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How was the author able to factor the expression from the left side to the expression on the right? $a_nb_n-LM=(a_n-L)(b_n-M)+M(a_n-L)+L(b_n-M)$

Thanks!

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    Oh okay thank you for all your help! I will continue to persist with my calculus course2011-10-03

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To verify that it's true, just multiply everything out.

To obtain the identity in the first place, the idea is that you want to make use of the assumption that $a_n \to L$ and $b_n \to M$, and that it's often easier to use the equivalent formulation $a_n - L \to 0$ and $b_n - M \to 0$. So one introduces new variables $x_n = a_n - L$ and $y_n = b_n -M$. Then $ a_n b_n - LM = (x_n + L)(y_n + M) - LM = \dots $ (Expand this expression and watch what happens!)

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    Wow thanks! That was really helpful, Hans. Now I can understand the proof better.2011-10-03