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

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:

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

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

1 person
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

Ray Vickson
Homework Helper
Dearly Missed

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
$$\left(\frac{3}{n}\right)^2 n,$$
but may be you meant
$$\left(\frac{3}{n}\right)^{2 n}.$$
If you meant the latter, write it as (3/n)^(2n).

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.

vanhees71
Gold Member
It's pretty easy to show that the sequence goes to 0 for $n \rightarrow \infty$, if you consider the logarithm of the expression.