Using the precise definition of a limit (epsilon & delta)

In summary, the conversation discusses the precise definition of a limit using epsilon and delta and provides an example involving a linear function. The conversation then moves on to a similar example involving a quadratic function and addresses a typo in the previous example. Finally, it is mentioned that quadratic functions require more effort to calculate limits and the limit involving 4x^2+2 and x->2 is discussed.
  • #1
Cal124
28
0
I'm trying to practise, precise definition of a limit (epsilon & delta)
Just to check I'm along the right lines here's a previous question to the one I'm stuck on

If epsilon > 0 then there is delta >0 ... All that introduction stuff, then
Lim x-> 2 (3x-1) =5
Hence
|x-2| < delta then |3x - 6| < epsilon
>> |x-2|< epsilon/2
>> delta = epsilon/2
>> epsilon = 3delta

On my other problem I have the same setup but f(x) = 4x^2 +2 & L=18
So
|4x^2 -16| < epsilon |x-2|<delta
I've tried /4 but
I'm just not sure how I can remove the squared from the left of epsilon
Any help would be great
 
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  • #2
Cal124 said:
I'm trying to practise, precise definition of a limit (epsilon & delta)
Just to check I'm along the right lines here's a previous question to the one I'm stuck on

If epsilon > 0 then there is delta >0 ... All that introduction stuff, then
Lim x-> 2 (3x-1) =5
Hence
|x-2| < delta then |3x - 6| < epsilon
>> |x-2|< epsilon/2 ##\quad \quad ## That should be epsilon/3
>> delta = epsilon/2
>> epsilon = 3delta

On my other problem I have the same setup but f(x) = 4x^2 +2 & L=18
So
|4x^2 -16| < epsilon |x-2|<delta
I've tried /4 but
I'm just not sure how I can remove the squared from the left of epsilon
Any help would be great
I should have mentioned the above typo.

Linear functions work out nicely. Quadratic functions require somewhat more effort.

I assume that for the limit involving 42 +2, you also have x → 2 .
 
  • #3
SammyS said:
I should have mentioned the above typo.

Linear functions work out nicely. Quadratic functions require somewhat more effort.

I assume that for the limit involving 42 +2, you also have x → 2 .
Awh sorry missed that. Yeah it's
Lim. (4x^2 +2) = 18
X--> 2
 

FAQ: Using the precise definition of a limit (epsilon & delta)

What is the precise definition of a limit using epsilon and delta?

The precise definition of a limit using epsilon and delta states that for a function f(x), the limit as x approaches a is equal to L if, for any positive value of epsilon, there exists a corresponding positive value of delta such that the distance between f(x) and L is less than epsilon whenever the distance between x and a is less than delta.

Why is the precise definition of a limit important?

The precise definition of a limit is important because it allows us to rigorously define and understand the concept of a limit in calculus. It also allows us to prove the existence of a limit and determine its value, which is essential in many applications of calculus.

How is the precise definition of a limit used in calculus?

The precise definition of a limit is used in calculus to prove the existence of a limit and determine its value. It is also used to understand and analyze the behavior of functions near a certain point, which is important in finding derivatives and integrals.

What is the role of epsilon and delta in the precise definition of a limit?

Epsilon and delta are used in the precise definition of a limit to represent the distance between the function f(x) and the limit L, and the distance between the input x and the limit point a, respectively. The goal is to find a delta value that ensures the distance between f(x) and L is less than epsilon for any chosen epsilon value.

What are some common misconceptions about the precise definition of a limit?

One common misconception is that the limit of a function is equal to its value at a certain point. However, the precise definition of a limit states that the limit is the value that the function approaches as the input approaches a certain point, not necessarily the value at that point. Another misconception is that delta must be a specific value, when in reality it can vary depending on the chosen epsilon value.

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