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Existence of a limit

  1. Feb 4, 2016 #1
    • Moved from a technical forum, so homework template missing
    what value of the constants a and b if the following limit exists
    lim (ax + |x + 1|)|x + b − 2| |x + 1|
    x→−1

    |x|= x for x≥ 0 and |x|= -x for x<0
    |x+1|= x+1 for x≥ -1 and |x+1|= -(x+1) for x<-1
    I don't know how to determine |x + b − 2| is positive or negative.

    i know that if limit exists, lim x→ −1 ^- f(x) =lim x→ −1 ^+ f(x)

    but i cannot cancell the factor x + 1
     
  2. jcsd
  3. Feb 4, 2016 #2
    1. The problem statement, all variables and given/known data
    http://holland.pk/nwhxy2ji [Broken]
    2. Relevant equations



    3. The attempt at a solution
    |x|= x for x≥ 0 and |x|= -x for x<0
    |x+1|= x+1 for x≥ -1 and |x+1|= -(x+1) for x<-1
    I don't know how to determine |x + b − 2| is positive or negative.

    i know that if limit exists, lim x→ −1 ^- f(x) =lim x→ −1 ^+ f(x)

    i tired to solve it,but i cannot cancell the factor x + 1
     
    Last edited by a moderator: May 7, 2017
  4. Feb 4, 2016 #3

    Samy_A

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    I've not solved the whole exercise, but you could start by simplifying the problem somewhat.
    Clearly the denominator is 0 when x=-1. So if the numerator isn't equal to 0 in x=-1, there is no way the limit will exist.
    What can you tell about a and/or b if you require the numerator to be 0 in x=-1?
     
    Last edited by a moderator: May 7, 2017
  5. Feb 4, 2016 #4
    (ax+ (x+1)) |x+b-2| =0 for x≥-1
    (ax+ (x+1))=0
    a=-(x+1)/x

    or |x+b-2| =0
    b=2-x

    (ax-(x+1)) |x+b-2| =0 for x<-1
    a=(x+1)/x

    or |x+b-2| =0
    b=2-x

    (x+1)/x =-(x+1)/x
    x=-1

    b=2-(-1)=3

    a=-1 and b=3 when limit will exist
    is it right ?
    thank you
     
  6. Feb 4, 2016 #5

    Samy_A

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    No, you made an error somewhere.

    You make it way too complicated. You don't (at this stage) have to distinguish between x≤-1 and x≥-1.
    All you have to do is set the numerator at x=-1 to 0.
    That will give you necessary conditions on a and/or b.

    After that, you still will have to check if the limit exists if a and/or b meet the conditions, and if the limits exists, compute it. This all turns out to be rather straightforward, though.
     
  7. Feb 4, 2016 #6
    at x=-1
    (ax+ (x+1)) |x+b-2| =0
    -a|b-3| =0
    a=0 , b=3 but i have a problem , can |b-3| = b-3?
    i put a=0 and b=3 to the lim
    and do it in left-hand limits and right-hand limits

    left-hand limits=right-hand limits=0 ,so it exists

    is it right?

    Samy_A thank you
     
  8. Feb 4, 2016 #7

    Samy_A

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    Yes, this is correct. At least if you meant that a=0 or b=3. These two conditions don't have both to be true (although they can both be true of course).

    Whether b-3 is positive or negative in the case a=0 doesn't matter: 0|b-3| will be 0 anyway.
    I'm not sure I understand what you did here. The limit isn't necessarily 0 though.
    I would treat the two cases (a=0, b=3) separately.

    For example, in the case a=0, the expression becomes:##\frac{|x+1||x+b-2|}{|x+1|}=|x+b-2|## for ##x \neq -1##. Calculating the limit for ##x \to-1## is then straightforward.
    Similarly for the case b=3.

    You are welcome.
     
  9. Feb 4, 2016 #8
    ##|x+b-2|## for ##x \neq -1## , i don't know what is your meaning
    if i just substitute -1 to the limit , then i get lim ##x \to-1## ##|x+b-2|## the next step is that do it in left-hand limits and right-hand limits?
    but i don't not have to distinguish |x+b-2|?
     
  10. Feb 4, 2016 #9

    Samy_A

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    Let's give the expression a name: ##\displaystyle f(x)=\frac{(ax+|x+1|)|x+b-2|}{|x+1|}## (for ##x \neq -1)##.
    You are asked to establish whether ##\displaystyle \lim_{x\rightarrow -1} f(x)## exists, and if so, what the limit is.

    We already know that the limit can only exist if a=0 or b=3 (or both).

    For a=0, we can simplify the expression for ##f(x)##:
    ##\displaystyle f(x)=\frac{(ax+|x+1|)|x+b-2|}{|x+1|}=\frac{(0x+|x+1|)|x+b-2|}{|x+1|}=\frac{|x+1||x+b-2|}{|x+1|}=|x+b-2|## (for ##x \neq -1##).

    So, in the case a=0, you have to compute ##\displaystyle \lim_{x\rightarrow -1} f(x)=\lim_{x\rightarrow -1}|x+b-2|##.
    That's an easy limit to compute, no need to distinguish left-hand and right-hand limits.

    And similarly for the other case, b=3.
     
  11. Feb 4, 2016 #10
    ##\lim_{x\rightarrow -1}|x+b-2| = |b-3|## but i don't know that value of b when a=0 ,how can i compute.
    similarly,when b=3, ##\lim_{x\rightarrow -1}(ax+|x+1|) = -a## ,it is same with the above one
    when a=0 ,b=3 ,##\displaystyle \lim_{x\rightarrow -1} f(x) = 0 ##
     
  12. Feb 4, 2016 #11

    Samy_A

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    Correct.
    When a=0, the limit is |b-3|. You can't say more. This limit exists for all values of b.
    Similarly, when b=3, the limit is -a. This limit exists for all values of a.
     
  13. Feb 4, 2016 #12
    thank you, you are a patient and good teacher.
    you have a full explanation ,thank you again.
     
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