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Math Help - another exponential limit

  1. #1
    Junior Member winterwyrm's Avatar
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    another exponential limit

    Thanks so much! The limit as n--> infinity of [1 + 7/n + 1/(n^4)]^n
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by winterwyrm View Post
    Thanks so much! The limit as n--> infinity of [1 + 7/n + 1/(n^4)]^n
    (1+\frac 7n+\frac 1{n^4})^n

    Taking the logarithm :

    n \ln(1+\frac 7n+\frac 1{n^4})

    We know that \lim_{x \to 0} \ln(1+x)=\lim_{x \to 0} x

    Here, \lim_{n \to \infty} \frac 7n+\frac 1{n^4}=0.

    So \lim_{n \to \infty} n \ln(1+\frac 7n+\frac 1{n^4})=\lim_{n \to \infty} n( \frac 7n+\frac 1{n^4})=\lim_{n \to \infty} 7+\frac 1{n^3}=7

    By taking back the exponential (because we took the logarithm) :

    \lim_{n \to \infty}(1+\frac 7n+\frac 1{n^4})^n=\lim_{n \to \infty} e^{n \ln(1+\frac 7n+\frac 1{n^4})}=e^7
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    Junior Member winterwyrm's Avatar
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