Using Epsilon Delta to prove a limit

In summary: I had no idea.In summary, Paul proves that the lim as x goes to 4 of x^2 + x - 11 = 9 by factoring x^2 + x - 11 and making an assumption that K/x-4/ < Epsilon. If x+5<K for x close enough to 4, then Epsilon is also less than K|x-4|.
  • #1
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Homework Statement


prove that the lim as x goes to 4 of x^2 + x -11 = 9
This is the example used on Paul's Online Notes on limits in calculus which can be found here http://tutorial.math.lamar.edu/Classes/CalcI/DefnOfLimit.aspx (I really like this resource.)

Homework Equations


Paul factors x^2 + x - 11 to get /x+5//x-4/ < Epsilon, then notes that if we can find a value K that is greater than /x+5/ we can say /x+5//x-4/ < K/x-4/

(I am using / for absolute value bars, is there a better convention?)

From there he makes the assumption that K/x-4/ < Epsilon
How does he make that assumption?

The Attempt at a Solution


To me it looks like this:
/x+5//x-4/ < Epsilon and /x+5//x-4/ < K/x-4/
Therefor Epsilon = K/x-4/ and /x-4/ = Epsilon/K

Which would be great except that if that is true, the rest of the proof doesn't work. What am I missing about the relationship of the inequalities?
 
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  • #2
Absolute value bars are written |x|; the | character is shift-\ on a US keyboard.

You seem to be misunderstanding what Paul does, since [tex]x^2 + x - 11 \neq (x + 5)(x - 4)[/tex]. In fact, what he appears to be doing is subtracting 9 from both sides and factoring [tex]x^2 + x - 20 = (x + 5)(x - 4)[/tex], then claiming that [tex]\lim_{x\to 4} x^2 + x - 20 = 0[/tex].

From there, the limit argument goes: Fix some [tex]\varepsilon > 0[/tex], and try to find [tex]\delta[/tex] so that whenever [tex]0 < |x - 4| < \delta[/tex] we have [tex]|x^2 + x - 20| < \varepsilon[/tex].

The argument with [tex]K[/tex] uses the fact that we only care about what happens "close to 4", because we can always make [tex]\delta[/tex] smaller without breaking anything in the proof. In a limit proof, you often want to focus attention only on the parts of your expression that vanish, by estimating the other parts. In this case, Paul says "suppose we could estimate [tex]x + 5[/tex] by a constant, that is, find some [tex]K > 0[/tex] so that -- at least for [tex]x[/tex] close enough to [tex]4[/tex] -- we have [tex]|x + 5| < K[/tex]. Then in order to prove [tex]|x^2 + x - 20| < \varepsilon[/tex], it would be enough to prove [tex]K|x - 4| < \varepsilon[/tex]."

You seem to be getting confused by the order of reasoning. [tex]\varepsilon[/tex] is always given at the beginning, and the whole chain of reasoning is based around getting your error to fit under [tex]\varepsilon[/tex] -- because what "limit" means is "no matter how small [tex]\varepsilon[/tex] is, you can manage to fit under it by making [tex]\delta[/tex] small enough". Then the inequality [tex]K|x - 4| < \varepsilon[/tex] is not an assumption, but a reduction. Paul is saying "because of our estimate [tex]|x + 5| < K[/tex] for [tex]x[/tex] close enough to 4, in order to prove what we want, it is good enough to prove [tex]K|x - 4| < \varepsilon[/tex]".

To make things more concrete, let's say that when [tex]|x - 4| < 1[/tex] we find [tex]|x + 5| < 10[/tex], so that [tex]K = 10[/tex]. Then we need to find [tex]\delta[/tex] small enough that when [tex]0 < |x - 4| < \delta[/tex] then [tex]10|x - 4| < \varepsilon[/tex] (and in any case [tex]\delta \leq 1[/tex] because we made that assumption already to set the value of [tex]K[/tex]). Well, that's simple enough: to get [tex]10|x - 4| < \varepsilon[/tex] all we need is [tex]|x - 4| < \varepsilon/10[/tex]; so finally we end up with a proof that, if [tex]\delta = \min\{1, \varepsilon/10\}[/tex], then [tex]0 < |x - 4| < \delta[/tex] implies [tex]|x^2 + x - 20| < \varepsilon[/tex].
 
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  • #3
Thank you. Explained that way, it made a lot of sense to me.

So, just to make sure I understand, I think you are saying that it is enough to to prove that
K|x-4|< Epsilon because |x+5||x-4| gets smaller as x approaches 4. If we replace |x+5| with a constant such that |x+5|< K, it will still get smaller. So, K|x-4| may not equal |x+5||x-4|, but it will be less than Epsilon.

The problem I was having was that it wasn't clear to me that when starting from
|x+5||x-4|< Epsilon and assigning a value to K such that |x+5|< K , that it necessarily followed that K|x-4|< Epsilon.

I was thinkning that if ab<c and a<d, it doesn't necessarily follow that db<c, right?

But after reading your point about focusing on the part of the expression that vanishes I think I understand.

In order to make sure I wasn't violating the inequality I explained it to myself this way:

We are assuming a range of x values close to 4, and K is found using the limit of that range, but the x in |x-4| will be within the range, never the limit of the range, so K|x-4| will still be less than Epsilon.

Does that make sense?

Thank you so much for your help! Also thanks for explaining where the absolute value bar is.
 

What is the Epsilon-Delta Definition of a Limit?

The Epsilon-Delta definition of a limit is a mathematical approach used to prove that a function has a limit at a specific point. It involves finding a suitable value for 'delta' (denoted as δ) that corresponds to a given value of 'epsilon' (denoted as ɛ), such that when the distance between the input and the limit point is less than δ, the distance between the output and the limit of the function is less than ɛ.

How is Epsilon-Delta used to evaluate limits?

Epsilon-Delta is used to evaluate limits by setting up a logical framework that ensures the function's output values get closer to the limit point as the input values get closer to the limiting value. This means that for any given input value, we can find a corresponding output value that is within a certain distance of the limit point, satisfying the definition of a limit.

What are the main steps involved in using Epsilon-Delta to prove a limit?

There are three main steps involved in using Epsilon-Delta to prove a limit:

  1. Identify the given limit and the limit point.
  2. Set up the Epsilon-Delta definition of a limit by choosing a suitable value for ɛ.
  3. Use algebraic manipulation and logical reasoning to find a suitable value for δ that satisfies the definition of a limit.

Can Epsilon-Delta be used to prove all limits?

Yes, Epsilon-Delta can be used to prove all limits, as long as the function satisfies the definition of a limit. However, in some cases, using other methods such as L'Hopital's rule may be more convenient or efficient.

How can Epsilon-Delta be applied in real-life scenarios?

Epsilon-Delta can be applied in real-life scenarios in fields such as physics, engineering, and economics, where limits and continuity play a crucial role. It can be used to analyze and optimize systems, predict future trends, and make precise calculations. For example, it can be used to determine the maximum capacity of a bridge or the optimal production level for a company.

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