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Let $E$ be a metric space, $f: E \rightarrow \mathbb{R}$ and $\bar{x} \in \text{cl}(E)$. We say that $\lim_{x \rightarrow \bar{x}} f(x) = L$ if:

$\textit{Heine's definition}$: for any sequence $x_n \rightarrow \bar{x}$ and for any $\epsilon >0$ there exists $N \in \mathbb{N}$ such that $|f(x_n)-L|<\epsilon$ for any $n>N$;

$\textit{Cauchy's definition}$: for any $\epsilon >0$ there exists $\delta >0$ such that $|f(x)-L|<\epsilon$ whenever $\text{d}(x,\bar{x})<\delta$.

How can I prove that Heine's definition implies Cauchy's?

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    It is called Heine's criterion. – 2017-02-14

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It is easy to prove that (not Cauchy) implies (not Heine): Assume the Cauchy criterion is not satisfied at $x$. Then there is $\epsilon>0$ such that for all $\delta=1/n$, $n\in \mathbb N$, there is $x_n$ with $d(x,x_n)<\delta$ and $|f(x_n)-L|>\epsilon$. Now Heine's criterion is not fulfilled, since $x_n\to x$ but $f(x_n)$ does not converge to $L$.

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    Not Heine means that there exists a sequence $x_n \rightarrow \bar{x}$ and an $\epsilon >0$ such that $|f(x_n)-L|>\epsilon$ for any $n$? – 2017-02-15
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    Heine is equivalent to: for all sequences $(x_n)$ with $x_n\to x$ it holds $f(x_n)\to L$. Not Heine: there is a sequence $x_n\to x$ such that $f(x_n)$ does not converge to $L$. – 2017-02-15