anonymous
  • anonymous
i need help finding the limit of the sequence (1 + 1/n) ^ n using l'hopitals rule
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
the limit as n goes to infinity I assume?
anonymous
  • anonymous
yes sir
anonymous
  • anonymous
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

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anonymous
  • anonymous
you almost did 75% of the problem
anonymous
  • anonymous
now you're ready to apply the l'hopital's rule, that's taking the derivative of both The numerator and denominator
anonymous
  • anonymous
so the denominator would be -1/x^2
anonymous
  • anonymous
you will get: \[\lim_{n \rightarrow \infty} {\ln (1+1/n) \over 1/n}=\lim_{n \rightarrow \infty} { {-1/n^2 \over 1+1/n} \over -1/n^2}=\lim_{n \rightarrow \infty}{1 \over 1+1/n}\]
anonymous
  • anonymous
which is equal to 1 as you can see.
anonymous
  • anonymous
so the top is 1 / 1 + 1/n and then you take the derivate of the inside of ln (1 + 1/n) right
anonymous
  • anonymous
but this is the limit of the function after taking the ln, so the limit of your original function is e^1
anonymous
  • anonymous
yes you're right!!
anonymous
  • anonymous
awesome thanks bud
anonymous
  • anonymous
no problem.. just be a fan ;)
anonymous
  • anonymous
AnwarA you still there I got another tought one its (1 + 1/sqrt(x))^x
anonymous
  • anonymous
x is approaching what?
anonymous
  • anonymous
infinty again
anonymous
  • anonymous
use the same procedure we did with the first one.
anonymous
  • anonymous
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
anonymous
  • anonymous
you should use parentheses.

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