# Sequence Convergence Proofs vs. Continuity Function Proofs

• Oct 6th 2007, 12:58 PM
tbyou87
Sequence Convergence Proofs vs. Continuity Function Proofs
Definition of converge:
For each epsilon > 0 there exists a number N such that n>N implies
|sn -s|<epsilon.

In this case the N you chose was usually a max {constant, f(epsilon)}.

Definition of continuity
For each epsilon > 0 there exists a delta>0 such that x element dom(f) and
|x-x0| < delta implies |f(x)-f(x0)|< epsilon.

In this case the delta you choose is usually a min{constant, f(epsilon)}

I was just wondering how to think about why for one you need a min and the other a max.

Thanks
• Oct 6th 2007, 01:16 PM
Plato
Frankly, I do not think what you wrote above is correct.
Perhaps you can fill out what you wrote with more detail.
There is no way to know what you mean by “$\displaystyle \left\{ \mbox{constant},f(\varepsilon ) \right\}$.
• Oct 6th 2007, 01:32 PM
tbyou87
Well to prove that 4n^3 + 3n / (n^3 - 6) = 4 you choose an
N = max{2,squareroot(54/epsilon)}

To prove that 2x^2 + 1 is continuous, you choose a
delta = min{1, epsilon/(2(2|x0|+1))}

Hopefully that's more clear.
• Oct 6th 2007, 02:01 PM
Plato
While those exact examples may not work in all cases. However, I do get the idea.

In the case of the sequence, the subscript is approaching infinity.
Therefore, we take the maximum N to insure all conditions hold.

The case of continuity is a bit more abstract. In choosing the $\displaystyle \delta$ we are creating a bound on x, we want to be able to say that $\displaystyle \left| {x - x_0 } \right| < \delta \quad \Rightarrow \quad \left| {f(x) - f\left( {x_0 } \right)} \right| < \varepsilon$.
Knowing that $\displaystyle \delta < 1$ means $\displaystyle x_0 - 1 < x < x_0 + 1$.
This means we can construct bounds on $\displaystyle f(x)$.
So we want $\displaystyle \delta$ to be at most 1 (not more).

I hope that helps somewhat.