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Thread: prove or disprove... involving sided limits

  1. #1
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    prove or disprove... involving sided limits

    How can I prove the following:
    $\displaystyle \lim_{x\to 0}f(x)=L$ implies $\displaystyle \lim_{x\to \infty}f(\frac{1}{x})=L$
    For that matter, how would I disprove this:
    $\displaystyle \lim_{x\to \infty}f(\frac{1}{x})=L$ implies $\displaystyle \lim_{x\to 0}f(x)=L$
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by dannyboycurtis View Post
    How can I prove the following:
    $\displaystyle \lim_{x\to 0}f(x)=L$ implies $\displaystyle \lim_{x\to \infty}f(\frac{1}{x})=L$
    For that matter, how would I disprove this:
    $\displaystyle \lim_{x\to \infty}f(\frac{1}{x})=L$ implies $\displaystyle \lim_{x\to 0}f(x)=L$
    $\displaystyle \forall~\epsilon>0$, $\displaystyle \exists~\delta>0$ such that $\displaystyle |y-0|=|y|<\delta \implies |f(y)-L|<\epsilon$

    Replace $\displaystyle y$ with $\displaystyle \frac{1}{x}$ and we have $\displaystyle |1/x|<\delta \implies |f(1/x)-L|<\epsilon$.

    So $\displaystyle \forall~\epsilon>0$, $\displaystyle \exists~\delta>0$ such that $\displaystyle x>1/\delta \implies |f(1/x)-L|$.

    This means $\displaystyle \lim_{x\to\infty}f(1/x)=L$.

    -------------

    I cannot think of a counterexample for the second one. I'm sure someone else will, though.
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