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Prove that if for every sequence ($x_n$) in A that converges to c such that $x_n \ne c$ for all $n \in \mathbb{N}$, the sequence $(f(x_n))$ converges to L, then $lim_{x\to c} f = L$.

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    $\begingroup$ Hint: Assume the contrary $\endgroup$ Apr 13, 2014 at 17:40

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If not, then there exists $\epsilon>0$ such that for all $\delta>0$ and $0<|x-c|<\delta$ we have

$$|f(x)-L|>\epsilon$$

In particular this holds for $\delta=1/n$. Now pick for every $n\in\mathbb{N}$ some $x_n$ with $0<|x_n-c|<\frac{1}{n}$. Then $x_n\not= c$ for all $n$ and $x_n\rightarrow x$ as $n\rightarrow\infty$. But still

$$|f(x_n)-L|>\epsilon$$

for all $n$. Contradiction, because we assumed that $f(x_n)\rightarrow L$.

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