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Prove limit by definition $\lim_{n \to \infty}\frac{2n^2-3n+1}{3n+5} = \infty$

The limit is $\infty$ by the definition if for all $M>0$ we can find an $N$ such for all $n>N$, $An>M$

Any suggestions for how to get $N$?

EDIT: I would like to know if can I isolate $n$ in inequality and choose $N$ by it. I mean from this inequality (after long polynomials division):

$\frac{2}{3}n-\frac{19}{9}+\frac{104}{9(3n+5)} > M$

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    $\begingroup$ Do long division on the polynomials. $\endgroup$
    – B. Goddard
    Apr 2, 2018 at 12:23

3 Answers 3

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Young's inequality is very useful to find exact bound for polynomials. The inequality is the following $$ a b \leq \frac{a^p}{p}+\frac{b^q}{q}, \quad a,b\geq 0, \frac{1}{p}+\frac{1}{q}=1. $$ Therefore we can estimate the upper polynomial $$ 2 n^2 -3 n + 1 \geq 2 n^2 - \left( \frac{3^2}{2} + \frac{n^2}{2} \right) +1 = \frac{3}{2}n^2 - \frac{7}{2}. $$ About the lower polynomial, if $n \geq 5$ $$ 3 n+5 \leq 4n. $$ Therefore $$ p(n) = \frac{2n^2-3n+1}{3n+5} \geq \frac{3n^2-7}{8n} = \frac{3n}{8} - \frac{7}{8n} \geq \frac{3n}{8}-1. $$ So fix $M>0$, if $$ \frac{3n}{8}-1 > M \quad \iff \quad n > (M-1)\frac{8}{3}, $$ then $p(n)>M$.

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  • $\begingroup$ +1: Nice use of (young-inequality) $\endgroup$ Apr 2, 2018 at 12:35
  • $\begingroup$ Great idea, but I have to do it by isolate $n$ in the inequality, and choose $N$ by it $\endgroup$
    – Oran Sherf
    Apr 2, 2018 at 13:03
  • $\begingroup$ Your $N$ will be $N = \inf_{n\in \mathbb{N}} \{n\geq (M-1)\frac{8}{3} \} = \lceil (M-1)\frac{8}{3} \rceil $. $\endgroup$ Apr 2, 2018 at 14:48
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  • For $n\ge 5$ then $3n+5\le 4n$
  • For $n\ge 3$ then $n(n-3)\ge 0\implies 2n^2-3n\ge n^2$

Thus for $n\ge 5$ we have $$\dfrac{2n^2-3n+1}{3n+5}\ge \dfrac{n^2}{4n}\ge \dfrac n4\to+\infty$$

Do not hesitate to simplify expressions with rough inequalities when you want to exhibit limits.

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Yes you can $$A_n=\frac{2n^2-3n+1}{3n+5}= \frac{2n}{3}+ \frac{1-\frac{19n}{3}}{3n+5}$$ $$= \frac{2n}{3}+ \frac{\frac{1}{n}-\frac{19}{3}}{3+\frac{5}{n}} > \frac{2n}{3}+\frac{-19/3}{3+5}>\frac{2n}{3}-1$$ Thus, $A_n>\frac{2n}{3}-1$

Now, for any $M>0$, chose $N=3(M+1)$ so that for any $n>N$, $$A_n>\frac{2}{3}n -1 >\frac{2}{3}\cdot 3(M+1) -1 = 2M+2-1>M$$ Thus, $\lim_{n\to\infty}A_n = \infty$

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