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Limit to Infinty

  1. Oct 22, 2012 #1
    Hi, I am a newbie here and a newbie too at calculus since I just started my first year in college. Anyway, there are some questions that I don't quite understand and I hope that anyone can help me with this.

    http://sphotos-a.ak.fbcdn.net/hphotos-ak-ash4/269471_368130629940171_945927165_n.jpg [Broken]

    I am sorry for the use of an image because I think that the questions are quite complex for me to write on keyboard.

    Thank you in advance. :)
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Oct 22, 2012 #2

    lurflurf

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    a) write
    [tex]\frac{\sqrt{4-x^2}}{x-2}=\sqrt{-1+\frac{4}{x-2}}[/tex]
    b) use the fact
    [tex]e=\lim_{x \rightarrow \infty} \left( 1+\frac{1}{n} \right) ^n[/tex]
     
    Last edited: Oct 22, 2012
  4. Oct 22, 2012 #3

    SammyS

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    Hello kuskus94. Welcome to PF !

    [itex]\displaystyle \frac{\sqrt{x^2-4}}{x-2}=\frac{\sqrt{(x-2)(x+2)}}{x-2}[/itex]
     
    Last edited by a moderator: May 6, 2017
  5. Oct 23, 2012 #4
    Dear lurflurf,

    I sill don't get how the question a) can me modified to be like that and I would like to ask you to break it down more for me.

    As for question b), I apologize that my lack of reading. It is my bad to not read the calculus book deeper. :blushing:

    Thank you for your reply.

    Dear SammyS,

    I thank you for welcoming me in this forum. But then I got a waring message from Mark44 (the admin to my suppose) beacuse I didn't state that I have tried doing this question. Well, it was my bad :tongue2:.

    For the question a), I still got [tex]{\sqrt{0}}[/tex] over 0. I don't think I have any other idea to workaround this question. I would be really glad if you can help me more. :biggrin:

    Thank you in advance.
     
  6. Oct 23, 2012 #5

    SammyS

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    You can cancel one of the factors in the numerator. What is [itex]\displaystyle \frac{\sqrt{u}}{u}\ [/itex] or [itex]\displaystyle \frac{\sqrt{3}}{3}\ ?[/itex]
     
  7. Oct 25, 2012 #6
    Ah, I see... Thank you very much for your help. :)
     
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