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Please help with these questions-limits etc.

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    (Original post by betapro)
    Yes for 2 i got 6) so L=1 and not a clue how to do 4
    Just by looking at Q4 I'd say I would test a base case of n=1 before evaluating what happens as n \rightarrow \infty as see if the inequality still holds, examining closely the denominators if two things tend to the same thing. Just a thought though.
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    For problem 4 i got my ansdwer as (a) is correct
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    (Original post by betapro)
    Found values for which absolute values vanish and split into three intervals and then tested each one
    I got a different result when x is less than -1/2.
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    (Original post by betapro)
    For problem 4 i got my ansdwer as (a) is correct
    What was wrong with (e)?
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    (Original post by EricPiphany)
    What was wrong with (e)?
    For e i agree with first step but i dont get how u get that to 1
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    (Original post by EricPiphany)
    I got a different result when x is less than -1/2.
    Could you send screenshot of working to see where i went wrong? ty <3
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    (Original post by betapro)
    For e i agree with first step but i dont get how u get that to 1
    You can always use a little cheeky algebra
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    (Original post by betapro)
    Could you send screenshot of working to see where i went wrong? ty <3
    For x&lt;-1/2, both 2x+1, 2x-1 &lt; 0, so LHS is (-2x+1)-(-2x-1)=2 \le 2...
    Unless I've done something really dumb somewhere...
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    (Original post by EricPiphany)
    For x&lt;-1/2, both 2x+1, 2x-1 &lt; 0, so LHS is (-2x+1)-(-2x-1)=2 \le 2...
    Unless I've done something really dumb somewhere...
    Right yea ur right so if you would say that the answer is 6?
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    (Original post by EricPiphany)
    You can always use a little cheeky algebra
    Can u show me this hidden magic of algebra xD
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    (Original post by betapro)
    Right yea ur right so if you would say that the answer is 6?
    I'll let you decide on that one...
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    (Original post by EricPiphany)
    I'll let you decide on that one...
    Well for most right interval u get -2 less than equal to 2 so u take that to be right, and from previous working we know that -infinity to -0.5 is right and also the interval between -0.5 and half is also right then yes? A confirmation would be great
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    (Original post by betapro)
    Can u show me this hidden magic of algebra xD
    consider  \dfrac{2n}{3n+3} = \dfrac{2}{3} \times \dfrac{n}{n+1}...
    or simply multiply by the denominator...
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    (Original post by betapro)
    Well for most right interval u get -2 less than equal to 2 so u take that to be right, and from previous working we know that -infinity to -0.5 is right and also the interval between -0.5 and half is also right then yes? A confirmation would be great
    https://www.desmos.com/calculator/7scpx7ack1
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    Graph doesnt break or anything so i see why its -infitinty to positive infinity thanks. Still requiring some further assistance with that question 4 with the getting to one situation xD
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    (Original post by betapro)
    Graph doesnt break or anything so i see why its -infitinty to positive infinity thanks. Still requiring some further assistance with that question 4 with the getting to one situation xD
    (Original post by EricPiphany)
    consider  \dfrac{2n}{3n+3} = \dfrac{2}{3} \times \dfrac{n}{n+1}...
    or simply multiply by the denominator...
    \dfrac{n}{n+1} is clearly less than 1, so \dfrac{2}{3} \times \dfrac{n}{n+1} &lt; \dfrac{2}{3}...
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    (Original post by EricPiphany)
    \dfrac{n}{n+1} is clearly less than 1, so \dfrac{2}{3} \times \dfrac{n}{n+1} &lt; \dfrac{2}{3}...
    Yea see what u mean but surely it could never be 1, or is that just the notation you are meant to use?
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    (Original post by betapro)
    Yea see what u mean but surely it could never be 1, or is that just the notation you are meant to use?
    \le means less than or equal to.
    It is true if &lt; or = or both.

    so if x&lt;y then x \le y
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    (Original post by EricPiphany)
    \le means less than or equal to.
    It is true if &lt; or = or both.

    so if x&lt;y then x \le y
    Derp yea...

    Thanks so much for your help, just to conclude for answers:
    q1-3
    q2-6
    q3-4
    q4-2
    q5-2
    q6-6

    Also is there a way i can give u rating? <3
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    (Original post by betapro)
    Derp yea...

    Thanks so much for your help, just to conclude for answers:
    q1-3
    q2-6
    q3-4
    q4-2
    q5-2
    q6-6

    Also is there a way i can give u rating? <3
    i can't confirm them

    and not really, but you can click the green thumbs up under a post by me
 
 
 
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