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

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

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

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

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

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

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

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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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 $\displaystyle \epsilon>0,\ a_n\to a\ \implies$ there is $\displaystyle N$ such that $\displaystyle |a_n-a|<\epsilon$ for all $\displaystyle n>N.$ The result follows immediately as $\displaystyle \left|(k+a_n)-(k+a)\right| = \left|a_n-a\right|.$
    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 $\displaystyle b_n=k+a_n$ and $\displaystyle b=k+a.$ Then you want that $\displaystyle \forall\,\epsilon>0,$ you can find some $\displaystyle N$ such that $\displaystyle n>N\ \implies\ \left|b_n-b\right|<\epsilon,$ right? That was what I proved.
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