Proof using Definition of Limit

[nugget]

Homework Statement

f(x) = (x³-11x²+43x-60)/(x-4)

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

Homework Equations

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

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

 

[Gib Z]

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.

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?

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

This isn't complete. State it fully.

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

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.

 

[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.

 

[Gib Z]

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

That is NOT what you want to show, in fact that's 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 .