November 13th, 2016, 05:58 AM  #1 
Member Joined: Aug 2016 From: South Korea Posts: 54 Thanks: 0  Simple Question from Limits
Can someone explain to me of why x + 1 was multiplied to 1/x1 to both numerator and denominator ?? sdgf.jpg 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. 
November 13th, 2016, 06:29 AM  #2 
Senior Member Joined: Sep 2007 From: USA Posts: 349 Thanks: 67 Math Focus: Calculus 
Well, $\frac{2}{x^21}=\frac{2}{(x+1)(x1)}$ as a result of factoring the denominator. In order to subtract that from $\frac{1}{x1}$ we need to multiply it by a special case of 1 in order to get the denominators the same. Therefore, $\frac{1}{x1}=\frac{x+1}{x+1}\cdot\frac{1}{x1}=\frac{x+1}{(x+1)(x1)}$. So, $\frac{x+1}{(x+1)(x1)}\frac{2}{(x+1)(x1)}=\frac{x1}{(x+1)(x1)}=\frac{1}{x+1}$.

November 13th, 2016, 06:38 AM  #3  
Member Joined: Aug 2016 From: South Korea Posts: 54 Thanks: 0  Quote:
 
November 13th, 2016, 06:46 AM  #4 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 896 
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 $\displaystyle (x1)(x+1)= x^2 1$, 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.


Tags 
limits, question, simple 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
kind of silly/simple question with probably a simple answer.  GumDrop  Math  4  October 4th, 2016 05:34 PM 
Simple limits questions  msman88  Calculus  4  February 17th, 2012 12:03 AM 
Simple, not so simple question about areas of triangles  jkh1919  Algebra  1  November 20th, 2011 09:14 AM 
A few (relatively) simple limits...  1040ez  Calculus  2  September 17th, 2008 08:28 PM 
need help with two 'simple' limits  ourubo  Calculus  4  November 21st, 2007 10:38 AM 