# Thread: Limit Proof

1. ## Limit Proof

A sequence (sn) is said to converge to the real number s provided that for each ε > 0 there exists a real number N such that for all n that are natural numbers, n > N implies that lsn - sl < ε. If (sn) converges to s, then s is called the limit of the sequence (sn). If a sequence does not converge to a real number, it is said to diverge.

Using only the above definition, prove that
lim (3n + 1)/(n + 2) = 3.

2. Originally Posted by bearej50
A sequence (sn) is said to converge to the real number s provided that for each ε > 0 there exists a real number N such that for all n that are natural numbers, n > N implies that lsn - sl < ε. If (sn) converges to s, then s is called the limit of the sequence (sn). If a sequence does not converge to a real number, it is said to diverge.

Using only the above definition, prove that
lim (3n + 1)/(n + 2) = 3.
by the definition, you need to show that there exists an $N$ such that $\Bigg| \frac {3n + 1}{n + 2} - 3 \Bigg| < \epsilon$ for some $\epsilon > 0$ and $n > N$

can you finish?

3. Not some epsilon. For any epsilon. No?

4. Originally Posted by mr. frege
Not some epsilon. For any epsilon. No?
yes

5. I am still having some difficulty constructing this proof...

6. 3n+1/n+2 - 3 is equal to 3n+1/n+2 - 3(n+2)/(n+2)=-5/n+2. So now since 1/n+2 converges, you're set.

7. $\left| \frac{3n + 1}{n + 2} - 3 \right| = \left| \frac{-5}{n+2}\right| = \frac{5}{n+2} \leq \frac{5}{n}$

To make $\frac{5}{n} < \varepsilon$ you need to choose $n > \frac{5}{\varepsilon}$ so take $N = \frac{5}{\varepsilon}$.