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

1. Sep 1, 2013

Isaac Wiebe

1. The problem statement, all variables and given/known data
lim (3/n)^(2n)
n→ ∞

2. Relevant 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

3. 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: Sep 1, 2013
2. Sep 1, 2013

2n * ln(3/n)
ln(1/x)=?

3. Sep 1, 2013

vela

Staff Emeritus
The Hospital rule doesn't apply here because you don't have an indeterminate form.

4. Sep 1, 2013

Enigman

Woops, didn't read the question carefully,,,I art iDiot

5. Sep 1, 2013

Ray Vickson

Please use parentheses: using standard parsing rules for mathematical expressions, what you wrote means
$$\left(\frac{3}{n}\right)^2 n,$$
but may be you meant
$$\left(\frac{3}{n}\right)^{2 n}.$$
If you meant the latter, write it as (3/n)^(2n).

6. Sep 1, 2013

Seydlitz

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. Sep 2, 2013

vanhees71

It's pretty easy to show that the sequence goes to 0 for $n \rightarrow \infty$, if you consider the logarithm of the expression.