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 $\displaystyle x_a \to x$ as $\displaystyle a \to \infty$

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

$\displaystyle |x_a-x| < \epsilon$

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

Now for any $\displaystyle a> N_1$

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

3. Thanks!

So would the other case be when B is 0?

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