# How do I prove the following: lim x^2 + x - 12 = 7x-> 3

1. Oct 8, 2005

### powp

How do I prove the following:

lim x^2 + x - 12 = 7
x-> 3 ------------
x-3

here is my solution so far but think it is wrong

Given E > 0 and D > 0 such that

|x^2 + x - 12 |
|------------ - 7| < E whenever 0 < |x-3| < D
| x-3 |

|(x-3)(x+4) |
|------------ - 7| < E whenever 0 < |x-3| < D
| x-3 |

|x + 4 - 7| < E

|x - 3| < E

Not sure where to go from here or if I am even close.

Thanks

Peter

PS I am new to proofs

2. Oct 8, 2005

### Edwin

I'll give it a try, but I've never really taken a formal proofs class, so you might want to double check my results to make sure that it is accurate.

-(x^2 + x - 12)/(x-3) < D < (x^2 +x - 12)/(x - 3)

-(x^2 + x -12) > D(x-3) > (x^2 + x - 12)

let u = D(x-3)

(u/D) = x - 3

(u/D) + 3 = x

So as x approaches 3, (u/D) must approach 0, which would only occur if u approaches 0.

D((u/D+3) - 3) = D(u/D) = u, so the inequality above becomes:

-((u/D + 3)^2 + (u/D +3) -12) > u > ((u/D + 3)^2 + (u/D +3) -12)

-u(u + 7*D)/ D^2 > u > u(u+7*D)/D^2

divide through by u, and you get:

-(u + 7*D)/D^2 > 1 > (u + 7*D) / D^2

multiply through by D^2 and you get:

-(u +7*D) > D^2 > u + 7*D
-u - 7*D > D^2 > u + 7*D
letting u go to 0, as x approaches 3, we have:

-7*D> D^2 > 7*D

Now -7*D can not be negative as D^2 is a positive number, so D be negative, so dividing through by D yields:

-7< D < 7

So, D < |7|.

Now the question I have is, does this prove:

lim x^2 + x - 12 = 7
x-> 3 ------------
...............x-3

3. Oct 8, 2005

### HallsofIvy

Staff Emeritus
Edwin, you forgot the "7" right from the start!

Peter, you are approaching it correctly. You want to show that for any positive $\epsilon$there exist a $\delta$ so that as long as 0<|x-3|< $\delta$.
$$|\frac{x^2+x-12}{x-3}-7|< \epsilon$$.
Yes, doing the algebra, the left side of that is
$$|\frac{x^2+ x-12- 7x+21}{x-7}|= |\frac{x^2-6x+9}{x-3}|=|\frac{(x-3)^2}{x-3}|$$
(I just noticed that you factored and canceled before subtracting the 7! Easier your way but the same thing!)
and because 0< |x-3| that denominator is not 0 so this is equal to
$$|x-3|$$
so that your original expression reduces to
$$|x-3|<\epsilon$$

It should be clear that taking $\delta$= $\epsilon$ works nicely.

Last edited: Oct 8, 2005
4. Oct 8, 2005

### powp

Thanks

Still trying to figure this whole thing out.

In my text it says that once I detemine the delta value I need to use it it the actual proof. doing the acutal proof just seems like do the reverse of what was done before. Is this correct?

How does this prove anything? I am a little confussed.

5. Oct 8, 2005

well, fundamentally when you work backwards through the proof on your "scrap paper" to find a delta value is "cheating". Thankfully we don't concern our selves with this little problem, as long as the formal proof holds.

As an example from math history, Gauss would pull extremely insightful delta vaules out of thin air to do proofs. In some cases it took over a hundred years for people to figure out where he got his delta values.

Cheers.

6. Oct 10, 2005

### Edwin

Oops! I forgot the 7. I just read a little more on formal definition of limits. Anyhow, I'll be taking some formal proof based classes shortly, so I'll keep an eye on these posts to see if I can pick up some tid-bits here and there.

Thanks!

Best Regards,

Edwin