# Thread: Need help understanding/proving formal definition of limits

1. ## Need help understanding/proving formal definition of limits

This is my first analysis class and I'm having trouble understanding how to prove these limits. Here's an example:

$\displaystyle a_{n}=\frac {1}{2n-3}$

Obviously, the sequence converges to 0. I'm just having trouble PROVING this using the formal definition, which states: A sequence $\displaystyle a_{n}$ converges to a real number A iff for each real number $\displaystyle \epsilon \geq 0$ there exists a positive integer n* such that
$\displaystyle |{a_{n}-A}| < \epsilon$ for all n $\displaystyle \geq$ n*.

Any help with this would be appreciated!

2. Just pick $\displaystyle n^* > \frac{{3\varepsilon + 1}}{{2\varepsilon }}$.

3. Originally Posted by paupsers
This is my first analysis class and I'm having trouble understanding how to prove these limits. Here's an example:

$\displaystyle a_{n}=\frac {1}{2n-3}$

Obviously, the sequence converges to 0. I'm just having trouble PROVING this using the formal definition, which states: A sequence $\displaystyle a_{n}$ converges to a real number A iff for each real number $\displaystyle \epsilon \geq 0$ there exists a positive integer n* such that
$\displaystyle |{a_{n}-A}| < \epsilon$ for all n $\displaystyle \geq$ n*.

Any help with this would be appreciated!
Need to show that $\displaystyle \forall~\epsilon>0, \exists~n^*$ such that $\displaystyle \left|\frac{1}{2n-3}-0\right|<\epsilon$.

So let's solve for $\displaystyle n$.
$\displaystyle \left|\frac{1}{2n-3}-0<\epsilon\right| \implies 2n-3>\frac{1}{\epsilon} \implies 2n>\frac{1}{\epsilon}+3 \implies n>\frac{\frac{1}{\epsilon}+3}{2} = \frac{3\epsilon+1}{2\epsilon}$

Therefore, choose $\displaystyle n>n^*=\frac{3\epsilon+1}{2\epsilon}$ and the condition $\displaystyle \left|\frac{1}{2n-3}-0\right|<\epsilon$ will be satisfied.