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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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Expand the right hand side. –  Mariano Suárez-Alvarez Oct 3 '11 at 7:43
    
If I expand the right side, I get the left side but how do I factor the left side? –  Juan de la John Oct 3 '11 at 7:51
    
There is no sensible sense in which this equality is the result of "factoring", really. The right hand side is a useful expression for the left hand side in the context of the proof---you will see, as you continue learning calculus, that this same trick is used very often, and then it will become natural... –  Mariano Suárez-Alvarez Oct 3 '11 at 8:09
    
Oh okay thank you for all your help! I will continue to persist with my calculus course –  Juan de la John Oct 3 '11 at 16:01

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up vote 3 down vote accepted

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. –  Juan de la John Oct 3 '11 at 8:01

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