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Picture of my solution

This is a fairly rough, disorganized proof I did for a homework assignment and I'm curious to see alternate solutions.

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    I know it is asking a lot, but please type all relevant information into the body of the question itself. This will get you more attention for your questions and also avoid an annoying situation if the link dies later. – 2017-01-20
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    Nothing much to add, i would solve it the same way if I had to. At some point you have to use that $a_n$ is bounded to get that one of the products converges to zero, but it follows easily from convergence. – 2017-01-20
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    Could go a more elementary route here. If the vectors converge then all of their components converge to the corresponding values. The dot product is just a simple sum and product of the components, so it converges. (your solution is right though, and probably better overall) – 2017-01-20
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    @spaceisdarkgreen I originally went with that route for the proof, however, my professor wanted a more in depth proof. I'm curious to see how one would prove it with the delta epsilon method! – 2017-01-20

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As you asked for alternative solutions, Cauchy-Schwarz inequality gives the shortest one: $$|\langle a_n-a,b_n-b\rangle|\leq ||a_n-a||\,||b_n-b||.$$

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In the comments you requested an 'epsilon-delta' proof. Your proof converts nicely into one. I will be fairly detailed.

We need to show that for any $\epsilon>0$ there is an $N$ such that $$ |a_n\cdot b_n - a\cdot b| \le \epsilon $$ for all $n>N.$

So let $\epsilon >0$ and we must show there exists such an $N$.

You established the inequality $$ |a_n\cdot b_n - a\cdot b| \le |a_n||b_n-b| + |b||a_n-a|.$$

As you noted, the fact that both $|b_n-b|$ and $|a_n-a|$ go to zero means the right hand side goes to zero. In epsilon-delta language that means you can make the right hand side less than $\epsilon$ by choosing $n$ large enough .

The second term is easiest to bound. Since $a_n\rightarrow a,$ we can find an $N_1$ so that $$|a_n-a| < \frac{\epsilon}{2|b|}$$ for any $n>N_1$ so that the second term is less than $\epsilon /2.$

The first term's a little trickier since $|a_n|$ changes with $n$ as well. But since it goes to a limit, it doesn't change much.

Since $a_n\rightarrow a,$ $|a_n|\rightarrow |a|$ . (To establish this, just use the reverse triangle inequality: $\lvert |a_n|-|a|\rvert \leq |a_n-a|.)$ Now let $l>0$ just be some number. The fact that $|a_n|\rightarrow |a|$ means that there is an $N_2$ such that $$ |a_n| < |a| + l$$ for all $n> N_2.$

Now that we've bounded $|a_n|,$ we can do the $|b_n-b|$ factor. Since $b_n\rightarrow b,$ we can pick an $N_3$ such that $$|b_n-b| < \frac{\epsilon}{2(|a|+l)}$$ for all $n>N_3.$

So if we let $N= \max(N_1,N_2,N_3)$ then for $n>N$ all the above inequalities hold and we have $$ \begin{eqnarray}|a_n\cdot b_n - a\cdot b| &\le& |a_n||b_n-b| + |b||a_n-a| \\&<& (|a|+l)\frac{\epsilon}{2(|a|+l)} + |b|\frac{\epsilon}{2b} \\&=& \epsilon \end{eqnarray}$$