Question

How do you evaluate the limit `(x^2-x-6)/(x+2)` as x approaches `oo`?

Answer 1

The limit of a sum is the sum of the limits, provided the individual limits exist. Similarly with product, quotient, etc.

Explanation:

Let's manipulate for the expression to a more convenient form `(x^2-x-6)/(x+2)=((x^2-x-6)/x)/((x+2)/x)=((x^2/x-x/x-6/x))/((x/x+2/x))=(x-1-6/x)/(1+2/x)`

Then `lim_(xrarroo) ((x^2-x-6)/(x+2)) = lim_(xrarroo) ((x-1-6/x)/(1+2/x)) = (oo -1 -0)/(1+0)=oo`