Limit of a sequence with floor function.











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How do I compute the following limit: $lim limits_{n to infty} frac{n +
lfloor sqrt[3]nrfloor^3}{n - lfloor sqrt{n+9}rfloor}$



Without the floor function this would be simple, but I never encountered it before so I have no idea what to do, maybe utilize the squeeze theorem somehow?






calculus limits floor-function







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  • What is the limit without floor function?
    – coffeemath
    yesterday















up vote
1
down vote

favorite
1












How do I compute the following limit: $lim limits_{n to infty} frac{n +
lfloor sqrt[3]nrfloor^3}{n - lfloor sqrt{n+9}rfloor}$



Without the floor function this would be simple, but I never encountered it before so I have no idea what to do, maybe utilize the squeeze theorem somehow?






calculus limits floor-function







share|cite|improve this question






















  • What is the limit without floor function?
    – coffeemath
    yesterday













up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





How do I compute the following limit: $lim limits_{n to infty} frac{n +
lfloor sqrt[3]nrfloor^3}{n - lfloor sqrt{n+9}rfloor}$



Without the floor function this would be simple, but I never encountered it before so I have no idea what to do, maybe utilize the squeeze theorem somehow?






calculus limits floor-function







share|cite|improve this question













How do I compute the following limit: $lim limits_{n to infty} frac{n +
lfloor sqrt[3]nrfloor^3}{n - lfloor sqrt{n+9}rfloor}$



Without the floor function this would be simple, but I never encountered it before so I have no idea what to do, maybe utilize the squeeze theorem somehow?






calculus limits floor-function




calculus limits floor-function






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asked yesterday









J. Lastin

153




153












  • What is the limit without floor function?
    – coffeemath
    yesterday


















  • What is the limit without floor function?
    – coffeemath
    yesterday
















What is the limit without floor function?
– coffeemath
yesterday




What is the limit without floor function?
– coffeemath
yesterday










1 Answer
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You can bound the floor function above and below by $n-1 leqlfloor n rfloor leq n$. This means you can bound your limit by



$$
lim frac{n+(n^frac{1}{3}-1)^3}{n-sqrt{n+9}} leq lim frac{n+lfloor n^frac{1}{3} rfloor^3}{n-lfloor sqrt{n+9}rfloor } leq lim frac{n+(n^frac{1}{3})^3}{n-sqrt{n+9}-1}
$$
as in the left most limit you have made the numerator smaller and the denominator bigger, while in the right most limit you have done the opposite.



If you compute the limits of these two sequences (and show they are equal), then you get your limit.






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    up vote
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    You can bound the floor function above and below by $n-1 leqlfloor n rfloor leq n$. This means you can bound your limit by



    $$
    lim frac{n+(n^frac{1}{3}-1)^3}{n-sqrt{n+9}} leq lim frac{n+lfloor n^frac{1}{3} rfloor^3}{n-lfloor sqrt{n+9}rfloor } leq lim frac{n+(n^frac{1}{3})^3}{n-sqrt{n+9}-1}
    $$
    as in the left most limit you have made the numerator smaller and the denominator bigger, while in the right most limit you have done the opposite.



    If you compute the limits of these two sequences (and show they are equal), then you get your limit.






    share|cite|improve this answer

























      up vote
      2
      down vote



      accepted










      You can bound the floor function above and below by $n-1 leqlfloor n rfloor leq n$. This means you can bound your limit by



      $$
      lim frac{n+(n^frac{1}{3}-1)^3}{n-sqrt{n+9}} leq lim frac{n+lfloor n^frac{1}{3} rfloor^3}{n-lfloor sqrt{n+9}rfloor } leq lim frac{n+(n^frac{1}{3})^3}{n-sqrt{n+9}-1}
      $$
      as in the left most limit you have made the numerator smaller and the denominator bigger, while in the right most limit you have done the opposite.



      If you compute the limits of these two sequences (and show they are equal), then you get your limit.






      share|cite|improve this answer























        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        You can bound the floor function above and below by $n-1 leqlfloor n rfloor leq n$. This means you can bound your limit by



        $$
        lim frac{n+(n^frac{1}{3}-1)^3}{n-sqrt{n+9}} leq lim frac{n+lfloor n^frac{1}{3} rfloor^3}{n-lfloor sqrt{n+9}rfloor } leq lim frac{n+(n^frac{1}{3})^3}{n-sqrt{n+9}-1}
        $$
        as in the left most limit you have made the numerator smaller and the denominator bigger, while in the right most limit you have done the opposite.



        If you compute the limits of these two sequences (and show they are equal), then you get your limit.






        share|cite|improve this answer












        You can bound the floor function above and below by $n-1 leqlfloor n rfloor leq n$. This means you can bound your limit by



        $$
        lim frac{n+(n^frac{1}{3}-1)^3}{n-sqrt{n+9}} leq lim frac{n+lfloor n^frac{1}{3} rfloor^3}{n-lfloor sqrt{n+9}rfloor } leq lim frac{n+(n^frac{1}{3})^3}{n-sqrt{n+9}-1}
        $$
        as in the left most limit you have made the numerator smaller and the denominator bigger, while in the right most limit you have done the opposite.



        If you compute the limits of these two sequences (and show they are equal), then you get your limit.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        Eric

        915




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