Limit of a function with absolute value of polynomial in a quotient

[UNknown 2010]

Homework Statement

Find:

Lim | x²+x-12 |-8 / (x-4)
x --> 4

Homework Equations

The Attempt at a Solution

My answer is 9.
It it right ?
or there is not a limit for F(x) when x --> 4

 

[HallsofIvy]

Homework Statement

Find:

Lim | x²+x-12 |-8 / (x-4)
x --> 4

Homework Equations

The Attempt at a Solution

My answer is 9.
It it right ?
or there is not a limit for F(x) when x --> 4

Can we assume you mean $\lim_{x\rightarrow 4} \left(\left|x^2+ x-12\right|- 8\right)/\left(x- 4\right)$?

Close to 4, $x^2+ x- 12$ is close to +8 so the absolute value is not needed.
$x^2+ x- 12- 8= x^2+ x- 20= (x- 4)(x+ 5)$ so $\left(\left|x^2+ x-12\right|- 8\right)/\left(x- 4\right)= (x-4)(x+5)/(x-4)$. That, of course, has limit 4+ 5= 9 at x= 4.

If, however, you meant
$$|x^2+ x- 12|- \frac{8}{x- 4}$$
which what you actually wrote, that has no limit at x= 4.

 

[CompuChip]

Of course, without a calculation or proof, no answer is really right :P
Did you sketch the graph, for instance? How can you see there whether there is a limit or not, as x --> 4?

I think it is 9 too, actually.[/size]

 

[tiny-tim]

Find:

Lim | x²+x-12 |-8 / (x-4)
x --> 4
…
My answer is 9.
It it right ?
or there is not a limit for F(x) when x --> 4

Hi UNknown 2010!

Yes, 9 is right …

I assume it's the | | that's worrying you?

But it makes no difference at x = 4 (beacuse it's nowhere near 0 there).

It would make a difference, and there would be no limit, if it were | x²+x-20 | / (x-4)