Limit proof:but there is something wrong

[maupassant]

Hi everybody,

Suppose one wants to prove that lim (3x²-x) = 10 as x approaches 2.

Taking the definition of a limit we should have:
"If 0 < |x-2| < d then it follows that |f(x) - 10| < e"

Proving this statement, we can say |f(x)-10| < e
iff |3x²-x-10| <e
iff |3x+5|.|x-2| <e
iff |x-2| < e/(|3X+5|)

The problem that arises now is: how can one make the "x" in the denominator disappear?
Because we need a delta in function of only an epsilon.

So we could say: set d<=1 then it follows
|x-2|<d<=1
|x-2|<1
-1<x-2<1
1<x<3

We know now that our x should lie between 1 and 3. But now comes the thing I fail to understand well: Why does one select 3 which is larger than the x should be. Why could we not just select 2 or 2.5 or something like that. So why should we have
|x-2| < e/(|3.3+5|)
|x-2| < e/(14)

Finally we can set d=min{1, e/(14)} and we should have a solution for

"If 0<|x-2|< d=e/(14) then |f(x) - 10|<e" and have proved that indeed
lim (3x²-x) = 10 as x approaches 2.

Could somebody tell me why I always should use that maximum border on x (i.c. 3) ? That's something that's been annoying me for about a week now!

Thank you very much!
M.

 

[HallsofIvy]

Hi everybody,

Suppose one wants to prove that lim (3x²-x) = 10 as x approaches 2.

Taking the definition of a limit we should have:
"If 0 < |x-2| < d then it follows that |f(x) - 10| < e"

Proving this statement, we can say |f(x)-10| < e
iff |3x²-x-10| <e
iff |3x+5|.|x-2| <e
iff |x-2| < e/(|3X+5|)

The problem that arises now is: how can one make the "x" in the denominator disappear?
Because we need a delta in function of only an epsilon.

So we could say: set d<=1 then it follows
|x-2|<d<=1
|x-2|<1
-1<x-2<1
1<x<3

We know now that our x should lie between 1 and 3. But now comes the thing I fail to understand well: Why does one select 3 which is larger than the x should be. Why could we not just select 2 or 2.5 or something like that.

You are taking the limit as x goes to 2. Further, since you have to take the limit "from both sides", x may well be larger than 3! The point is that x must, eventually, be "close" to 2 but it may be above or below 2. For x sufficiently close to 2, yes, it will be less than 3, but that is also true of any number you might pick. We can't select 2 because we could not be sure that x is "< 2". A number arbitrarily close to 2 might still be larger than 2.

We certainly could use 2.5 - it's just that 1 tends to be easier to do arithmetic with than 1/2! If require that |x-2|< .5, then -.5< x- 2< .5. Adding 2 to both sides, 1.5< x< 2.5. Multiplying by 3, 4.5< 3x< 7.5. Adding 5, 9.5< 3x+ 5< 12.5. If 9.5< |3x+ 5|, then 1/|3x+5|< 1/9.5 so \epsilon/|3x+5|&lt; \epsilon/9.5. It's just easier to use 1 than 0.5.

So why should we have
|x-2| < e/(|3.3+5|)
|x-2| < e/(14)

Finally we can set d=min{1, e/(14)} and we should have a solution for

"If 0<|x-2|< d=e/(14) then |f(x) - 10|<e" and have proved that indeed
lim (3x²-x) = 10 as x approaches 2.

Could somebody tell me why I always should use that maximum border on x (i.c. 3) ? That's something that's been annoying me for about a week now!

Thank you very much!
M.

Homework Statement

Homework Equations

The Attempt at a Solution

 

[maupassant]

Thanks for the answer!