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Simple prob pls solve

  1. Jan 16, 2007 #1
    simple prob....pls solve

    Plz some one explan How...

    lim x->1 [ 1 / 1-x - 3 / 1-x^2 ] = lim x->1 [1+x-3 /1-x^2]
  2. jcsd
  3. Jan 16, 2007 #2
    I think you should put the terms in bracket so that we can understand which numerator belongs to which denominator. It's quite hard to see it like this.
  4. Jan 16, 2007 #3


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    Well, here you are asked to simplify [tex]\frac{1}{(1-x)}-\frac{3}{(1-x^2)}=\frac{1}{(1-x)}-\frac{3}{(1+x)(1-x)}[/tex]

    There's a hint, but please note we cannot do your homework for you; you must attempt it first!

    Edit: As lkh1986 says, it would be clearer to bracket it. (However, it's clear that the expression I wrote above is correct).
    Last edited: Jan 16, 2007
  5. Jan 16, 2007 #4
    Thanks cristo..............i know this step plz do one step more......
  6. Jan 16, 2007 #5


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    How about you try the next (and last) step, and I'll help you if you get stuck- I'm not going to do it for you!

    What do you need to multiply the left hand term by in order to make the denominator 1-x^2?
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