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Homework Help: Cancellation Limit

  1. Jul 25, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the limit.

    [tex]\lim_{x\rightarrow0}\frac{\frac{1}{x+1}-1}{x}[/tex]

    2. Relevant equations



    3. The attempt at a solution

    I have to do this analytically. Although, I know that the limit is supposed to be -1 from a graphing approach. When you substitute in 0 for x you get 0/0. How do I get this function into a determinate form? I cant factor and I'm not sure what to multiply the top and bottom by.

    I tried [tex]\frac{\frac{1}{x+1}+1}{\frac{1}{x+1}+1}[/tex]

    and I got

    [tex]\frac{1-(x^2+2x+4)}{x(x+1)(x+2)}[/tex] which is still in indeterminate form.
     
  2. jcsd
  3. Jul 25, 2010 #2

    eumyang

    User Avatar
    Homework Helper

    I wouldn't do that. Try multiplying the original fraction by
    [tex]\frac{x+1}{x+1}[/tex]
    instead. Something should eventually cancel.


    69
     
  4. Jul 25, 2010 #3

    Mute

    User Avatar
    Homework Helper

    Do you know, or can you use, L'Hopital's rule?

    In this case, L'Hopital's rule would take the form

    If
    [tex]\lim_{x\rightarrow a} \frac{f(x)}{g(x)} = \frac{0}{0}[/tex]
    but
    [tex]\lim_{x\rightarrow a} \frac{f'(x)}{g'(x)} = L[/tex]
    where L is a finite number,
    then
    [tex]\lim_{x\rightarrow a} \frac{f(x)}{g(x)} = \lim_{x\rightarrow a} \frac{f'(x)}{g'(x)} = L[/tex]
     
  5. Jul 25, 2010 #4
    You can get a common denominator with the terms in the numerator and simplify, then you should get a fraction that's not indeterminate anymore.
     
  6. Jul 25, 2010 #5
    No I dont, I'm just starting calculus, but that looks interesting. I'm sure I'll learn it later.



    Thanks, that works perfectly. I know where I went wrong.
     
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