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Math Help - Proving that lim(k + a_{n}) = k + lim(a_{n})

  1. #1
    Senior Member chella182's Avatar
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    Proving that lim(k + a_{n}) = k + lim(a_{n})

    Sorry for the naff title, don't really know how else to put it. The question goes...

    Let a_{n} be a sequence tending to a and let k be a real number. Give an \epsilon-N proof that \lim_{n\rightarrow\infty}(k+a_{n})=k+\lim_{n\right  arrow\infty}a_{n}.

    Looks simple enough, but I can't get a handle on it at all.
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by chella182 View Post
    Sorry for the naff title, don't really know how else to put it. The question goes...

    Let a_{n} be a sequence tending to a and let k be a real number. Give an \epsilon-N proof that \lim_{n\rightarrow\infty}(k+a_{n})=k+\lim_{n\right  arrow\infty}a_{n}.

    Looks simple enough, but I can't get a handle on it at all.
    Hi chella182.

    Given \epsilon>0,\ a_n\to a\ \implies there is N such that |a_n-a|<\epsilon for all n>N. The result follows immediately as \left|(k+a_n)-(k+a)\right| = \left|a_n-a\right|.
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  3. #3
    MHF Contributor chisigma's Avatar
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    In general if we have two sequence a_{n} and  b_{n} with...

    \lim_{n \rightarrow \infty} a_{n}=a , \lim_{n \rightarrow \infty} b_{n}=b

    ... and f(x,y) is a continuos function \forall [x,y] \in \mathbb{R}^{2} is...

    \lim_{n \rightarrow \infty} f(a_{n},b_{n}) = \lim_{n \rightarrow \infty} f(a_{n}, b) = \lim_{n \rightarrow \infty} f(a,b_{n}) = f(a,b)

    In your case is f(x, y) = x + y ...

    Kind regards

    \chi \sigma
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  4. #4
    Senior Member chella182's Avatar
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    Quote Originally Posted by TheAbstractionist View Post
    Hi chella182.

    Given \epsilon>0,\ a_n\to a\ \implies there is N such that |a_n-a|<\epsilon for all n>N. The result follows immediately as \left|(k+a_n)-(k+a)\right| = \left|a_n-a\right|.
    Is that it? That the solution? :S
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  5. #5
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by chella182 View Post
    Is that it? That the solution? :S
    Hi chella182.

    Yes. Let b_n=k+a_n and b=k+a. Then you want that \forall\,\epsilon>0, you can find some N such that n>N\ \implies\ \left|b_n-b\right|<\epsilon, right? That was what I proved.
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