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Limit problem

  1. Feb 17, 2016 #1
    1. The problem statement, all variables and given/known data
    For what value of the constants a and b such that the following limit exists?

    lim {(ax+|x+1|)|x+b-2|}/|x+1|
    x->-1


    help me ,thx

    2. Relevant equations


    3. The attempt at a solution

    first, I know that I should cancel the absolute value at denominator of x+1. but i don't how to do. also x+1 if x>1. -(x+1) if x<1
     
    Last edited: Feb 17, 2016
  2. jcsd
  3. Feb 17, 2016 #2

    Samy_A

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    Hello Steven, :welcome:

    You have to show an "attempt at a solution" before we are allowed to help.

    To help you start: what is the first thing you try when you have to evaluate such a limit? And why doesn't it work in this case?
     
  4. Feb 17, 2016 #3

    Samy_A

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    What value must the expression in the numerator have in x=-1 in order for the limit for x → -1 to possibly exist?
     
  5. Feb 17, 2016 #4
    i do not understand
    have some tip?
     
  6. Feb 17, 2016 #5

    Samy_A

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    Say you have to establish whether the following limit exists:
    ##\displaystyle \lim_{x\rightarrow -1} \frac{x+7}{|x+1|}##.
    How would you do that, and what is the answer?

    Then look at the limit you have to compute:
    ##\displaystyle \lim_{x\rightarrow -1} \frac{(ax+|x+1|)|x+b-2|}{|x+1|}##
    What value must the numerator ##(ax+|x+1|)|x+b-2|## have for x=-1 in order for this limit to possibly exist?
     
  7. Feb 17, 2016 #6
     

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  8. Feb 17, 2016 #7

    Samy_A

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    I don't understand how you get the 2 or -2 in the denominators as ##|1+(-1)|=0##.

    Just plugging in ##x=-1## in ##\frac{x+7}{|x+1|}## gives you ##\frac{6}{0}##, obviously not a very nice limit.
    See how the problem is not only that the denominator is 0 in ##x=-1##, but also that the numerator is 6. If the numerator was also 0, then maybe (just maybe), the limit could exist.

    That should give you a huge clue about what should happen with the numerator ##(ax+|x+1|)|x+b-2|## at ##x=-1## for the limit to maybe (just maybe) exist.
     
  9. Feb 17, 2016 #8
    (-a)(b-3)>0
    b-3>0 or -a>0
    b>3 0r a <0
    that right?
     
  10. Feb 17, 2016 #9

    Samy_A

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    (Please always explain how you get a result, else it may be difficult to help you along.)

    I assume you got ##(-a)(b-3)## by setting ##x=-1## in the numerator ##(ax+|x+1|)|x+b-2|##.
    That actually gives ##-a|b-3|##. Why do you want that numerator to be larger than 0? How does >0 help you to establish that a limit may exist?
     
    Last edited: Feb 17, 2016
  11. Feb 17, 2016 #10
    so, -a|b-3| if put in x=-1
    then, a is all real number, b is 3 ?
     
  12. Feb 17, 2016 #11

    Samy_A

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    I have asked the key question a number of times: what value must the numerator have in ##x=-1## in order for the limit to possibly exist?
    You now know that the numerator is ##-a|b-3|## when ##x=-1##.
    If the numerator is, say, equal to 7, you get ##\frac{7}{0}## when ##x=-1##. Can the limit then exist?

    What is the only value for the numerator ##-a|b-3|## that may makes it possible for a limit to exist for ##x\to -1##?
     
  13. Feb 17, 2016 #12
    i need some time to think,because i learn limit on 3 days
     
  14. Feb 17, 2016 #13

    Samy_A

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    The more theoretical point is as follows.
    Say you have two continuous functions ##f:\mathbb R \to \mathbb R## and ##g:\mathbb R \to \mathbb R##.

    Does ##\displaystyle \lim_{x\rightarrow -1} \frac{f(x)}{g(x)}## exist, and if so, what is the limit?

    If ##g(-1) \neq 0##, the answer is easy: ##\displaystyle \lim_{x\rightarrow -1} \frac{f(x)}{g(x)}=\frac{f(-1)}{g(-1)}##.

    It get's more complicated if ##g(-1) = 0##. Then just plugging in ##x=-1## would give ##\displaystyle \lim_{x\rightarrow -1} \frac{f(x)}{g(x)}=\frac{f(-1)}{0}##.
    Now dividing a real number by 0 is not possible, so if, for example, ##f(-1)=7##, we can conclude that ##\displaystyle \lim_{x\rightarrow -1} \frac{f(x)}{g(x)}## doesn't exist (or maybe is ##\pm \infty##).
    It is only when plugging in ##x=-1## gives ##\frac{0}{0}## (that's called an indeterminate form), that a finite limit may exist. Notice the may. It's not sure the limit will exist, but it is possible. It will depend on the precise behavior of ##f## and ##g## near ##x=-1##.
     
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