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

Hence:

To find the limit of (sqrt(x+6)-x)/(x^3-3x^2) as x approaches negative infinity, we can use the concept of limits.

First, let's simplify the expression by multiplying both the numerator and denominator by the conjugate of the numerator, which is (sqrt(x+6)+x). This will help us eliminate the square root in the numerator.

After simplifying, we get: (sqrt(x+6)-x)/(x^3-3x^2) * (sqrt(x+6)+x)/(sqrt(x+6)+x)

This simplifies further to: (x+6 - x^2)/(x^3-3x^2)

Now, as x approaches negative infinity, we can observe that the highest power of x in the denominator is x^3, and the highest power of x in the numerator is x^2.

Since the degree of the numerator is less than the degree of the denominator, the limit as x approaches negative infinity will be 0.

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