# for each of following sequences state whether it converges and if it does to what limit explain your answer to the below An =(6n-3n^3)/(2n^3+4n^2) (n=1,2,3,...)

## An = 3 n (2- n^2) / [ 2 n^2 (n+2) ]

An = (3/2) (1/n) (-1) (n^2-2) /(n+2)

as n gets big, 2 looks small

An = (-3/2) (1/n) (n^2/n)

An = -3/2 (n^2/n^2)

An = -3/2

## To determine whether the sequence converges and, if it does, to what limit, we need to analyze the behavior of the terms of the sequence as n approaches infinity.

Let's rewrite the given sequence:

An = (6n - 3n^3)/(2n^3 + 4n^2)

Now, let's simplify the expression further by factoring out the highest power of n in the numerator and denominator:

An = (n^3 (6/n^2 - 3))/(n^3 (2/n + 4/n^2))

Simplifying inside the brackets:

An = (6/n^2 - 3)/(2/n + 4/n^2)

As n approaches infinity, we can see that the terms involving 1/n will approach zero:

lim (n→∞) (2/n) = 0

lim (n→∞) (4/n^2) = 0

Therefore, the sequence can be simplified to:

An ≈ (6/n^2)/(0) = undefined

Since the denominator becomes zero as n approaches infinity, the sequence is undefined. This means that the sequence does not converge to any specific limit.

In conclusion, the given sequence (An = (6n - 3n^3)/(2n^3 + 4n^2)) does not converge to any limit as n approaches infinity.