# Thread: Epsilon-Delta Proof

1. ## Epsilon-Delta Proof

Q. Xa is a convergent sequence in R^n where Xa->X as a-> infinity. Use epsilon-delta proof to prove that, in two possibly cases, BXa -> BX as a-> infinity (where B is in R).

I tried using the Squeeze Theorem. (Since Xa-X -> 0 as a -> infinity,
B(Xa-X) -> 0 as a -> infinity). But don't know how there are two cases and whether this proof can be expressed as epsilon-delta proof.

Any help would be greatly appreciated.

Thanks.

2. Originally Posted by 6DOM
Q. Xa is a convergent sequence in R^n where Xa->X as a-> infinity. Use epsilon-delta proof to prove that, in two possibly cases, BXa -> BX as a-> infinity (where B is in R).

I tried using the Squeeze Theorem. (Since Xa-X -> 0 as a -> infinity,
B(Xa-X) -> 0 as a -> infinity). But don't know how there are two cases and whether this proof can be expressed as epsilon-delta proof.

Any help would be greatly appreciated.

Thanks.
Since $x_a \to x$ as $a \to \infty$

This means that for every $\epsilon > 0$ there is a $N \in \mathbb{N}$ such that for all $a > N$

$|x_a-x| < \epsilon$

Since this is true for every epsilon set $\epsilon ' =\frac{\epsilon}{B}$ and pick a new $N_1 \in \mathbb{N}$ such that when $a > N_1 \implies$ $|x_a-x|< \epsilon '$

Now for any $a> N_1$

$|Bx_a-Bx|=|B||x_a-x|<|B|\epsilon ' =|B|\frac{\epsilon}{B}=\epsilon$

3. Thanks!

So would the other case be when B is 0?

4. I suppose the cases are when $B=0$ and when $B \neq 0$. The first one is trivial for the second, let $\epsilon$ and $\delta$ be such that if $a> \delta$ then $\Vert x_a - x \Vert < \frac{ \epsilon }{ \vert B \vert }$ (this can be done since $x_a$ converges and $B$ is a constant) then if $a> \delta$ $\Vert Bx_a - Bx \Vert < \epsilon$