Question

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

Answer 1

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

Explanation:

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.