Limit of a sequence with floor function.
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1
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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
asked Nov 19 at 15:23
J. Lastin
294
add a comment |
up vote
1
down vote
favorite
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
asked Nov 19 at 15:23
J. Lastin
294
What is the limit without floor function?
– coffeemath
Nov 19 at 15:31
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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
asked Nov 19 at 15:23
J. Lastin
294
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
asked Nov 19 at 15:23
J. Lastin
294
asked Nov 19 at 15:23
J. Lastin
294
asked Nov 19 at 15:23
J. Lastin
294
asked Nov 19 at 15:23
J. Lastin
294
asked Nov 19 at 15:23
J. Lastin
294
294
What is the limit without floor function?
– coffeemath
Nov 19 at 15:31
add a comment |
What is the limit without floor function?
– coffeemath
Nov 19 at 15:31
What is the limit without floor function?
– coffeemath
Nov 19 at 15:31
What is the limit without floor function?
– coffeemath
Nov 19 at 15:31
add a comment |
1 Answer
1
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2
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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.
answered Nov 19 at 15:54
Eric
1816
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
answered Nov 19 at 15:54
Eric
1816
add a comment |
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.
answered Nov 19 at 15:54
Eric
1816
add a comment |
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.
answered Nov 19 at 15:54
Eric
1816
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.
answered Nov 19 at 15:54
Eric
1816
answered Nov 19 at 15:54
Eric
1816
answered Nov 19 at 15:54
Eric
1816
answered Nov 19 at 15:54
Eric
1816
1816
add a comment |
add a comment |
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What is the limit without floor function?
– coffeemath
Nov 19 at 15:31