Evaluate the limit of the fraction f(n+1)/f(n) if n-->infinite and f(x)= x^3+3x+2.

First, let's write f(n+1) and f(n), substituting x by n+1
and n:

f(n+1) = (n+1)^3 + 3(n+1) +
2

f(n+1) = n^3 + 1 + 3n(n+1) + 3(n+1) +
2

f(n+1) = n^3 + 3(n+1)(n+1) +
3

f(n+1) = n^3 + 3(n+1)^2 +
3

f(n+1) = n^3 + 3n^2 + 6n + 3 +
3

f(n+1) = n^3 + 3n^2 + 6n +
6

f(n) = n^3 + 3n + 2

Now,
we'll evaluate the limit of the ratio:

lim f(n+1)/f(n)  =
lim (n^3 + 3n^2 + 6n + 6)/(n^3 + 3n + 2)

We'll factorize by
n^3:

lim f(n+1)/f(n) = lim n^3(1 + 3/n + 6/n^2 +
6/n^3)/n^3(1 + 3/n^2 + 2/n^3)

We'll simplify and we'll
get:

lim f(n+1)/f(n) = lim (1 + 3/n + 6/n^2 + 6/n^3)/(1 +
3/n^2 + 2/n^3)

Since lim 3/n = 3/infinite =
0

lim 6/n^2 = 6/infinte =
0

lim 6/n^3 = 6/infinte =
0

lim 3/n^2 = 3/infinite =
0

lim 2/n^3 = 2/infinte =
0

we'll get;

lim f(n+1)/f(n) =
lim (1+0+0+0)/(1+0+0)

lim f(n+1)/f(n) = lim
1

lim f(n+1)/f(n) = 1, if n ->
+infinite