# What is the limit of the fraction f(x)/f(x+1), if x tends to infinite and f(x)=x^3+x?

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### 2 Answers

We have f(x) = x^3 + x.

The value of lim x-->inf. [f(x)/f(x +1)]

=> lim x-->inf. [(x^3 + x)/((x + 1)^3 + x +1)]

=> lim x-->inf. [(x^3 + x)/((x^3 + 1 + 3x^2 + 3x + x +1)]

=> lim x-->inf. [(x^3 + x)/((x^3 + 2 + 3x^2 + 4x)]

divide the numerator and denominator by x^3

=> lim x-->inf. [(1 + 1/x^2)/((1 + 3/x + 4/x^2 + 2/x^3)]

For x --> inf. , 1/x --> 0

=> [1 + 0]/[1 + 0 + 0 + 0]

=> 1/1

=> 1

**The required limit is 1.**

We'll calculate the denominator of the fraction:

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

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

We'll combine like terms:

f(x+1) = x^3 + 3x^2 + 4x + 2

We'll evaluate the limit:

lim f(x)/f(x+1) = lim (x^3 + x)/(x^3 + 3x^2 + 4x + 2)

We'll force the factor x^3 at numerator and denominator:

lim f(x)/f(x+1) = lim x^3 (1+1/x^2)/x^3(1+3/x+4/x^2+2/x^3)

Since the following limits approaches to zero if x approaches to infinite, we'll get:

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

lim f(x)/f(x+1) = 1

**The requested limit of the fraction f(x)/f(x+1), if x approaches to infinite, is lim f(x)/f(x+1) = 1.**