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Please help with these questions-limits etc. Watch

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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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