can anyone help please?

Thanks

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  • $\begingroup$ This is false. The series diverges. I think there is a typo. Either you mean $5n^2$ or $4n^3$. $\endgroup$ – Cornman Nov 5 '17 at 16:31
  • $\begingroup$ You are right, fixed the equation $\endgroup$ – kal pola Nov 5 '17 at 16:38
  • $\begingroup$ Recall that a sequence converges when $\forall \epsilon > 0$ there is some N such that the limit holds for all n > N. Try to find a series of inequalities that let you seperate n so that you can find an n in terms of epsilon. Sort of like you would with the limit of a regular function when doing delta-epsilon proofs. $\endgroup$ – InsigMath Nov 5 '17 at 16:40
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Let $\varepsilon > 0$. Pick $n_0 \in \mathbb{N}$ such that $n_0 > \frac{23}{16\varepsilon}$.

For $n \ge n_0$ we have: \begin{align} \left|\frac{5n^3-3n^2+1}{4n^3+n+2} - \frac54\right| &= \left|\frac{4(5n^3-3n^2+1)-5(4n^3+n+2)}{4(4n^3+n+2)}\right|\\ &= \left|\frac{-12n^2-5n-6}{4(4n^3+n+2)}\right|\\ &= \frac{12n^2+5n+6}{4(4n^3+n+2)}\\ &\le \frac{12n^2+5n^2+6n^2}{4(4n^3 + 0 + 0)}\\ &= \frac{23n^2}{16n^3}\\ &= \frac{23}{16}\cdot\frac1n\\ &< \varepsilon \end{align}

Therefore $$\lim_{n\to\infty} a_n = \frac54$$

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$|f(n)|:=|\dfrac{5n^3-3n^2+1}{4n^3+n+2} -5/4| =$

$|\dfrac{-12n^2 -5n-6}{4(4n^3+n+2)}| \lt$

$|\dfrac{12n^2 +5n+6}{16n^3}| \lt$

$|\dfrac{12n^2+5n^2+6n^2}{n^3}| =$

$|\dfrac{23n^2}{n^3}| = |\dfrac{23}{n}|.$

Let $\epsilon >0$ be given:

Choose $M \gt 23/\epsilon,$ with $M$, real.

There is a $n_0 \gt M$.(Archimedes)

Then for $n \ge n_0:$

$|f(n)| \lt 23/n \le 23/n_0 \lt 23/M $

$\lt \epsilon.$

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