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anonymous

  • 5 years ago

i need help finding the limit of the sequence (1 + 1/n) ^ n using l'hopitals rule

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  1. anonymous
    • 5 years ago
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    the limit as n goes to infinity I assume?

  2. anonymous
    • 5 years ago
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    yes sir

  3. anonymous
    • 5 years ago
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    I understand taking the ln of it so you get n ln (1+1/n) then you have ln(1+1/n) / 1/n but then i am lost

  4. anonymous
    • 5 years ago
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    you almost did 75% of the problem

  5. anonymous
    • 5 years ago
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    now you're ready to apply the l'hopital's rule, that's taking the derivative of both The numerator and denominator

  6. anonymous
    • 5 years ago
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    so the denominator would be -1/x^2

  7. anonymous
    • 5 years ago
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    you will get: \[\lim_{n \rightarrow \infty} {\ln (1+1/n) \over 1/n}=\lim_{n \rightarrow \infty} { {-1/n^2 \over 1+1/n} \over -1/n^2}=\lim_{n \rightarrow \infty}{1 \over 1+1/n}\]

  8. anonymous
    • 5 years ago
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    which is equal to 1 as you can see.

  9. anonymous
    • 5 years ago
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    so the top is 1 / 1 + 1/n and then you take the derivate of the inside of ln (1 + 1/n) right

  10. anonymous
    • 5 years ago
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    but this is the limit of the function after taking the ln, so the limit of your original function is e^1

  11. anonymous
    • 5 years ago
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    yes you're right!!

  12. anonymous
    • 5 years ago
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    awesome thanks bud

  13. anonymous
    • 5 years ago
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    no problem.. just be a fan ;)

  14. anonymous
    • 5 years ago
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    AnwarA you still there I got another tought one its (1 + 1/sqrt(x))^x

  15. anonymous
    • 5 years ago
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    x is approaching what?

  16. anonymous
    • 5 years ago
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    infinty again

  17. anonymous
    • 5 years ago
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    use the same procedure we did with the first one.

  18. anonymous
    • 5 years ago
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    ok so I got to that part I got ( -1/2x ^ 3/2)/ 1 + 1/sqrt(x) / -1/x^2 but it doesnt look good to simplify to me

  19. anonymous
    • 5 years ago
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    you should use parentheses.

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