# 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.