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Let $A \subset \mathbb{R}^n$ and let $f,g \rightarrow \mathbb{R}^m$ be two functions on $A$ such that $f$ is bounded. Let $c$ be the limit point of $A$. Show that if $\lim \limits_{x \to c}g(x)=0$ then $\lim \limits_{x \to c}(g(x)f(x))=0$

My attempt:

Since $f$ is bounded, there exists $M>0$ such that $|f(x)|\leq M$ for all $x \in A$

Let $\epsilon>0$. Since $\lim_{x\rightarrow c}g(x)=0$, there exists a $\delta_1>0$ such that $$ |x-a|<\delta_1\Rightarrow |g(x)|<\epsilon $$

Not sure where to go from here

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    $\begingroup$ Hint: choose $\delta$ such that $|g(x)| < \frac{\epsilon}{M}$. $\endgroup$ – MathematicsStudent1122 Feb 23 '17 at 16:17
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    $\begingroup$ Just note that $|g(x)f(x)|\le |g(x)\cdot |f(x)|\le M|g(x)|$ $\endgroup$ – Hagen von Eitzen Feb 23 '17 at 16:17
  • $\begingroup$ specifically, i dont know what the epsilon delta limit definition for $\lim \limits_{x \to c}(g(x)f(x))=0$ looks like and i think that's where i am having trouble $\endgroup$ – combo student Feb 23 '17 at 17:03
  • $\begingroup$ Is it just for $\epsilon >0$ there exists $\delta >0$ such that $|x-a|<\delta \Rightarrow |g(x)f(x)|<\epsilon$? $\endgroup$ – combo student Feb 23 '17 at 17:09
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    $\begingroup$ @combostudent Yes you're right, except that the $a$ is supposed to be a $c$. Please let me know if you need more clarification. $\endgroup$ – Ovi Feb 25 '17 at 3:44
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From your attempt, for any $\epsilon/ M>0$ there exists $\delta >0$ such that $|x-c|<\delta \Rightarrow |g(x)|<\epsilon/M$. Since we know that $|f(x)|<M$, we multiply both sides by $|g(x)|$ to get $$|f(x)g(x)|=|f(x)||g(x)|<M|g(x)|<\frac {\epsilon}{M}M=\epsilon$$ so long as $|x-c|<\delta$ .

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