1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Cancellation Limit

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


    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


    User Avatar
    Homework Helper

    I wouldn't do that. Try multiplying the original fraction by
    instead. Something should eventually cancel.

  4. Jul 25, 2010 #3


    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

    [tex]\lim_{x\rightarrow a} \frac{f(x)}{g(x)} = \frac{0}{0}[/tex]
    [tex]\lim_{x\rightarrow a} \frac{f'(x)}{g'(x)} = L[/tex]
    where L is a finite number,
    [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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook