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Thread: Finding a constant that satisfies a limit

  1. #1
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    Finding a constant that satisfies a limit

    Hi, I was given the following question, and I am not entirely sure how to approach it.

    Find the constant a such that

    lim x->infinity ((x+a)/(x-a))^x=e


    I have tried applying l'hopitals rule, and substituting x=1/y so that the limit becomes

    lim y->0 ((1+ay)/1-ay))

    However, I keep getting stuck. Guidance would be much appreciated.

    Thank you very much!
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by KelvinScc View Post
    Hi, I was given the following question, and I am not entirely sure how to approach it.

    Find the constant a such that

    lim x->infinity ((x+a)/(x-a))^x=e


    I have tried applying l'hopitals rule, and substituting x=1/y so that the limit becomes

    lim y->0 ((1+ay)/1-ay))

    However, I keep getting stuck. Guidance would be much appreciated.

    Thank you very much!
    Observe that $\displaystyle \dfrac{x+a}{x-a}=\dfrac{x-a+2a}{x-a}=1+\dfrac{2a}{x-a}$

    So now keep in mind the fact that $\displaystyle \lim\limits_{u\to\infty}\left(1+\dfrac{1}{u}\right )^u=e$.

    Messing around with the original limit, we get

    $\displaystyle \begin{aligned}\lim\limits_{x\to\infty}\left(1+\df rac{1}{(x-a)/(2a)}\right)^x &=\lim\limits_{x\to\infty}\left(1+\dfrac{1}{(x-a)/(2a)}\right)^{2a[(x-a)/(2a)]+a}\\ &=\left[\lim\limits_{x\to\infty}\left(1+\dfrac{1}{(x-a)/(2a)}\right)^{(x-a)/(2a)}\right]^{2a}\cdot\lim\limits_{x\to\infty}\left(1+\dfrac{1 }{(x-a)/(2a)}\right)^a\\ &= e^{2a}\cdot 1\end{aligned}$

    Thus, for $\displaystyle \lim\limits_{x\to\infty}\left(\dfrac{x+a}{x-a}\right)^x=e$, we must have $\displaystyle e^{2a}=e\implies a=\frac{1}{2}$.

    Does this make sense?
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  3. #3
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    Thanks!~

    Yes, it makes complete sense. Thank you so much
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