Question

Question #55f2d

Answer 1

`lim_(xrarr2) (x^3-3x^2+4)/(x^3-x^2-8x+12) = color(green)(3/5)`

Explanation:

The problem is that both:
`color(white)("XXX")x^3-3x^2+4` and
`color(white)("XXX")x^3-x^2-8x+12`
are equal to `0` if `x=2`

...but this means that both of these expressions are divisible by `(x-2)`

and using either synthetic or long division we can get:
`color(white)("XXX")(x^3-3x^2+4)/(x^3-x^2-8x+12)=(cancel(""(x-2))(x^2-x-2))/(cancel(""(x-2))(x^2+x-6)`

Unfortunately both
`color(white)("XXX")x^2-x-2` and
`color(white)("XXX")x^2+x-6`
are also both equal to `0` if `x=2`

...but (again) this means that both of these expressions are divisible by `(x-2)`

and we can get
`color(white)("XXX")(x^2-x-2)/(x^2+x-6)=(cancel(""(x-2))(x+1))/(cancel(""(x-2))(x+3))`

So we have
`color(white)("XXX")lim_(xrarr2) (x^3-3x^2+4)/(x^3-x^2-8x+12) =lim_(xrarr2) (x+1)/(x+3)`

`color(white)("XXXXXXXXXXXXXXXXX")=(2+1)/(2+3)=3/5`

Answer 2

I found: `3/5`

Explanation:

Try this: