Class example: limit of a function using definition

In summary, the conversation discusses finding the limit of a function using the definition of a limit. The homework statement provides an example of using the definition to show that limx→2(x2)=4. The attempt at a solution involves finding the appropriate value for delta (δ) using the given information and then showing that this value satisfies the definition. The use of "min" in the definition ensures that the value for delta chosen will work for all possible values of epsilon (ε). There may be multiple ways to choose delta, but the "min" function helps to ensure that it will always satisfy the definition.
  • #1
PirateFan308
94
0
I am having trouble understanding how to find the limit of a function (using the definition of a limit). I have a class example, and was wondering if anyone could walk me through the steps.

Homework Statement


Using the definition of the limit to show that limx→2(x2)=4
f(x) = x2
c=2
L=4

Given an arbitrary ε>0, take δ=min{1,ε/5}
If x≠2 and |x-2|<δ then |x-2|<1 and |x-2|< ε/5
|f(x)-L| = |x2-4| = |(x-2)(x+2)| = |x-2||x+2|
|x-2|<1 => 1<x<3 => 3<x+2<5 => |x+2|<5
|x-2||x+2| < (ε/5)(5) = ε so |f(x)-L|<ε


Homework Equations


We say that lim f(x)x→c=L if:
[itex]\forall[/itex]ε>0 [itex]\exists[/itex]δ>0 [itex]\forall[/itex]x[itex]\in[/itex]dom f if x≠c and |x-c|<δ then |f(x)-ε|<L


The Attempt at a Solution


The biggest thing I am confused about is how the professor got δ? Did he have to do the later work first and then went back and plugged in the answer he got?

Also, in the definition, it says that then |f(x)-ε|<L but we ended up getting |f(x)-L|<ε. Why is this? I understand that we can rearrange the equation, but then doesn't this mess up the absolute value signs?
 
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  • #2
PirateFan308 said:
I am having trouble understanding how to find the limit of a function (using the definition of a limit). I have a class example, and was wondering if anyone could walk me through the steps.

Homework Statement


Using the definition of the limit to show that limx→2(x2)=4
f(x) = x2
c=2
L=4

Given an arbitrary ε>0, take δ=min{1,ε/5}
If x≠2 and |x-2|<δ then |x-2|<1 and |x-2|< ε/5
|f(x)-L| = |x2-4| = |(x-2)(x+2)| = |x-2||x+2|
|x-2|<1 => 1<x<3 => 3<x+2<5 => |x+2|<5
|x-2||x+2| < (ε/5)(5) = ε so |f(x)-L|<ε


Homework Equations


We say that lim f(x)x→c=L if:
[itex]\forall[/itex]ε>0 [itex]\exists[/itex]δ>0 [itex]\forall[/itex]x[itex]\in[/itex]dom f if x≠c and |x-c|<δ then |f(x)-ε|<L


The Attempt at a Solution


The biggest thing I am confused about is how the professor got δ? Did he have to do the later work first and then went back and plugged in the answer he got?

Also, in the definition, it says that then |f(x)-ε|<L but we ended up getting |f(x)-L|<ε. Why is this? I understand that we can rearrange the equation, but then doesn't this mess up the absolute value signs?
"The biggest thing I am confused about is how the professor got δ? Did he have to do the later work first and then went back and plugged in the answer he got?"
Your professor likely did some scratch work, starting with |x2-4|<ε, and then getting his result for δ.​

"in the definition, it says that then |f(x)-ε|<L but we ended up getting |f(x)-L|<ε"

It should be |f(x)-L|<ε in the definition.​
 
  • #3
|f(x)-epsilon|<L is a typo. |f(x)-L|<epsilon is the correct form. And yes, the professor figured out a delta using the later work and then went back and plugged it in.
 
  • #4
Another question, is there more than one δ that will prove this?
Say, Given an arbitrary ε>0, take δ=min{2,ε/6}
If x≠2 and |x-2|<δ, then |x-2|<2 and |x-2|<ε/6
|f(x)-L| = |x2-4| = |(x+2)(x-2)| = |x+2||x-2|
|x-2|<2 => -2<x-2<2 => 0<x<4 => 2<x+2<6 => |x+2|<6
|x-2||x+2| < (6)(ε/6) = ε so |f(x)-L|<ε
 
  • #5
PirateFan308 said:
Another question, is there more than one δ that will prove this?
Say, Given an arbitrary ε>0, take δ=min{2,ε/6}
If x≠2 and |x-2|<δ, then |x-2|<2 and |x-2|<ε/6
|f(x)-L| = |x2-4| = |(x+2)(x-2)| = |x+2||x-2|
|x-2|<2 => -2<x-2<2 => 0<x<4 => 2<x+2<6 => |x+2|<6
|x-2||x+2| < (6)(ε/6) = ε so |f(x)-L|<ε

Sure, that choice works just as well.
 
  • #6
Yes, there are many ways to come up with δ .
 
  • #7
Why is it that I must say d=min{1,ε/5}? Would it also work if I said that δ=1,ε/5. I'm a bit fuzzy on how the "min" makes this true, or the absence of "min" makes it false.
 
  • #9
PirateFan308 said:
Why is it that I must say d=min{1,ε/5}? Would it also work if I said that δ=1,ε/5. I'm a bit fuzzy on how the "min" makes this true, or the absence of "min" makes it false.

If ε > 5, then if you say that δ > ε/5, the proof won't work.

Added in Edit:
Let's say ε = 10.

Then the claim would be that δ = 2 will satisfy the definition.
But if x=3.9, then f(3.99)=15.21, so |f(3.99)-2| = 13.21 > 10
 
Last edited:
  • #10
SammyS said:
If ε > 5, then if you say that δ > ε/5, the proof won't work.

So is it standard procedure to always take δ=min if there is more than one condition? Will it ever be wrong for me to make δ=min ?
 
  • #11
PirateFan308 said:
Why is it that I must say d=min{1,ε/5}? Would it also work if I said that δ=1,ε/5. I'm a bit fuzzy on how the "min" makes this true, or the absence of "min" makes it false.

In the proof you used that d<=1 AND d<=epsilon/5. min(1,epsilon/5) is less than or equal to both of them. d=1 doesn't work if you pick a small epsilon. d=epsilon/5 doesn't work if you pick a large epsilon. Try it.
 
  • #12
Thank you! This makes so much more sense now!
 

What is the definition of a limit of a function?

The limit of a function is a fundamental concept in calculus that describes the behavior of a function as its input approaches a certain value or point. It is denoted by the symbol "lim" and is often used to analyze the behavior of functions at points where they are not defined or are discontinuous.

How do you find the limit of a function using the definition?

To find the limit of a function using the definition, you must evaluate the function at values approaching the given point from both the left and right sides. If the values approach the same number, then that number is the limit. If the values approach different numbers, then the limit does not exist.

Why is the limit of a function important?

The limit of a function is important because it allows us to understand and analyze the behavior of complex functions. It helps us determine the continuity and differentiability of functions, as well as understand the behavior of a function at points where it is not defined. It is also crucial in many real-world applications, such as determining maximum and minimum values.

What are some common misconceptions about the limit of a function?

One common misconception is that a function must be defined at the point in question in order for the limit to exist. This is not true, as the limit describes the behavior of the function as the input approaches the point, not necessarily at the point itself. Another misconception is that the limit of a function must always be a number, when in fact it can also be infinity or negative infinity.

How is the limit of a function used in real-life applications?

The limit of a function is used in many real-life applications, particularly in physics, engineering, and economics. For example, it can be used to determine the maximum and minimum values of a function, such as finding the maximum profit or minimum cost in business. It is also used in physics to calculate instantaneous velocity and acceleration. In engineering, it is used to optimize the performance of systems and in economics to model supply and demand.

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