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Suppose that $f$ is defined on a deleted nbh of $x_0$. Denote by $D(f)$ the domain of $f$. Show that the following statements are equivalent:

(i) $\lim_{x\to x_0} f(x) = \ell$

(ii) whenever $\{u_k\}$ is a sequence in $D(f)-\{x_0\}$ that converges to $x_0$, then the sequence $\{f(u_k)\}$ converges to $l$.

I know that (i) means that for all $\varepsilon > 0$ there is some $\delta > 0$ such that $|x - x_0| < \delta \Rightarrow |f(x) - \ell| < \varepsilon $

also know that $\{u_k\} \rightarrow x_0$ means that for all $\varepsilon > 0$ there is some $N \in \mathbb N$ such that $|u_k - x_0| < \varepsilon$ for all $k > N$.

Can I not just make this second $\varepsilon$ smaller than $\delta$? That seems to simple and not that solid though.

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  • $\begingroup$ You could just let it be equal. $\endgroup$ – Eoin Nov 10 '14 at 18:42
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Suppose i) is true. Then $\forall \varepsilon >0$, $\exists \delta >0$ such that $|x-x_0|<\delta \implies |f(x)-l|<\varepsilon$. Let $\varepsilon >0$ be given. So there is some $\delta$ for this particular epsilon. Now set our new $\varepsilon$ equal to this $\delta$. Then for a convergent sequence $\{x_n\}$ which converges to $x_0$, we have $\exists N\in \mathbb{N}$ for this particular $\varepsilon$ (our $\delta$). Since for $n\geq N$ we have $|x_n-x_0|<\delta$ it follows that $|f(x_n)-l|<\varepsilon$, and therefore $\{f(x_n)\}$ converges to $l$.

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  • $\begingroup$ Great! Do you know how I could show that (ii) implies (i)? Or is this solid enough to show that both are equivalent? $\endgroup$ – Jane Nov 10 '14 at 18:56
  • $\begingroup$ proving two statements are equivalent generally means proving two implications. $\endgroup$ – Matt A Pelto Nov 10 '14 at 18:57
  • $\begingroup$ @Jane It is still required to show ii) implies i), as Pelto mentioned. You can do this direction by applying the definition of limit as $x_n\rightarrow x_0$ of a sequence. $\endgroup$ – Eoin Nov 10 '14 at 19:21

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