Webpage title: The Importance of Correct Order in the Definition of Limit

[CStudent]

 

[Opalg]

Hi CStudent, and welcome to MHB!

The condition $$*\ \ \forall N\in\Bbb{N}\ \exists\epsilon>0 \text{ s.t. } \forall n>N \text{ we get } |a_n-L|<\epsilon$$ does not imply that \(\displaystyle \lim_{n\to\infty}a_n = L.\)

For an example of how this can go wrong, suppose that $a_n = 1$ for all $n$ and that $L=0$. If you choose $\epsilon = 2$, the condition $|a_n-L|<\epsilon$ becomes $|1-0|<2$. That condition certainly holds for all $n$, but it is not true that \(\displaystyle \lim_{n\to\infty}a_n = L.\) In fact, \(\displaystyle \lim_{n\to\infty}a_n = 1\), which is different from $L$.

That example illustrates that in definitions like this it is very important to put things in the correct order. In the definition of limit, the choice of $\epsilon$ comes first, and then the choice of $N$ (which will usually depend on $\epsilon$) comes next. But in the condition $*$, you have put the $N$ before the $\epsilon$. That changes the whole meaning of the statement.

Your other condition $*$ is also not equivalent to the definition of limit.

But your final condition (the statement that \(\displaystyle b = |a_n-L| \to0\ \Longrightarrow\ \lim_{n\to\infty}a_n = L\)) is correct.

 

[CStudent]

Hi CStudent, and welcome to MHB!

The condition $$*\ \ \forall N\in\Bbb{N}\ \exists\epsilon>0 \text{ s.t. } \forall n>N \text{ we get } |a_n-L|<\epsilon$$ does not imply that \(\displaystyle \lim_{n\to\infty}a_n = L.\)

For an example of how this can go wrong, suppose that $a_n = 1$ for all $n$ and that $L=0$. If you choose $\epsilon = 2$, the condition $|a_n-L|<\epsilon$ becomes $|1-0|<2$. That condition certainly holds for all $n$, but it is not true that \(\displaystyle \lim_{n\to\infty}a_n = L.\) In fact, \(\displaystyle \lim_{n\to\infty}a_n = 1\), which is different from $L$.

That example illustrates that in definitions like this it is very important to put things in the correct order. In the definition of limit, the choice of $\epsilon$ comes first, and then the choice of $N$ (which will usually depend on $\epsilon$) comes next. But in the condition $*$, you have put the $N$ before the $\epsilon$. That changes the whole meaning of the statement.

Your other condition $*$ is also not equivalent to the definition of limit.

But your final condition (the statement that \(\displaystyle b = |a_n-L| \to0\ \Longrightarrow\ \lim_{n\to\infty}a_n = L\)) is correct.

Great, thank you1