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November 13th, 2016, 05:58 AM   #1
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Simple Question from Limits

Can someone explain to me of why x + 1 was multiplied to 1/x-1 to both numerator and denominator ??

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Was that rationalizing?


And another question I've got from another problem is this..




How did the numerator of the upper fraction became 2 - x?


Sorry, Im just confused to alot of things in Calculus @.@

Last edited by SlayedByMath; November 13th, 2016 at 06:31 AM.
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November 13th, 2016, 06:29 AM   #2
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Well, $\frac{2}{x^2-1}=\frac{2}{(x+1)(x-1)}$ as a result of factoring the denominator. In order to subtract that from $\frac{1}{x-1}$ we need to multiply it by a special case of 1 in order to get the denominators the same. Therefore, $\frac{1}{x-1}=\frac{x+1}{x+1}\cdot\frac{1}{x-1}=\frac{x+1}{(x+1)(x-1)}$. So, $\frac{x+1}{(x+1)(x-1)}-\frac{2}{(x+1)(x-1)}=\frac{x-1}{(x+1)(x-1)}=\frac{1}{x+1}$.
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November 13th, 2016, 06:38 AM   #3
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Well, $\frac{2}{x^2-1}=\frac{2}{(x+1)(x-1)}$ as a result of factoring the denominator. In order to subtract that from $\frac{1}{x-1}$ we need to multiply it by a special case of 1 in order to get the denominators the same. Therefore, $\frac{1}{x-1}=\frac{x+1}{x+1}\cdot\frac{1}{x-1}=\frac{x+1}{(x+1)(x-1)}$. So, $\frac{x+1}{(x+1)(x-1)}-\frac{2}{(x+1)(x-1)}=\frac{x-1}{(x+1)(x-1)}=\frac{1}{x+1}$.
so when I got another given problem similar to this, I can do something like multiplying like that to be able to get a similar denominators? Is there some kind of rules about this? Like when can I only do that and when can I not?
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November 13th, 2016, 06:46 AM   #4
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Yes, there is "some kind of rule about this" that, hopefully, you learned at some time in elementary school- that to add and subtract fractions you must get a "common denominator". You also should have learned that, if you multiply the denominator of a fraction by something, in order to have that new fraction represent the same number, you must multiply the numerator by the same thing. That is being done here. One of the denominators is x- 1, the other is $\displaystyle x^2- 1$. Since $\displaystyle (x-1)(x+1)= x^2- 1$, we can make the denominators the same (get the "common denominator") by multiplying the denominator (and of course the numerator) of the first fraction by x+ 1.
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