Compute $lim_{xtoinfty} x lfloor frac{1}{x} rfloor$
I'm working out a limit and I'm not sure if my assumption is considered rigorous
$$lim_{xtoinfty} xleftlfloorfrac1xrightrfloor$$
I supposed that $0leq xleftlfloorfrac1xrightrfloor leq leftlfloorfrac1xrightrfloor$ since $x$ is approaching $infty$ $($thus $x > 1$$)$ and to get the answer $0$.
Any mistakes here?
calculus limits floor-function
edited Dec 24 at 23:17
Xander Henderson
14.1k103554
add a comment |
I'm working out a limit and I'm not sure if my assumption is considered rigorous
$$lim_{xtoinfty} xleftlfloorfrac1xrightrfloor$$
I supposed that $0leq xleftlfloorfrac1xrightrfloor leq leftlfloorfrac1xrightrfloor$ since $x$ is approaching $infty$ $($thus $x > 1$$)$ and to get the answer $0$.
Any mistakes here?
calculus limits floor-function
edited Dec 24 at 23:17
Xander Henderson
14.1k103554
add a comment |
I'm working out a limit and I'm not sure if my assumption is considered rigorous
$$lim_{xtoinfty} xleftlfloorfrac1xrightrfloor$$
I supposed that $0leq xleftlfloorfrac1xrightrfloor leq leftlfloorfrac1xrightrfloor$ since $x$ is approaching $infty$ $($thus $x > 1$$)$ and to get the answer $0$.
Any mistakes here?
calculus limits floor-function
edited Dec 24 at 23:17
Xander Henderson
14.1k103554
I'm working out a limit and I'm not sure if my assumption is considered rigorous
$$lim_{xtoinfty} xleftlfloorfrac1xrightrfloor$$
I supposed that $0leq xleftlfloorfrac1xrightrfloor leq leftlfloorfrac1xrightrfloor$ since $x$ is approaching $infty$ $($thus $x > 1$$)$ and to get the answer $0$.
Any mistakes here?
calculus limits floor-function
calculus limits floor-function
edited Dec 24 at 23:17
Xander Henderson
14.1k103554
edited Dec 24 at 23:17
Xander Henderson
14.1k103554
edited Dec 24 at 23:17
Xander Henderson
14.1k103554
edited Dec 24 at 23:17
Xander Henderson
14.1k103554
edited Dec 24 at 23:17
Xander Henderson
14.1k103554
14.1k103554
asked Dec 24 at 12:53
user531476
add a comment |
add a comment |
3 Answers
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No mistakes. But, can you see it in terms of sequences instead? That will be much easier. $limlimits_{x rightarrow infty}$ can be viewed as a sequence $leftlbrace x_n rightrbrace$, where $forall M > 0, exists N in mathbb{N}$ such that $x_n > M$. This tells us something special for the choice of $M geq 1$. Once we choose $M geq 1$, we will get a stage after which $x_n > 1$ and hence $leftlfloor{dfrac{1}{x_n}}rightrfloor = 0$. Hence, the image sequence is evetually zero and $limlimits_{x rightarrow infty} x leftlfloor{dfrac{1}{x}}rightrfloor = 0$.
edited Dec 24 at 15:41
BPP
2,170927
answered Dec 24 at 13:00
Aniruddha Deshmukh
899418
Thank you, I see what you did there. (right click the equation and under "Show Math As" select Tex or MathML ;)
– user531476
Dec 24 at 13:34
add a comment |
You make it sound like the reason that $xlfloorfrac{1}{x}rfloorleqlfloorfrac{1}{x}rfloor$ is true for positive $x$ is because $xyleq y$ for positive $x$ and $y$, but this is not so. Instead, the reason it is true is because if $xleq 1$ then $xyleq y$ for positive $y$ and if $x>1$ then $lfloorfrac{1}{x}rfloor=0$ so that $xlfloorfrac{1}{x}rfloor=0$ too.
It is far simpler just to note that if $x>1$ then $lfloorfrac{1}{x}rfloor=0$ and hence $xlfloorfrac{1}{x}rfloor=0$, giving us $lim_{xtoinfty}xlfloorfrac{1}{x}rfloor=0$.
answered Dec 24 at 13:03
Ben W
1,363513
add a comment |
The function is $0$ for $x>1$. Alternatively, substitute $m=1/x$. This limit is the same as $displaystylelim_{mto0^+}frac{lfloor mrfloor}m$, which easily evaluates to $0$.
answered Dec 24 at 13:33
Shubham Johri
3,826716
How does it easily evaluate to 0?
– user531476
Dec 24 at 13:39
$displaystylelim_{mto0^+}frac{lfloor mrfloor}m=displaystylelim_{mto0^+}frac{0}m=0$. This is due to the fact that $lfloor mrfloor$ is exactly $0$, but $m$ only tends to $0$
– Shubham Johri
Dec 24 at 13:41
Oh, I thought we take that as $frac{0}{0}$. Thank you!
– user531476
Dec 24 at 14:28
add a comment |
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3 Answers
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active
oldest
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3 Answers
3
active
oldest
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No mistakes. But, can you see it in terms of sequences instead? That will be much easier. $limlimits_{x rightarrow infty}$ can be viewed as a sequence $leftlbrace x_n rightrbrace$, where $forall M > 0, exists N in mathbb{N}$ such that $x_n > M$. This tells us something special for the choice of $M geq 1$. Once we choose $M geq 1$, we will get a stage after which $x_n > 1$ and hence $leftlfloor{dfrac{1}{x_n}}rightrfloor = 0$. Hence, the image sequence is evetually zero and $limlimits_{x rightarrow infty} x leftlfloor{dfrac{1}{x}}rightrfloor = 0$.
edited Dec 24 at 15:41
BPP
2,170927
answered Dec 24 at 13:00
Aniruddha Deshmukh
899418
Thank you, I see what you did there. (right click the equation and under "Show Math As" select Tex or MathML ;)
– user531476
Dec 24 at 13:34
add a comment |
No mistakes. But, can you see it in terms of sequences instead? That will be much easier. $limlimits_{x rightarrow infty}$ can be viewed as a sequence $leftlbrace x_n rightrbrace$, where $forall M > 0, exists N in mathbb{N}$ such that $x_n > M$. This tells us something special for the choice of $M geq 1$. Once we choose $M geq 1$, we will get a stage after which $x_n > 1$ and hence $leftlfloor{dfrac{1}{x_n}}rightrfloor = 0$. Hence, the image sequence is evetually zero and $limlimits_{x rightarrow infty} x leftlfloor{dfrac{1}{x}}rightrfloor = 0$.
edited Dec 24 at 15:41
BPP
2,170927
answered Dec 24 at 13:00
Aniruddha Deshmukh
899418
Thank you, I see what you did there. (right click the equation and under "Show Math As" select Tex or MathML ;)
– user531476
Dec 24 at 13:34
add a comment |
No mistakes. But, can you see it in terms of sequences instead? That will be much easier. $limlimits_{x rightarrow infty}$ can be viewed as a sequence $leftlbrace x_n rightrbrace$, where $forall M > 0, exists N in mathbb{N}$ such that $x_n > M$. This tells us something special for the choice of $M geq 1$. Once we choose $M geq 1$, we will get a stage after which $x_n > 1$ and hence $leftlfloor{dfrac{1}{x_n}}rightrfloor = 0$. Hence, the image sequence is evetually zero and $limlimits_{x rightarrow infty} x leftlfloor{dfrac{1}{x}}rightrfloor = 0$.
edited Dec 24 at 15:41
BPP
2,170927
answered Dec 24 at 13:00
Aniruddha Deshmukh
899418
No mistakes. But, can you see it in terms of sequences instead? That will be much easier. $limlimits_{x rightarrow infty}$ can be viewed as a sequence $leftlbrace x_n rightrbrace$, where $forall M > 0, exists N in mathbb{N}$ such that $x_n > M$. This tells us something special for the choice of $M geq 1$. Once we choose $M geq 1$, we will get a stage after which $x_n > 1$ and hence $leftlfloor{dfrac{1}{x_n}}rightrfloor = 0$. Hence, the image sequence is evetually zero and $limlimits_{x rightarrow infty} x leftlfloor{dfrac{1}{x}}rightrfloor = 0$.
edited Dec 24 at 15:41
BPP
2,170927
answered Dec 24 at 13:00
Aniruddha Deshmukh
899418
edited Dec 24 at 15:41
BPP
2,170927
edited Dec 24 at 15:41
BPP
2,170927
edited Dec 24 at 15:41
BPP
2,170927
2,170927
answered Dec 24 at 13:00
Aniruddha Deshmukh
899418
answered Dec 24 at 13:00
Aniruddha Deshmukh
899418
answered Dec 24 at 13:00
Aniruddha Deshmukh
899418
899418
Thank you, I see what you did there. (right click the equation and under "Show Math As" select Tex or MathML ;)
– user531476
Dec 24 at 13:34
add a comment |
Thank you, I see what you did there. (right click the equation and under "Show Math As" select Tex or MathML ;)
– user531476
Dec 24 at 13:34
Thank you, I see what you did there. (right click the equation and under "Show Math As" select Tex or MathML ;)
– user531476
Dec 24 at 13:34
Thank you, I see what you did there. (right click the equation and under "Show Math As" select Tex or MathML ;)
– user531476
Dec 24 at 13:34
add a comment |
You make it sound like the reason that $xlfloorfrac{1}{x}rfloorleqlfloorfrac{1}{x}rfloor$ is true for positive $x$ is because $xyleq y$ for positive $x$ and $y$, but this is not so. Instead, the reason it is true is because if $xleq 1$ then $xyleq y$ for positive $y$ and if $x>1$ then $lfloorfrac{1}{x}rfloor=0$ so that $xlfloorfrac{1}{x}rfloor=0$ too.
It is far simpler just to note that if $x>1$ then $lfloorfrac{1}{x}rfloor=0$ and hence $xlfloorfrac{1}{x}rfloor=0$, giving us $lim_{xtoinfty}xlfloorfrac{1}{x}rfloor=0$.
answered Dec 24 at 13:03
Ben W
1,363513
add a comment |
You make it sound like the reason that $xlfloorfrac{1}{x}rfloorleqlfloorfrac{1}{x}rfloor$ is true for positive $x$ is because $xyleq y$ for positive $x$ and $y$, but this is not so. Instead, the reason it is true is because if $xleq 1$ then $xyleq y$ for positive $y$ and if $x>1$ then $lfloorfrac{1}{x}rfloor=0$ so that $xlfloorfrac{1}{x}rfloor=0$ too.
It is far simpler just to note that if $x>1$ then $lfloorfrac{1}{x}rfloor=0$ and hence $xlfloorfrac{1}{x}rfloor=0$, giving us $lim_{xtoinfty}xlfloorfrac{1}{x}rfloor=0$.
answered Dec 24 at 13:03
Ben W
1,363513
add a comment |
You make it sound like the reason that $xlfloorfrac{1}{x}rfloorleqlfloorfrac{1}{x}rfloor$ is true for positive $x$ is because $xyleq y$ for positive $x$ and $y$, but this is not so. Instead, the reason it is true is because if $xleq 1$ then $xyleq y$ for positive $y$ and if $x>1$ then $lfloorfrac{1}{x}rfloor=0$ so that $xlfloorfrac{1}{x}rfloor=0$ too.
It is far simpler just to note that if $x>1$ then $lfloorfrac{1}{x}rfloor=0$ and hence $xlfloorfrac{1}{x}rfloor=0$, giving us $lim_{xtoinfty}xlfloorfrac{1}{x}rfloor=0$.
answered Dec 24 at 13:03
Ben W
1,363513
You make it sound like the reason that $xlfloorfrac{1}{x}rfloorleqlfloorfrac{1}{x}rfloor$ is true for positive $x$ is because $xyleq y$ for positive $x$ and $y$, but this is not so. Instead, the reason it is true is because if $xleq 1$ then $xyleq y$ for positive $y$ and if $x>1$ then $lfloorfrac{1}{x}rfloor=0$ so that $xlfloorfrac{1}{x}rfloor=0$ too.
It is far simpler just to note that if $x>1$ then $lfloorfrac{1}{x}rfloor=0$ and hence $xlfloorfrac{1}{x}rfloor=0$, giving us $lim_{xtoinfty}xlfloorfrac{1}{x}rfloor=0$.
answered Dec 24 at 13:03
Ben W
1,363513
answered Dec 24 at 13:03
Ben W
1,363513
answered Dec 24 at 13:03
Ben W
1,363513
answered Dec 24 at 13:03
Ben W
1,363513
1,363513
add a comment |
add a comment |
The function is $0$ for $x>1$. Alternatively, substitute $m=1/x$. This limit is the same as $displaystylelim_{mto0^+}frac{lfloor mrfloor}m$, which easily evaluates to $0$.
answered Dec 24 at 13:33
Shubham Johri
3,826716
How does it easily evaluate to 0?
– user531476
Dec 24 at 13:39
$displaystylelim_{mto0^+}frac{lfloor mrfloor}m=displaystylelim_{mto0^+}frac{0}m=0$. This is due to the fact that $lfloor mrfloor$ is exactly $0$, but $m$ only tends to $0$
– Shubham Johri
Dec 24 at 13:41
Oh, I thought we take that as $frac{0}{0}$. Thank you!
– user531476
Dec 24 at 14:28
add a comment |
The function is $0$ for $x>1$. Alternatively, substitute $m=1/x$. This limit is the same as $displaystylelim_{mto0^+}frac{lfloor mrfloor}m$, which easily evaluates to $0$.
answered Dec 24 at 13:33
Shubham Johri
3,826716
How does it easily evaluate to 0?
– user531476
Dec 24 at 13:39
$displaystylelim_{mto0^+}frac{lfloor mrfloor}m=displaystylelim_{mto0^+}frac{0}m=0$. This is due to the fact that $lfloor mrfloor$ is exactly $0$, but $m$ only tends to $0$
– Shubham Johri
Dec 24 at 13:41
Oh, I thought we take that as $frac{0}{0}$. Thank you!
– user531476
Dec 24 at 14:28
add a comment |
The function is $0$ for $x>1$. Alternatively, substitute $m=1/x$. This limit is the same as $displaystylelim_{mto0^+}frac{lfloor mrfloor}m$, which easily evaluates to $0$.
answered Dec 24 at 13:33
Shubham Johri
3,826716
The function is $0$ for $x>1$. Alternatively, substitute $m=1/x$. This limit is the same as $displaystylelim_{mto0^+}frac{lfloor mrfloor}m$, which easily evaluates to $0$.
answered Dec 24 at 13:33
Shubham Johri
3,826716
answered Dec 24 at 13:33
Shubham Johri
3,826716
answered Dec 24 at 13:33
Shubham Johri
3,826716
answered Dec 24 at 13:33
Shubham Johri
3,826716
3,826716
How does it easily evaluate to 0?
– user531476
Dec 24 at 13:39
$displaystylelim_{mto0^+}frac{lfloor mrfloor}m=displaystylelim_{mto0^+}frac{0}m=0$. This is due to the fact that $lfloor mrfloor$ is exactly $0$, but $m$ only tends to $0$
– Shubham Johri
Dec 24 at 13:41
Oh, I thought we take that as $frac{0}{0}$. Thank you!
– user531476
Dec 24 at 14:28
add a comment |
How does it easily evaluate to 0?
– user531476
Dec 24 at 13:39
$displaystylelim_{mto0^+}frac{lfloor mrfloor}m=displaystylelim_{mto0^+}frac{0}m=0$. This is due to the fact that $lfloor mrfloor$ is exactly $0$, but $m$ only tends to $0$
– Shubham Johri
Dec 24 at 13:41
Oh, I thought we take that as $frac{0}{0}$. Thank you!
– user531476
Dec 24 at 14:28
How does it easily evaluate to 0?
– user531476
Dec 24 at 13:39
How does it easily evaluate to 0?
– user531476
Dec 24 at 13:39
$displaystylelim_{mto0^+}frac{lfloor mrfloor}m=displaystylelim_{mto0^+}frac{0}m=0$. This is due to the fact that $lfloor mrfloor$ is exactly $0$, but $m$ only tends to $0$
– Shubham Johri
Dec 24 at 13:41
$displaystylelim_{mto0^+}frac{lfloor mrfloor}m=displaystylelim_{mto0^+}frac{0}m=0$. This is due to the fact that $lfloor mrfloor$ is exactly $0$, but $m$ only tends to $0$
– Shubham Johri
Dec 24 at 13:41
Oh, I thought we take that as $frac{0}{0}$. Thank you!
– user531476
Dec 24 at 14:28
Oh, I thought we take that as $frac{0}{0}$. Thank you!
– user531476
Dec 24 at 14:28
add a comment |
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