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Let $A \subseteq \mathbb{R}^n$, let $f:A \rightarrow \mathbb{R}^m$, and let $c$ be a limit point of $A$. $L$ and $L'$ are both limits of $f$ as $x$ approaches $c$. Show $L=L'$.

My attempt:

$\lim\limits_{x \to c}f(x)=L$ and $\lim\limits_{x \to c}f(x)=L'$. Therefore: For any $\epsilon >0$, we can find a $\delta_1,\delta_2 >0$ such that if:

$$0<|x-c|< \delta_1 \text{ and } 0<|x-c|< \delta_2$$ then $$0<|f(x)-L|<\epsilon \text{ and } 0<|f(x)-L'|<\epsilon$$.

Not sure where to go from here...

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    Hint: Choose $\epsilon$ small enough so that $\epsilon<\frac{1}{2}|L-L'|$. Then, derive a contradiction if $L\not=L'$. Use the triangle inequality. – 2017-02-02
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    when in doubt, triangle inequality – 2017-02-02
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    I think you need this http://meta.math.stackexchange.com/questions/3399/why-should-we-accept-answers – 2017-02-02

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You are almost there. Take $$\delta=\min\{\delta_1,\delta_2\}.$$ Because $c$ is a limit point of $A$, we can find $x$ in the domain of $f$ for which $0<|x-c|<\delta$. Because $\delta\leq\delta_1$ and $\delta\leq\delta_2$, we get $$\begin{align} |L-L'|&=|(L-f(x))+(f(x)-L')|\\ &\leq|L-f(x)|+|f(x)-L'|\\ &=|f(x)-L|+|f(x)-L'|\\ &<\epsilon+\epsilon=2\epsilon. \end{align}.$$ By arbitrariness of $\epsilon$, we conclude that $|L-L'|=0$ and so $L=L'$.

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    where does $|L-L'|$ come from? – 2017-02-02
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    $ 0 \leq |(f(x) - L) - (f(x) - L')| = |L - L'| \leq |L - f(x)| + |f(x) - L'| \leq 2 \epsilon$. @combostudent – 2017-02-02
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    That's part of the trick. The idea is to come up with $$|L-L'|<2\epsilon$$ for all $\epsilon>0$. With this are forced to have $$|L-L'|\leq 0.$$ But abosulte value is non negative and so we get $$|L-L'|=0.$$ – 2017-02-02