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Math Help - Epsilon-Delta Proof

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by 6DOM View Post
    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
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  3. #3
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    Thanks!

    So would the other case be when B is 0?
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  4. #4
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    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
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