Given the sequence x(n+1)=((xn)^5+3xn)/4, show limit x(n+2)/xn=9/16, x go to inff?

subscripts n,n+1,n+2

### 1 Answer | Add Yours

`x_(n+2)=((x_(n+1))^5+3x_(n+1))/4`

`x_(n+2)=((((x_n)^5+3x_n)/4)^5+3(((x-n)^5+3x_n)/4))/4`

`x_(n+2)=((((x_n)^5+3x_n)/4)^5+(3(x_n)^5+9x_n)/4)/4`

`x_(n+2)=(((x_n)^5+3x_n)^5/4^5+(4^4(3(x_n)^5+9x_n))/4^5)/4`

`x_(n+2)=(((x_n)^5+3x_n)^5+4^4(3(x_n)^5+9x_n))/4^6`

`x_(n+2)/x_n=(((x_n)^5+3x_n)^5+4^4(3(x_n)^5+9x_n))/(4^6x_n)`

The limit of this function is 9/16 as `x_n-gt0` , not as `x_n-gtoo` . As `x_n-gtoo` this function's limit is `oo` :

`x_(n+2)/x_n=(((0)^5+3(0))^5+4^4(3((0)^5+9x_n)))/(4^6x_n)`

`x_(n+2)/x_n=4^4(9x_n)/(4^6x_n)=9/16`

**Sources:**

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes