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Infinite Limit question (3/x)^2x, proof it goes to 0

  • #1

Homework Statement


lim (3/n)^(2n)
n→ ∞


Homework Equations



L'hopital's rule: lim F(a)/G(a) is indeterminate form, then the limit can be written as lim F'(a)/G'(a)
x → a x→ a



The Attempt at a Solution



lim e^[ln(3/n)^(2n)] = lim e^(2n * ln(3/n))
n→ ∞ n → ∞

2n * ln(3/n) =

[2 * ln(3/n)] / (n^-1 )

and applying L.h., [2 * (n/3) * (-3/n^2)]/(-n^-2)

reduces to, 2 * n, and clearly, plugging in n = infinity is wrong. Where did I go wrong, and what can I do to fix this problem?
 
Last edited:

Answers and Replies

  • #2
662
307
2n * ln(3/n)
ln(1/x)=?
 
  • #3
vela
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The Hospital rule doesn't apply here because you don't have an indeterminate form.
 
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  • #4
662
307
The Hospital rule doesn't apply here because you don't have an indeterminate form.
Woops, didn't read the question carefully,,,I art iDiot
 
  • #5
Ray Vickson
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Homework Statement


lim (3/n)^2n
n→ ∞


Homework Equations



L'hopital's rule: lim F(a)/G(a) is indeterminate form, then the limit can be written as lim F'(a)/G'(a)
x → a x→ a



The Attempt at a Solution



lim e^[ln(3/n)^2n] = lim e^(2n * ln(3/n))
n→ ∞ n → ∞

2n * ln(3/n) = [2 * ln(3/n)] / n^-1 and applying L.h., [2 * (n/3) * (-3/n^2)]/(-n^-2)

reduces to, 2 * n, and clearly, plugging in n = infinity is wrong. Where did I go wrong, and what can I do to fix this problem?
Please use parentheses: using standard parsing rules for mathematical expressions, what you wrote means
[tex] \left(\frac{3}{n}\right)^2 n,[/tex]
but may be you meant
[tex] \left(\frac{3}{n}\right)^{2 n}.[/tex]
If you meant the latter, write it as (3/n)^(2n).
 
  • #6
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4
You can see that the the denominator will increase faster than the nominator when ##n>2##. The limit is positive and is bounded by 0. So you just need to find a function which bounds your sequence and converges to 0 also when the limit goes to infinity. In this case I suggest having a function in form of geometric sequence. If that's so then you can prove that this series reaches 0.
 
  • #7
vanhees71
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It's pretty easy to show that the sequence goes to 0 for [itex]n \rightarrow \infty[/itex], if you consider the logarithm of the expression.
 

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