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    f(x)= \dfrac{|x+1||x-1|}{x^2 -1}

    Calculating the limit as x\Rightarrow -1^-, -1^+, 1^-, 1^+

    So mod mean you can do x+1 and 1-x cus that's how mod works.

    So you can get the answers 1 and -1 only out f(x) ensuring that the original f(x) is exactly the same as 1 or -1

    however i'm not sure as how to determine the limits.
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    (Original post by will'o'wisp2)
    f(x)= \dfrac{|x+1||x-1|}{x^2 -1}

    Calculating the limit as x\Rightarrow -1^-, -1^+, 1^-, 1^+

    So mod mean you can do x+1 and 1-x cus that's how mod works.

    So you can get the answers 1 and -1 only out f(x) ensuring that the original f(x) is exactly the same as 1 or -1

    however i'm not sure as how to determine the limits.
    You can split the range into 3 sections; (-\infty,-1) \cup (-1,1) \cup (1, \infty)

    Then for (-\infty, -1) you have f(x)=\frac{(x+1)(x-1)}{x^2-1}

    Then for (-1,1) you have ... ?

    etc...

    Then you can determine the limits quite simply.
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    (Original post by RDKGames)
    You can split the range into 3 sections; (-\infty,-1) \cup (-1,1) \cup (1, \infty)

    Then for (-\infty, -1) you have f(x)=\frac{(x+1)(x-1)}{x^2-1}

    Then for (-1,1) you have ... ?

    etc...

    Then you can determine the limits quite simply.
    how do you get f(x)=\frac{(x+1)(x-1)}{x^2-1} for (-\infty, -1) i'm not even sure what you do to the modulus because i dont' know what to do for -1 and 1

    so far there's 3 different things i can do for the modulused terms

    for |x+1||x-1|

    you can have the variations of
    positive, positive
    positive, negative(swapping the order doesn't matter since they multiplied together then you can take the factor of -1 out)
    negative, negative

    now all i gotta do is piece it together so it seems to me that simply leaving them as is as you have done for the example you gave me is the positive one

    so for negative infinity to -1 you can simply rearrange f(x) as 1 which appears to be the positive positive pattern

    so for the negative negative that seems to be the 1 to infinity which is then 1 (cus -1x-1 is jus 1)

    so the last must be -1 and 1 which is the positive negative pattern so i should get -1 (-1x1=-1)

    other than that i don't really know how to work these out "properly"
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    (Original post by will'o'wisp2)
    how do you get f(x)=\frac{(x+1)(x-1)}{x^2-1} for (-\infty, -1) i'm not even sure what you do to the modulus because i dont' know what to do for -1 and 1
    Your values for the domains at the end are correct.

    For the one I worked out, it's quite simply the case of taking |x+1| and |x-1| individually, and seeing what equation without the modulus they are equivalent to on the domain x \in (-\infty, -1). Clearly, you (should) know that |x+1|=-(x+1) and |x-1|=-(x+1), and so the product of the two must just be |x+1||x-1|=(x+1)(x-1), and that's the numerator for that domain. Then just repeat similarly for the other two to deduce the answers you found.
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    (Original post by RDKGames)
    Your values for the domains at the end are correct.

    For the one I worked out, it's quite simply the case of taking |x+1| and |x-1| individually, and seeing what equation without the modulus they are equivalent to on the domain x \in (-\infty, -1). Clearly, you (should) know that |x+1|=-(x+1) and |x-1|=-(x+1), and so the product of the two must just be |x+1||x-1|=(x+1)(x-1), and that's the numerator for that domain. Then just repeat similarly for the other two to deduce the answers you found.
    so whatever sign the 2 defined ranges of where x can be that's what sign i stick in front?
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    (Original post by will'o'wisp2)
    so whatever sign the 2 defined ranges of where x can be that's what sign i stick in front?
    No... do you know how the graphs of |x+1| and |x-1| look like? You must've sketched them at A-Level at least.
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    (Original post by RDKGames)
    No... do you know how the graphs of |x+1| and |x-1| look like? You must've sketched them at A-Level at least.
    yup, they look like V and the x-1 is a V a bit more to the right
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    (Original post by will'o'wisp2)
    yup, they look like V and the x-1 is a V a bit more to the right
    Yeah but you do realise that |x-a| is the same thing as saying x-a for x\geq a, and -(x-a) for x<a?

    Same thing applied here.
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    (Original post by RDKGames)
    Yeah but you do realise that |x-a| is the same thing as saying x-a for x\geq a, and -(x-a) for x<a?

    Same thing applied here.
    Yup ok so how the -1,1 work?
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    (Original post by will'o'wisp2)
    Yup ok so how the -1,1 work?
    Well.. what's is the function |x+1| equivalent to on this interval? What about |x-1|..?
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    (Original post by RDKGames)
    Well.. what's is the function |x+1| equivalent to on this interval? What about |x-1|..?
    0 and 0 if i stick -1 and 1 in?
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    (Original post by will'o'wisp2)
    0 and 0 if i stick -1 and 1 in?
    OK... not quite... Got absolutely NO idea why you're sticking -1 and 1 in there... (unless I misunderstood what you've asked)

    On (-1,1) we have |x+1|=x+1 and |x-1|=-(x-1)

    So f(x)=\frac{-(x-1)(x+1)}{x^2-1}=-1 on x \in (-1,1) as you found.

    Honestly don't know how much more basic I can go on this.

    Having that, you can clearly see what the limits of x \rightarrow (-1)_- and x\rightarrow (-1)_+ are already.
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    (Original post by RDKGames)
    OK... not quite... Got absolutely NO idea why you're sticking -1 and 1 in there... (unless I misunderstood what you've asked)

    On (-1,1) we have |x+1|=x+1 and |x-1|=-(x-1)

    So f(x)=\frac{-(x-1)(x+1)}{x^2-1}=-1 on x \in (-1,1) as you found.

    Honestly don't know how much more basic I can go on this.
    It's just this line which i don't get.
    Do i need to like look at a graph of it or something?
    the graph of |x+1| between -1 and 1 is oh i see?

    so if i make whatevers inside || and make it 0 then i can see what the equation is?
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    (Original post by will'o'wisp2)
    It's just this line which i don't get.
    Do i need to like look at a graph of it or something?
    the graph of |x+1| between -1 and 1 is oh i see?

    so if i make whatevers inside || and make it 0 then i can see what the equation is?
    A graph would help understand, yes.

    Also going back to the definition of the modulus; you have \displaystyle |x+1|= \begin{cases}x+1, & x \in (-1, \infty) \\ -(x+1), & x \in (-\infty,-1] \end{cases}

    Since x \in (-1,1) \subset (-1,\infty), you have |x+1|=x+1 in this interval.

    Get it?
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    (Original post by RDKGames)
    A graph would help understand, yes.

    Also going back to the definition of the modulus; you have \displaystyle |x+1|= \begin{cases}x+1, & x \in (-1, \infty) \\ -(x+1), & x \in (-\infty,-1] \end{cases}

    Since x \in (-1,1) \subset (-1,\infty), you have |x+1|=x+1 in this interval.

    Get it?
    Ye thanks man
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    (Original post by RDKGames)
    Your values for the domains at the end are correct.

    For the one I worked out, it's quite simply the case of taking |x+1| and |x-1| individually, and seeing what equation without the modulus they are equivalent to on the domain x \in (-\infty, -1). Clearly, you (should) know that |x+1|=-(x+1) and |x-1|=-(x+1), and so the product of the two must just be |x+1||x-1|=(x+1)(x-1), and that's the numerator for that domain. Then just repeat similarly for the other two to deduce the answers you found.
    hol on a sec how come the |x+1||x-1| in the interval -infinity and -1 are both the same thing? wouldn't it be -(x-1) and -(x+1) ??
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    (Original post by will'o'wisp2)
    hol on a sec how come the |x+1||x-1| in the interval -infinity and -1 are both the same thing? wouldn't it be -(x-1) and -(x+1) ??
    Sorry, my mistake, |x-1|=-(x-1) in that interval, as you say. Their product still remains positive as stated.
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    (Original post by RDKGames)
    Sorry, my mistake, |x-1|=-(x-1) in that interval, as you say. Their product still remains positive as stated.
    https://cdn.discordapp.com/attachmen...95/unknown.png

    so these notes from the teacher wrong too?
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    (Original post by will'o'wisp2)
    https://cdn.discordapp.com/attachmen...95/unknown.png

    so these notes from the teacher wrong too?
    Yes, that highlighted part with |x+1| is incorrect. I recommend dropping them an email about it so they can amend it as required. The rest looks fine.
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    (Original post by RDKGames)
    Yes, that highlighted part with |x+1| is incorrect. I recommend dropping them an email about it so they can amend it as required. The rest looks fine.
    Ok thanks for helping as always
 
 
 
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