# Evaluate the limit of f(x)-x, if x goes to -infinite and f(x)=(x^3+3x)/(x^2+1)?

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### 1 Answer

We'll calculate the expression f(x) - x:

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

f(x) - x = [(x^3+3x) - x*(x^2+1)]/(x^2+1)

We'll remove the brackets:

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

We'll eliminate like terms inside brackets from numerator;

f(x) - x = (2x)/(x^2+1)

Now, we'll evaluate the limit, if x approaches to -infinite:

lim [f(x) - x] = lim (2x)/(x^2+1)

We'll force the factor x^2, at denominator:

lim (2x)/(x^2+1) = lim 2x/x^2*(1 + 1/x^2)

We'll simplify:

lim 2/x*(1 + 1/x^2) = lim (2/x)*lim [1/(1 + 1/x^2)]

Since the limit of 2/x approaches to zero, if x approaches to - infinite and the limit of 1/x^2 approaches to zero, if x approaches to - infinite, we'll get:

lim (2/x)*lim [1/(1 + 1/x^2)] = 0*1/(1+0) = 0

**The requested limit of the difference f(x) - x, if x approaches to -infinite is lim [f(x) - x] = 0.**