# Limit proof:but there is something wrong

• maupassant
In summary, the conversation discusses how to prove that the limit of (3x²-x) is equal to 10 as x approaches 2. The conversation explains the logic behind selecting a maximum border for x, and how to manipulate the equation to prove the limit statement.
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.

maupassant said:
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 $\epsilon/|3x+5|< \epsilon/9.5$. 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.

## The Attempt at a Solution

Last edited by a moderator:

## What is a limit proof?

A limit proof is a mathematical method used to prove that a function approaches a certain value as its input approaches a specific value.

## Why is there something wrong in the limit proof?

There could be various reasons for something to be wrong in a limit proof. Some common reasons include incorrect assumptions, mistakes in calculations, or using an incorrect approach.

## How do you identify the mistake in a limit proof?

To identify the mistake in a limit proof, you should carefully examine each step of the proof and check for any mistakes in the assumptions, calculations, or approach. You can also try redoing the proof using a different method or seeking help from a colleague or teacher.

## Can a limit proof have multiple errors?

Yes, a limit proof can have multiple errors. It is important to thoroughly check each step of the proof to ensure that all errors are identified and corrected.

## What are some tips for avoiding mistakes in a limit proof?

Some tips for avoiding mistakes in a limit proof include double-checking your assumptions, using different methods to verify your answer, and seeking help from others. It is also important to take your time and carefully review each step of the proof.

• Calculus and Beyond Homework Help
Replies
2
Views
496
• Calculus and Beyond Homework Help
Replies
10
Views
731
• Calculus and Beyond Homework Help
Replies
5
Views
991
• Calculus and Beyond Homework Help
Replies
2
Views
590
• Calculus and Beyond Homework Help
Replies
4
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
437
• Calculus and Beyond Homework Help
Replies
5
Views
433
• Calculus and Beyond Homework Help
Replies
5
Views
849
• Calculus and Beyond Homework Help
Replies
3
Views
916
• Calculus and Beyond Homework Help
Replies
1
Views
560