Find limit L. Then find δ > 0 such that |f(x) - L| < 0.01 (f(x) = x^2 -3)

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In summary, the conversation discusses finding the limit of a function and then finding a value for δ that satisfies the given condition for ε = 0.01. The range (1,3) is assumed for x in the function δ(ε) = ε / |x+2|, which allows for the calculation of δ = 0.002. This range is chosen because it is valid for the given ε value.
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Homework Statement


Find limit L. Then find δ > 0 such that |f(x) - L| < 0.01

Limit when x approaches 2 of x^2 -3

Homework Equations


0 < |x-c| < δ
0 < |f(x) - L| < ε

The Attempt at a Solution


function is continuous at D = ℝ so limit is f(2) = 1

0 < |x-2| < δ and

0 < |x^2 -3 -1| < 0.01 ⇔ 0 < |x^2 -4| <0.01 ⇔ 0 < |x + 2|*|x-2| < 0.01 which is good because:
0 < |x-2| = 0.01 / |x+2|

I've seen the solution and i see that I'm supposed to assume a range for x (like (1,3) ). I can imagine that because the function isn't linear a range has to be assumed.

They say that assuming this range gives δ = 0.01 / 5 = 0.002 witch seems to be the smallest of 0.01 / 3 and 0.01 / 5. That makes sense to me.

But why this chosen range? Why does this range apply to ε = 0.01? Choosing a different range gives a different δ.

Who can help me with this I'm really trying to understand this.
 
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  • #2
hi ialink! :smile:
ialink said:
Find limit L. Then find δ > 0 such that |x - L| < δ => |f(x) - L| < 0.01

They say that assuming this range gives δ = 0.01 / 5 = 0.002 witch seems to be the smallest of 0.01 / 3 and 0.01 / 5. That makes sense to me.

But why this chosen range? Why does this range apply to ε = 0.01? Choosing a different range gives a different δ.

exactly! :smile:

δ depends on ε

(we could write it δ(ε) )

these proofs all involve showing that whatever ε we choose, we can always find a δ :wink:

(usually δ gets smaller and smaller, just like ε)
 
  • #3
Hey Tiny Tim

If i Would express δ(ε) I'd say δ(ε) = ε / |x+2|. Calculating δ with this function is possible for D = ℝ (for both ε and x) so a logical conclusion is that the limit exists that's what you mean right?

but ε = 0.01 is given to find the appropriate δ. Something has to be assumed for x in δ(ε) = ε / |x+2|. They assume (1,3) and find δ = 1/5*ε or δ = 1/3*ε and therefore conclude that δ = 1/5 * 0.01 = 0.002. That i don't get. What's the relation between the chosen range (1,3) an ε = 0.01? Why is that valid?
 
Last edited:

1. What is the purpose of finding the limit in this equation?

The limit in this equation represents the value that the function approaches as x gets closer and closer to a certain value. It can help determine the behavior of the function and its overall trend.

2. How is the limit calculated in this equation?

In this equation, the limit is found by plugging in the given value for x and evaluating the function. If the function approaches a finite value, then that value is the limit. If the function approaches infinity or negative infinity, then the limit does not exist.

3. What does the symbol "δ" represent in this equation?

The symbol "δ" represents the distance between the given value of x and the value that x is approaching as it gets closer and closer. It is used to find the range of values for x that will result in a difference of less than 0.01 between the function and the limit.

4. Why is it important to find the value of δ in this equation?

Finding the value of δ is important because it ensures that the function and the limit are within a specific range of each other. This helps to accurately determine the behavior of the function and its overall trend near the given value of x.

5. How is the value of δ determined in this equation?

The value of δ is determined by trial and error, starting with a small value and gradually increasing it until the difference between the function and the limit is less than 0.01. This process is repeated until a suitable value for δ is found.

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