Question

How do you evaluate the limit `(x^2-9)/(x^3-27)` as x approaches `3`?

Answer 1

`lim_(xrarr3) (x^2-9)/(x^3-27) =color(green)(2/9)`

Explanation:

Two general equations to remember:
`color(white)("XXX")a^2-b^2 = (a-b)(a+b)`
`color(white)("XXX")a^3-b^3=(a-b)(a^2+ab+b^2)`

Therefore
`color(white)("XXX")(x^2-9)/(x^3-27) = ((x-3)(x+3))/((x-3)(x^2+3x+9))`

As long as `x!=3` (which is true as long as `x` is only approaching `3`)
the factors `(x-3)` can be cancelled out of the numerator an denominator.

So
`color(white)("XXX")lim_(xrarr3) (x^2-9)/(x^3-27)`

`color(white)("XXXXXX") = lim_(xrarr3) (x+3)/(x^2+3x+9)`

`color(white)("XXXXXX")=(3+3)/(9+9+9)`

`color(white)("XXXXXX")=6/27`

`color(white)("XXXXXX")=2/9`