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Math Help - Limit of e

  1. #1
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    Limit of e

    Ok. I know that the lim (n to inf) of the sequence [1+(1/n)]^n=e.
    How would I find the lim (n to inf) of [1+(2/n)]^n using the above limit? I know the answer is e^2, but how would I show it?
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  2. #2
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    Quote Originally Posted by JimmyT View Post
    Ok. I know that the lim (n to inf) of the sequence [1+(1/n)]^n=e.
    How would I find the lim (n to inf) of [1+(2/n)]^n using the above limit? I know the answer is e^2, but how would I show it?
    Hi,
    this one looks like a trick - but it works:

    \left(1+\frac{2}{n}\right)^n = \left(1+\frac{1}{\frac{n}{2}}\right)^n =
    \left( \left(1+\frac{1}{\frac{n}{2}}\right)^\frac{n}{2}\r  ight)^2

    Now substitute n/2 = y. Then you get:


    \left( \left(1+\frac{1}{y}\right)^y\right)^2

    \lim_{y\rightarrow \infty}{\left(1+\frac{1}{y}\right)^y}=e

    Therefore \left( \left(1+\frac{1}{\frac{n}{2}}\right)^\frac{n}{2}\r  ight)^2=e^2

    EB
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  3. #3
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    Quote Originally Posted by JimmyT View Post
    Ok. I know that the lim (n to inf) of the sequence [1+(1/n)]^n=e.
    How would I find the lim (n to inf) of [1+(2/n)]^n using the above limit? I know the answer is e^2, but how would I show it?
    If you want full rigour you will have to fill in the detail yourself but:

    Put m=n/2, then

    f(n)=[1+(2/n)]^n=[1+m]^(2m) = {[1+m]^m}^2

    so

    lim(n->infty) [1+(2/n)]^n = lim(m->infty) {[1+m]^m}^2
    ...................................= {lim(m->infty) [1+m]^m}^2=e^2.

    RonL
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  4. #4
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    Along the same lines, how would you manipulate the lim (n to inf) of
    (3n)^(1/(2n)) to get 1?

    Jim
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    If you want full rigour you will have to fill in the detail yourself but:
    Actually what Earboth did was rigorous. He used the limit composition rule for sequences.
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  6. #6
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    Quote Originally Posted by JimmyT View Post
    Along the same lines, how would you manipulate the lim (n to inf) of
    (3n)^(1/(2n)) to get 1?

    Jim
    I am not going to write infinity because all limits of sequences are taken long infinite.

    You have,
    \lim (3n)^{\frac{1}{2n}}=\lim 3^{\frac{1}{2n}}n^{\frac{1}{2n}}=\lim \sqrt{3}^{1/n}\cdot \sqrt{n}^{1/n}
    We divide the problem into two limit problems,

    1) \lim \sqrt{3}^{1/n}=\sqrt{3}^0=1 (sequence composition rule).

    2) 1\leq \sqrt{n}^{1/n}\leq n^{1/n}
    Squeeze theorem, the limit,
    \lim n^{1/n}=1 is well known.

    Thus the answer is 1\cdot 1 =1
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