How do you find the limit of # (x^2 +6) / (6x^3 +x^2 -1)# as x approaches #-oo#?

Question

Emilia Aguilar · Accepted Answer

#lim_(x->-oo) frac (x^2+6) (6x^3+x^2-1) = 0#

As the denominator is of higher grade in #x#,

#lim_(x->-oo) frac (x^2+6) (6x^3+x^2-1) = 0#

We can see that by rewriting the expression as such:

#frac (x^2+6) (6x^3+x^2-1) = frac (x^2) (6x^3+x^2-1) + frac 6 (6x^3+x^2-1) = frac 1 (frac (6x^3+x^2-1) (x^2))+ frac 6 (6x^3+x^2-1) = frac 1 (6x+1-1/x^2) + frac 6 (6x^3+x^2-1) #

For both terms of the sum, the numerator is finite, while the denominator grows in absolute value as #x->-oo#, so that the quotient decreases.

Elijah Abram · Answer

undefined To find the limit of (x^2 + 6) / (6x^3 + x^2 - 1) as x approaches negative infinity, we can analyze the highest power terms in the numerator and denominator. In this case, the highest power term in the numerator is x^2 and the highest power term in the denominator is 6x^3.

As x approaches negative infinity, the term 6x^3 dominates the expression. Therefore, we can simplify the expression by dividing both the numerator and denominator by x^3.

(x^2 + 6) / (6x^3 + x^2 - 1) simplifies to (1/x + 6/x^3) / (6 + 1/x - 1/x^3).

As x approaches negative infinity, 1/x and 1/x^3 both approach 0. Thus, the expression further simplifies to 0/6, which is equal to 0.

Therefore, the limit of (x^2 + 6) / (6x^3 + x^2 - 1) as x approaches negative infinity is 0.