# Proof using Definition of Limit

1. Apr 30, 2010

### nugget

1. The problem statement, all variables and given/known data

f(x) = (x3-11x2+43x-60)/(x-4)

Prove directly from the definition that limx->4f(x)=3.

2. Relevant equations

This requires an epsilon-delta proof, I think... (will refer to epsilon as E and delta as D)

3. The attempt at a solution

Firstly I simplified the numerator of f(x) to (x-4)(x2-7x+15), which enabled me to cancel the denominator.

After simplifying; f(x) = (x-4)(x-3)+3.

I want to prove that for all E>0, there exists a D>0.

Hence if |x-4|<D, then |x-4|.|x-3|+3<E.

I here assume that D≤1

Hence |x-4|≤1.

Now; |x-3| = |x-4+1|
|x-3|≤|x-4|+1

Hence |x-3|<2.

Finally,

|x-4|.|x-3|<2.|x-4|<2D

and |x-4|.|x-3|<E,

which means we choose D=min{1,E/2} i.e. delta is 1 unless E/2 is less than 1, in which case it is equal to E/2.

I want to know if i've done this correctly, and where i've gone wrong if not, also let me know if there are certain statements i should be making or ones i'm making incorrectly.

Thanks

2. Apr 30, 2010

### Gib Z

Careful, that isn't the same function as the original one. It's another function that happens to coincide in value for every point other than x=4. It's important to understand that, and also why can we use it in E-D proofs. Can you explain in words why it is sufficient to use the new function when doing these limits?

This isn't complete. State it fully.

No, to show that the limit of f(x) as x goes to a is L, we must show that for any chosen E>0, there exists a value D such that if |x-a|< D, then |f(x) - L| < E.

So in this hence, you want to show that there exists some value D such that if |x-4| < D, |f(x) - 3| = | (x-4)(x-3)+3 - 3| = |(x-4)(x-3)| < E.

3. May 2, 2010

### nugget

Hey thanks, think I've got it down now

I want to show that for E>0, there exists a D>0 such that if

|x-4|>0, then |(x-4)(x-3)|>0

The simplified function is f(x) = (x-4)(x-3)+3, where x≠4. It is sufficient to use this instead of the given one because we are looking for the limit of f(x) as x approaches four, hence x is never equal to four, and the denominator of the original function wouldn't equal zero under these conditions.

4. May 3, 2010

### Gib Z

That is NOT what you want to show, in fact thats true most of the time anyway.

You want to show that for every value of E>0 we choose, we can find some value D>0 such that if |x-4|< D, then |(x-4)(x-3)|< E .