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Thread: is f continuous

  1. #1
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    is f continuous

    If f is define as $\displaystyle f(x)=\displaystyle \lim_{n \to \infty} \frac{1}{n} \log(x^{n}+e^{n})$ then is f continuous.
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  2. #2
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    Is $\displaystyle n\in\mathbb{Z}?$
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  3. #3
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    hi

    n \in \mathbb{N}
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  4. #4
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    Thanks for the reply. I'm just trying to wrap my mind around the domain of the function. You have an $\displaystyle \infty/\infty$ situation there. l'Hospital's Rule applied once gives you

    $\displaystyle \displaystyle{f(x)=\lim_{n\to\infty}\dfrac{e^{n}+x ^{n}\ln(x)}{e^{n}+x^{n}}}.$

    If you divide everything through by $\displaystyle e^{n}$, you could simplify as follows:

    $\displaystyle \displaystyle{f(x)=\lim_{n\to\infty}\dfrac{1+(\fra c{x}{e})^{n}\ln(x)}{1+(\frac{x}{e})^{n}}}.$

    Now, the domain we can see to be $\displaystyle (0,\infty).$

    The behavior of this function will change depending on whether $\displaystyle x\le e$ or not. For $\displaystyle 0<x<e,$ the function is going to return 1. For $\displaystyle x=e,$ the function will return 1. For $\displaystyle x>e,$ I think you could apply l'Hospital's rule again.

    That's one approach.
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  5. #5
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    So what do u conclude from there. Is f continuous and 1 always
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  6. #6
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    No, I don't think so. It's 1 for the interval (0,e]. But if you apply l'Hospital's rule once more, I think you'll find the behavior changes considerably for x>e.
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  7. #7
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    I think no need to apply Hospital's a second time.
    If we look at the second form provided by Ackbeet:
    if $\displaystyle x<e$, we get 1 which is continuous
    if $\displaystyle x=e$, we get $\displaystyle \frac{1+ln(x)}{2}$ also continuous.
    if $\displaystyle x>e$, we get: $\displaystyle \displaystyle{f(x)=\lim_{n\to\infty}\dfrac{1+(\fra c{x}{e})^{n}\ln(x)}{1+(\frac{x}{e})^{n}} = \lim_{n\to\infty}\dfrac{(\frac{x}{e})^{n}\ln(x)}{( \frac{x}{e})^{n}} = ln(x)} $. which is clearly continuous.

    hope this helps
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