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How can you show that if $f:\mathbb{R}\rightarrow \mathbb{R} $ and $g:\mathbb{R}\rightarrow \mathbb{R} $ are both functions continuous at the point $c\in \mathbb{R}$ then the product $f\cdot g $ is continuous at the point c. I have $$|f(x)g(x)-f(x)g(c)+f(x)g(c)-f(c)g(c)|\leq |f(x)||g(x)-g(c)|+|g(c)||f(x)-f(c)| $$ But I'm having a hard time figuring out what I should make $|f(x)-f(c)|$ and $|g(x)-g(c)|$ less than (as a function of $\epsilon $) so that my final line shows that $|f(x)g(x)-f(c)g(c)|<\epsilon $. [I've missed out some of the details that I would actually include to keep this as short as possible].

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    Possible duplicate: http://math.stackexchange.com/questions/1593789/prove-that-the-product-of-two-continuous-functions-is-continuous – 2017-01-09
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    You could expand $f(x)$ into $f(x)-f(c)+f(c)$ and use the triangle inequality again. – 2017-01-09

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Let me give a hint. Of course we can make $|f(x)-f(c)|$ and $|g(x)-g(c)|$ as small as we want. No problem with $|g(c)|$ (which is constant). Because $f$ is continuous, it is bounded on a neighbourhod on $c$ which allows us to make the term on the right as small as we want.

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You want to make each of the two terms smaller than $\epsilon/2$. That means, first of all, that $|f(x)-f(c)|$ must be made smaller than $\frac{\epsilon}{2|g(c)|}$, unless $g(c)=0$, in which case that term doesn't matter. The other term, you first need to make sure that $|f(x)|<|f(c)|+1$. Then you can make $|g(x)-g(c)|<\frac{\epsilon}{2(|f(c)|+1)}$. Now pick $\delta$ to be the most limiting of all these factors, and you're good.

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    How do we know that we can make $|f(x)|<|f(c)|+1 $? Am I missing something? – 2017-01-09
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    @Ben By continuity. We can find a $\delta' > 0$ such that $|f(x)-f(c)| < 1$. This means that $|f(x)| < |f(c)| + 1$. – 2017-01-09
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Let $d>0$ such that $|x-c|

Let $M=1+\max (|f(c)|,|g(c)|).$ So $M>|g(c)|$ And also $|x-c||f(x)|.$

For $\epsilon >0$ let $\delta \in (0,d)$ such that $|x-c|<\delta \implies (\;|f(x)-f(c)|<\epsilon /2M\; \land \;|g(x)-g(c)|<\epsilon /2M \;).$ Now we have $$|x-c|<\delta \implies |f(x)g(x)-f(c)g(c)|= $$ $$=|f(x)(g(x)-g(c))\;+\; g(c)(f(x)-f(c))|\leq $$ $$\leq |f(x)|\cdot |g(x)-g(c)|+|g(c)|\cdot |f(x)-f(c)|\leq $$ $$\leq M\cdot (|g(x)-g(c)|+|f(x)-f(c)|)<\epsilon.$$