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One-Sided Limits

  1. Mar 23, 2008 #1
    [SOLVED] One-Sided Limits

    1. The problem statement, all variables and given/known data
    I've been having problems solving for these one-sided limits.

    1.[tex]\stackrel{lim}{x\rightarrow\pi^-}\frac{sinx}{1-cosx}[/tex]

    2.[tex]\stackrel{lim}{x\rightarrow\O^+}xln(x)[/tex]

    3.[tex]\stackrel{lim}{x\rightarrow\frac{\pi}{2}^-}(sec (x)-tan(x))[/tex]



    2. Relevant equations



    3. The attempt at a solution

    I solved #1 as =0/1 =0



    I solved #2 as [tex]\stackrel{lim}{x\rightarrow\O^+}\frac{ln(x)}{\frac{1}{x}}[/tex]

    =[tex]\frac{\frac{1}{x}}{1}[/tex] = [tex]\frac{\frac{1}{0}}{1}[/tex]

    =0


    I solved #3 as [tex]\stackrel{lim}{x\rightarrow\frac{\pi}{2}^-}(sec (x)-tan(x))[/tex]

    =[tex]\stackrel{lim}{x\rightarrow\frac{\pi}{2}^-}(\frac{1}{cosx}-\frac{sinx}{cosx})[/tex]


    =[tex]\stackrel{lim}{x\rightarrow\frac{\pi}{2}^-}(\frac{1-sinx}{cosx})[/tex] and I solved this down to =[tex]\stackrel{lim}{x\rightarrow\frac{\pi}{2}^-}tan(x)[/tex]


    = undefined

    I'm missing some steps in these problems. Any help from you is greatly appreciated!
     
  2. jcsd
  3. Mar 23, 2008 #2

    HallsofIvy

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    How did you get 0/1? What is cos([itex]\pi[/itex])?


    and why would 1/0 = 0??


    Since you used L'Hopital's rule in #2, why not here?
    [tex]\lim_{x\rightarrow \frac{\pi^-}{2}} \frac{1- sin(x)}{cos(x)}= \lim_{x\rightarrow \frac{\pi^-}{2}}\frac{-cos(x)}{sin(x)}[/itex]

    By the way- don't use "stackrel" in that way.
    [tex]\lim_{x\rightarrow \frac{\pi^-}{2}}[/tex]
    is much easier to read than
    [tex]\stackrel{lim}{x\rightarrow\frac{\pi}{2}^-}[/tex]
     
  4. Mar 23, 2008 #3

    HallsofIvy

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    And why would 1/0 be 0?


    Since you used L'Hopital's rule in #2, why not here?
    [tex]\lim_{x\rightarrow\frac{\pi^-}{2}}\frac{1- sin x}{cos x}= \lim_{x\rightarrow\frac{\pi^-}2}}\frac{cos x}{sin x}[/tex]
    What is that limit?

    By the way "stackrel" makes that very hard to read.
    [tex]\lim_{x\rightarrow\frac{\pi^-}{2}}[/tex]
    is much better than
    [tex]\stackrel{lim}{x\rightarrow\frac{\pi}{2}^-}[/tex]
    Click on the LaTex to see the code.
     
  5. Mar 23, 2008 #4
    #1. I should have gotten 0/2 for #1, because cos(pi)=1 and 0/(1-(-1))=0/2
    I thought that 0/(any nymber greater than or less than 0) is equal to 0, but if the fraction was 2/0, there would be an infinite, positive for the limit approaching from the right and negative infinity for numbers approaching from the left?

    #2. I reworked through the problem several times and I keep getting (1/0)/1 which is, after going over my notes several times, = to infinity/1 which is = infinity

    #3. I got = 0, after following L'Hopital's as you said. I originally -0/1, which would be = 0.

    I apologize about the laytex from my previous post. I was unable to check the box for quoting your reply and when I clicked to preview my post on this reply I wasn't seeing any latex images, so I just used the harder to read way of expressing my answers--sorry for not using latex.

    EDIT:
    Is this what you were trying to get me to understand? That 1/0 in limits is equal to an infinite answer and that 0/1 in limits is equal to 0?
     
    Last edited: Mar 23, 2008
  6. Mar 23, 2008 #5
    Am I going about the solutions correctly now?
     
  7. Mar 23, 2008 #6
    I found my error on problem #2. It should be equal to 0 after rewriting the form of the limit because it was indeterminant. I got it finalized and came up with 0 for the answer to number 2. Thanks HallsofIvy.
     
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