Rigorous delta-epsilon limit proof?

[quincyboy7]

I'm trying to prove the limit as x approaches 3 of sqrt(3x-5)=2.

Call delta D and epsilon E

So we have 0<abs(x-3)<D and must prove abs(sqrt(3x-5)-2)<E.
abs(sqrt(3x-5)-2)=abs(sqrt (3(x-3)+4)-2)<E if we choose delta to be ((E+2)^2-4)/3, which simplifies to (E^2+4E)/3

My questions arise because I am not sure if this is a valid form in which to express delta, even though it seems to me that any positive number can be squared and added to 4 times itself and divided by 3 to produce a positive number. Also, why DOESN'T the expression work for, let's say, proving the limit actually equals 3? And shouldn't my delta be in a min(1, form of epsilon) form like in other proofs? Any thoughts would be appreciated.

EDIT: I think I have found an error and messed up the placement of absolute values inside/outside the radical expression...now I'm totally lost. Any idea at all how to solve this limit?

 

[Maxter]

Regarding the limit equaling 3. The three here is being approached, so I'm guessing you mean that. When discussing limit, you are not allowed to look AT the limit point, only near the limit point. This is because some functions may not be defined at the limit point. To compare what happens near 3 versus what happens at 3 is to start talking about continuity at 3.

Technically, when you say the "limit equals 3", then you are talking about the function values and the L would change, and the abs(f(x)-L) will change form, and typically so will your delta (unless you have a uniformly continuous function). I don't think you mean that?

For delta, typically, if your delta is that complicated, there is a way to simplify it. You need to be able to work from left to right, starting with the abs(f(x)-L).

You won't need the delta=min(1,-) here because you are dealing with a different function here than the other problems.

For this problem, you'll want to mulitiply top and bottom by the conjugate and then bound it above to throw away some term(s) in the denom. The form for delta in this problem, at least without writing down the details, I think will look fairly nice.

 

[quincyboy7]

Alright, trying again...

Let D= (blank)

If 0<abs(x-3)<D, then abs(sqrt (3x-5)-2)=abs(3x-9)/abs(sqrt(3x-5)+2)=3abs(x-3)/abs(sqrt(3x-5)+2)<3delta/abs(sqrt(3x-5)+2)...I'm still left with a nasty expression on the bottom. Any tips?

EDIT: Continuing, We can get 3delta/abs(sqrt(3x-5)+2)<=3delta/2=E when delta=2epsilon/3??!

I think that's right, a confirmatory check would be great

 

[Maxter]

You need to find a way to get rid of that square root term in the denom, since it has a remaining x.

Can you make that entire fractional expression larger somehow?

 

[quincyboy7]

You need to find a way to get rid of that square root term in the denom, since it has a remaining x.

Can you make that entire fractional expression larger somehow?

Well, the bottom can, at the least, be equal to 2 (when the square root filth is equal to 0), so the expression has to be less than or equal to just 3D/2, correct?

 

[Maxter]

Looks good to me, that's what I got for delta also :)