show that..........

bgbgbgbg · Hodge Conjecture Offline Joined: 17 Sep 2009 Posts: 83

show that..........
lim[x^3+1]/[3x^3+x] =1/3 as x goes to infinity by definition of limit

**Posted:** Sun Nov 01, 2009 3:32 am

centry57 · P versus NP Offline Joined: 03 Jul 2009 Posts: 40

$\lim_{x\to \infty} \frac{x^{3}+1}{3x^{3}+x}=\lim_{x\to \infty} \frac{x^{-3}+1}{x^{-2}+3}=\frac{1}{3}$

$\lim_{x\to \infty} \frac{1}{x^2}=0$

$\lim_{x\to \infty} \frac{1}{x}=0$

**Posted:** Sun Nov 01, 2009 4:15 am

bgbgbgbg · Hodge Conjecture Offline Joined: 17 Sep 2009 Posts: 83

please i want the proof by dwfinition

**Posted:** Sun Nov 01, 2009 5:00 am

J.Y.Choi

Let's suppose there exists $M > 0$ for all $\varepsilon > 0$ such that $\varepsilon = \frac {1}{3M^3}$ .

When $x > M$ , $\left|\frac {x^3 + 1}{3x^3 + x} - \frac {1}{3}\right| < \left|\frac {x^3 + 1}{3x^3} - \frac {1}{3}\right| = \frac {1}{3x^3...$ . By definition of the limit, $\lim_{x\rightarrow{\infty}}\frac {x^3 + 1}{3x^3 + x} = \frac {1}{3}$ .

**Posted:** Sun Nov 01, 2009 7:41 am

bgbgbgbg · Hodge Conjecture Offline Joined: 17 Sep 2009 Posts: 83

thanksssssssssssssssssssssss

**Posted:** Sun Nov 01, 2009 11:26 am