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

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### 2 Answers

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

f(x) = x^3+3x+2.

Therefore f(n+1)/f(n) = {(n+1)^3+3(n+1)+2}/(n^3+3n+2).

f(n+1)/f(n) = {(n^3+3n^+3n+1)+3(n+1)+2}/(3n^3+n+1).

f(n+1)/f(n) = {n^3+3n^2+6n+6)/(n^3+3n+1). We divide each term in the numerator and denominator by n^3 and then take lt as an-> infinity.

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

Therefore Ltn-> infinity f(n+1)/f(n) = ltn-> infinity {1+3/n+6/n^2+6/n^3)}/(1+1/n^2+1n^3) = 1, as terms like 1/n, 1/n^2 and 1/n^3 become zero in limit.

Therefore Lt n--> infinity {(n+1)^3+3(n+1)+2}/(n^3+3n+2) = 1.

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**