Limit of 2x^2 as x Approaches 3: Using Epsilon-Delta Definition"

In summary, the student is having difficulty starting the limit problem and needs help understanding what they are trying to do and how to approach the problem. After understanding the basics of the limit, they are able to find a sufficient interval and prove that the limit is 18.
  • #1
jmed
46
0

Homework Statement


evaluate lim2x^2 as x approaches 3 using formal definition (epsilon and delta) of limit


Homework Equations





The Attempt at a Solution

 
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  • #2
jmed said:

Homework Statement


evaluate lim2x^2 as x approaches 3 using formal definition (epsilon and delta) of limit


Homework Equations





The Attempt at a Solution

What have you attempted?
 
  • #3
I am having a hard time starting it. I'm not sure how to prove an exponential.
 
  • #4
You could start by showing what you intend to prove in this problem.
 
  • #5
epsilon> 0 and (|x-1|) <delta...delta equals ??
 
  • #7
jmed said:
epsilon> 0 and (|x-1|) <delta...delta equals ??

Why |x - 1|? x is not approaching 1 in this limit problem.
 
  • #8
delta= a fraction of epsilon so that it is less than x^2 - 1?
 
  • #9
so it would be |x-3|...
 
  • #10
jmed said:
delta= a fraction of epsilon so that it is less than x^2 - 1?
x^2 - 1 has nothing to do with this problem.

What is the function you're dealing with, what value is x approaching, and what is the limit that you are trying to prove?
 
  • #11
evaluate lim2x^2 as x approaches 3 using formal definition (epsilon and delta) of limit
 
  • #12
And what is the limit that you are trying to prove? IOW
[tex]\lim_{x \to 3} 2x^2 = ?[/tex]

You need to know what the limit is before you can prove that this value is the limit using delta and epsilon.

After you answer my question above, show me the limit definition in the context of your function and its limit. You will need to use delta and epsilon in this definition.
 
  • #13
2x^2 is a continuous function, so this limit is easily computable (to a trivial level)
Once you know this limit, you may apply the concept of the limit: as f(x) gets closer to this limit, its values should lie somewhere in between (L - epsilon , L+ epsilon). You may find appropriate "deltas" by means of algebra (factor..et c), and you'll probably have to bound your interval (for example, say delta cannot be larger than 1)
 
  • #14
limit is 18...I am just confused with this whole process...
 
  • #15
jmed said:
limit is 18...I am just confused with this whole process...

You know that the limit is 18, now you have to find a sufficient interval for your "inputs" x so that any x you pick out of that interval (a - delta , a + delta ) or in this case ( 3 - delta, 3 + delta) will be "close enough" to your limit 18. We only know that the limit is 18 because we know that the function gets closer and closer to 18 as you get closer to 3. So appropriately, find an interval where you CAN see values that are extremely close to 18.

How can you do this? well, just play around with algebra and inequalities.. You know that if 18 is the limit, then you can find that interval I talked about eariler, so for x in (a - delta, a + delta) , applying the function to such an x should yield |f(x) - 18 | < epsilon (remember that |a - b | just denotes distance and epsilon is just the length of an interval, this is the interval for which values are sufficiently close to 18 in a way you can call it the "limit" ).

Now you know that |2x^2 - 18 | < epsilon must be true, from this true statement you may try to derive another true statement (algebraically). In layman terms, try to get the absolute value thing on the left look like |x - 3 | , so that you'll get | x - 3 | < something. Now you know that for x in some interval ( 3 - [something] , 3 + [something] ), |2x^2 - 18 | will be less than epsilon. Also, keep in mind that epsilon is a variable, it can be anything.
 
  • #16
I also found it very difficult to understand this epsilon-delta version of limits.

Not because of me, but because of the explanations I've found always seem to beat around the bush. It's such as imple idea but the way people talk about it really obscures the intuition behind it.

This is called the Weierstrass definition of a Limit, and it's such a smart way to look at the idea of a limit when you want to approach or inch up to a point that doesn't exist along a function.

This video, http://www.youtube.com/watch?v=-ejyeII0i5c&feature=youtube_gdata
and the next one in the playlist after it, part 2, should show you exactly what is going on.

Also, here is a .pdf with a few examples of how to apply this technique in practice.

http://docs.google.com/viewer?a=v&q=cache:_OYvmsulbDIJ:www.ocf.berkeley.edu/~yosenl/math/epsilon-delta.pdf+epsilon-delta+limit+example&hl=en&sig=AHIEtbQijZifL9dG46lTjmQMCpKpcrrY1g

If you study these two sources fully you should have no problem with ε - δ limits for a while, maybe real analysis is different, idk yet :tongue2:
 
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  • #17
Thanks! youtube was a GREAT help!
 

What is the limit of 2x^2 as x approaches 3 using epsilon-delta definition?

The limit of 2x^2 as x approaches 3 using epsilon-delta definition is 18. This means that as x gets closer and closer to 3, the value of 2x^2 will approach 18.

What is the epsilon-delta definition of a limit?

The epsilon-delta definition of a limit is a mathematical method used to precisely define the concept of a limit. It states that the limit of a function f(x) as x approaches a is L if, for any positive number ε, there exists a positive number δ such that if the distance between x and a is less than δ, then the distance between f(x) and L is less than ε.

How is the epsilon-delta definition used to find limits?

To find the limit of a function using the epsilon-delta definition, we first choose a value for ε (the desired distance between f(x) and L). Then, we use algebraic manipulation to find a corresponding value for δ (the distance between x and a) that satisfies the definition. This value of δ will ensure that the distance between f(x) and L is less than ε for all x within that distance of a.

Why is the use of epsilon-delta definition important in finding limits?

The use of epsilon-delta definition is important because it provides a rigorous and precise way to define and calculate limits, especially in more complicated functions where other methods may fail. It also allows for a deeper understanding of the concept of a limit and its relationship to a function's behavior near a particular point.

Can the epsilon-delta definition be used to prove the existence of a limit?

Yes, the epsilon-delta definition can be used to prove the existence of a limit. If we can find a value for δ that satisfies the definition for any given value of ε, then we can conclude that the limit exists. However, it is important to note that this method does not necessarily provide a numerical value for the limit, but rather proves its existence.

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