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From Wikipedia:

For functions on the real line, one way to define the limit of a function is in terms of the limit of sequences. In this setting: $$ \lim_{x\to a}f(x)=L $$ if and only if for all sequences $x_n$ converging to $a$ the sequence $f(x_n)$ converges to $L$.

I was considering the case when $\lim_{x\to a}f(x) \neq f(a)$ and $x_n=a$ for any $n \geq n_0$ and some $n_0 \in \mathbb{N}$. Isn't this case a counterexample to both "if" and "only if" parts? If yes, how shall the quote be modified?

Thanks and regards!

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The quote should specify that the sequence takes values on $X-\{a\}$. – Arturo Magidin Feb 20 '12 at 2:58
I've posted on the talk page of that page to discuss this issue. – Arturo Magidin Feb 20 '12 at 3:08
@ArturoMagidin: Thanks for confirmation here and discussion there! – Tim Feb 20 '12 at 3:11
up vote 1 down vote accepted

It has been confirmed by someone who watches the page that this was an issue at some point, but that this section seems to have slipped through the cracks when the problem was corrected/definition changed to the usual non-Bourbaki definition (as was recently discussed in the comments in this answer, Bourbaki defines two different kinds of limits; one in which $x$/the sequence is allowed to take the value $a$, and a "punctured limit" in which it isn't. The latter definition is the standard in the U.S. and elsewhere).

I've changed the sections in Wikipedia to specify that the sequence $x_n$ must have $x_n\neq a$ for all $n$, or equivalently that it must be a sequence in $X-\{a\}$ that converges to $a$.

Well spotted!

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Oops; I thought I had marked this as community wiki from the very beginning... My apologies! – Arturo Magidin Feb 20 '12 at 18:05

It is not a counterexample since you only found one sequence for which $lim_{n\to\infty}f(x_n)=f(a)$. You can construct a sequence as follows: since $lim_{x\to a}f(x)\ne f(a)$ then there is an $\epsilon>0$ such that for all $\delta>0$ there is an $x$ so that $|x-a|<\delta$ and $|f(x)-f(a)|\geq \epsilon$. Now, for every $n$ there is an $x_n$ so that $|x_n-a|<\frac 1 n$ so that $|f(x_n)-f(a)|\geq \epsilon$. It is easy to see that the sequence $x_n$ converges to $a$ and $lim_{n\to\infty} x_n\ne f(a)$.

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I think Tim's point is the following: suppose that $f$ has a removal discontinuity at $a$, with $\lim\limits_{x\to a}f(x) = L$. Under the definition given, if we take the sequence $\{x_n\}$ with $x_n=a$ for all $n$, then $\lim_{n\to\infty}f(x_n)=\lim_{n\to\infty}f(a) = f(a)\neq L$; so, under the definition given, since there is a sequence converging to $a$ for which $f(x_n)$ does not converge to $L$, then $\lim_{x\to a}f(x)$ cannot equal $L$... contrary to our assumption. – Arturo Magidin Feb 20 '12 at 3:07

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