Epsilon in the limit definition

In summary: Consider ##a_n = \frac {n - 1} n, n \ge 1##. It's easy to show that ##\lim_{n \to \infty}a_n = 1##. However, if ##\epsilon = 0##, it's not possible to find a specific number N for which ##|a_n - 1| = 0##, for all ##n \ge N##.With ##\epsilon > 0##, all that has to happen is to force the terms in the tail of the sequence arbitrarily close to L, not necessarily making them exactly equal to it.
  • #1
Shlomi93
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in the limit definition of a sequence, why does epsilon has to be greater than 0 and not greater or equal to 0?

thanks in advance.
 
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  • #2
Shlomi93 said:
in the limit definition of a sequence, why does epsilon has to be greater than 0 and not greater or equal to 0?

thanks in advance.
It doesn't really matter with sequences of real numbers. You could take both as you can always find another epsilon that is slightly smaller. In general, however, one speaks of open neighborhoods around the limit point as they are the defining element of general (topological) spaces. And open translates to smaller than, whereas smaller or equal includes the boundaries, and as such are closed sets. So the restriction to smaller than is somehow simply consequent, even if not needed (and it's available on the keyboard).
 
  • #3
Choose ##\epsilon = 0##, then in order for ##a_n \to L## we need to find a ##N## such that ##a_n = L## for all ##n>N##.
So under this definition the only sequences that converge are those that are eventually constant.
 
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  • #4
pwsnafu said:
Choose ##\epsilon = 0##, then in order for ##a_n \to L## we need to find a ##N## such that ##a_n = L## for all ##n>N##.
So under this definition the only sequences that converge are those that are eventually constant.
Yes, you're right. I confused it with the condition ##\,\vert \,a_n -L\,\vert \, < \varepsilon## where you could take ##\leq## instead.
Of course ##\varepsilon = 0## would make no sense as there would be only constant sequences left over.
 
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  • #5
fresh_42 said:
Yes, you're right. I confused it with the condition ##\,\vert \,a_n -L\,\vert \, < \varepsilon## where you could take ##\leq## instead.
Of course ##\varepsilon = 0## would make no sense as there would be only constant sequences left over.
pwsnafu said:
Choose ##\epsilon = 0##, then in order for ##a_n \to L## we need to find a ##N## such that ##a_n = L## for all ##n>N##.
So under this definition the only sequences that converge are those that are eventually constant.

thank to both of you!
 
  • #6
Shlomi93 said:
in the limit definition of a sequence, why does epsilon has to be greater than 0 and not greater or equal to 0?
Consider ##a_n = \frac {n - 1} n, n \ge 1##. It's easy to show that ##\lim_{n \to \infty}a_n = 1##. However, if ##\epsilon = 0##, it's not possible to find a specific number N for which ##|a_n - 1| = 0##, for all ##n \ge N##.

With ##\epsilon > 0##, all that has to happen is to force the terms in the tail of the sequence arbitrarily close to L, not necessarily making them exactly equal to it.
 

What is the limit definition of epsilon?

The limit definition of epsilon, also known as the epsilon-delta definition, is a mathematical concept used to define the limit of a function. It states that a limit exists at a certain point if, for any given small positive value of epsilon, there exists a corresponding value of delta such that the distance between the input and the limit of the function is less than epsilon whenever the distance between the input and the point of evaluation is less than delta.

Why is the epsilon-delta definition important?

The epsilon-delta definition is important because it provides a rigorous and precise way to define the concept of a limit in calculus. It allows us to mathematically prove that a limit exists at a certain point, and it also helps us to understand the behavior of functions near that point.

How is the epsilon-delta definition used to solve limit problems?

The epsilon-delta definition is used to solve limit problems by first setting a value for epsilon, then finding a corresponding value for delta using the definition. This value of delta is then used to determine the range of inputs for which the distance between the input and the limit of the function is less than epsilon. By choosing a smaller and smaller value for epsilon, we can get closer and closer to the actual limit of the function.

What are some common misconceptions about the epsilon-delta definition?

One common misconception about the epsilon-delta definition is that it is only applicable to continuous functions. However, this definition can also be used for discontinuous functions, as long as the limit exists at a certain point. Another misconception is that the value of delta must be smaller than epsilon, but in reality, it can be larger as long as it satisfies the definition.

How can I improve my understanding of the epsilon-delta definition?

To improve your understanding of the epsilon-delta definition, it is important to practice solving limit problems using this method. You can also read about different examples and applications of this concept, and seek help from a teacher or tutor if needed. It is also helpful to have a strong understanding of basic algebra and calculus principles.

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