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Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces,$E \subseteq X$ and $f:E \rightarrow Y$ a function. If $x_0 \in \overline E$ we define $$ \lim_{x \rightarrow x_0;x \in E} f(x) = L $$ iff $$ \exists L \in Y \forall \epsilon > 0 \exists \delta > 0 \forall x\in E: d_X(x,x_0) < \delta \rightarrow d_Y(f(x),L) < \epsilon \qquad (1) $$

He then gives the following exercise: Let $x_0 \in E$. Then $\lim_{x \rightarrow x_0; x \in E} f(x)$ exists $\iff$ $\lim_{x \rightarrow x_0; x \in E \setminus \{x_0\}} f(x)$ exists. Further we have that if $\lim_{x \rightarrow x_0; x \in E} f(x)$ exists, then the limit equals $f(x_0)$.

First the implications: From left to right is trivial. But what if $x \in E \setminus \{x_0\}$ ? Why does then (1) also hold for $x_0$ itself ?

I got some edits: The implication $\Leftarrow )$ also requires that the limit equals $f(x_0)$. Sorry for that. Then the equivalence is true.

Further I see that Tao doesn't intend that for example the function $f(x)= 1$ if $x = x_0$ and $f(x) = 0$ otherwise should have a limit where $E = \mathbb R$. First, if we consider $E \setminus \{x_0\}$ we have that the limit equals $0$. In fact if the limit exists and $x_0 \in E$ then the function must be continous.

  • If $x_0 \in E$, then $\lim\limits_{x\to x_0;, x \in E\setminus{x_0}} f(x)$ can exist when $\lim\limits_{x\to x_0;, x \in E} f(x)$ doesn't. Consider $f(x) = 0$ if $x \neq x_0$, and $f(x_0) = 1$. – Daniel Fischer Jul 12 '13 at 15:43
  • I dunno, but there seems to be something wrong with the definition, I think it should read, instead: $\exists L \in Y; \forall \epsilon > 0; \exists \delta > 0; \forall x\in E: \color{red}{0<}d_X(x,x_0) < \delta \rightarrow d_Y(f(x),L) < \epsilon$. Omitting the part '0<' it's the definition for continuous functions. – user49685 Jul 12 '13 at 15:48
  • Yes. But this is why Tao gives this exercise I guess. I am a bit confused. And this definition which I gave is explicitly stated like that in Tao's book. –  Jul 12 '13 at 15:52
  • @DanielFischer: You are right. This proves that the equivalence in the exercise can't be true. I will thus adopt the definition that $x \neq x_0$. –  Jul 13 '13 at 12:08

1 Answers1

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As indicated in the comments above, I don't think the definition of the limit that you've given above is the right one. Assuming that we're using the definition of the limit suggested by @user49684 in the comments (which I think is the appropriate definition), then I think it would be obvious to you why the two limits are equal.

Amitesh Datta
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  • So Tao's definition is not equivalent to the suggested definition ? I guess this is (due to this exercise) only true if $x \in E$. –  Jul 12 '13 at 15:55
  • @André I don't know what Terry Tao's definition for this is but I'll take your word for it. However, I don't think that's a good definition of the limit. The reason is that the limit at $x_0$ concerns behaviour of a function near but not equal to $x_0$. If we're considering behaviour at $x_0$ as well, then that's just the definition of continuity of $f$ at $x_0$. (Exercise: Can you see that in your definition, if $x_0\in E$, then $L=f(x_0)$?) However, the exercise that you have stated is certainly true if we adapt user49684's definition ... – Amitesh Datta Jul 12 '13 at 15:59