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quincyboy7
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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?
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?
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