Find the limit with greatest integer function
$begingroup$
Find the following limit
$$lim_{x to 0}frac{[x]}{x}$$
Where $[x]$ is the greatest integer function.
I tried using the squeeze theorem on this but couldn’t come up with the appropriate functions, can someone help me out?
Edit: I can now see that this limit will not exist.
But what can we say about
$$lim_{x to 0}x[1/x]$$
Where again [x] is the greatest integer function, how can i use squeeze theorem to find this?
calculus limits
asked Jan 15 at 13:54
user601297user601297
41719
$endgroup$
add a comment |
$begingroup$
Find the following limit
$$lim_{x to 0}frac{[x]}{x}$$
Where $[x]$ is the greatest integer function.
I tried using the squeeze theorem on this but couldn’t come up with the appropriate functions, can someone help me out?
Edit: I can now see that this limit will not exist.
But what can we say about
$$lim_{x to 0}x[1/x]$$
Where again [x] is the greatest integer function, how can i use squeeze theorem to find this?
calculus limits
asked Jan 15 at 13:54
user601297user601297
41719
$endgroup$
$begingroup$
There's no limit. If $x>0$ is really small the term is zero. If $x<0$ is really small we get $-frac{1}{x}$ which goes to $-infty$.
$endgroup$
– Yanko
Jan 15 at 13:55
1
$begingroup$
For your edit: Well now it seems like the limit is $1$. You need to calculate the limit for $xrightarrow 0$ but $x>0$ and for $xrightarrow 0$ but $x<0$.
$endgroup$
– Yanko
Jan 15 at 14:01
$begingroup$
Are you sure you didn't mean least integer function?
$endgroup$
– Cuhrazatee
Jan 15 at 14:36
add a comment |
$begingroup$
Find the following limit
$$lim_{x to 0}frac{[x]}{x}$$
Where $[x]$ is the greatest integer function.
I tried using the squeeze theorem on this but couldn’t come up with the appropriate functions, can someone help me out?
Edit: I can now see that this limit will not exist.
But what can we say about
$$lim_{x to 0}x[1/x]$$
Where again [x] is the greatest integer function, how can i use squeeze theorem to find this?
calculus limits
asked Jan 15 at 13:54
user601297user601297
41719
$endgroup$
Find the following limit
$$lim_{x to 0}frac{[x]}{x}$$
Where $[x]$ is the greatest integer function.
I tried using the squeeze theorem on this but couldn’t come up with the appropriate functions, can someone help me out?
Edit: I can now see that this limit will not exist.
But what can we say about
$$lim_{x to 0}x[1/x]$$
Where again [x] is the greatest integer function, how can i use squeeze theorem to find this?
calculus limits
calculus limits
asked Jan 15 at 13:54
user601297user601297
41719
asked Jan 15 at 13:54
user601297user601297
41719
edited Jan 15 at 13:59
user601297
asked Jan 15 at 13:54
user601297user601297
41719
asked Jan 15 at 13:54
user601297user601297
41719
asked Jan 15 at 13:54
user601297user601297
41719
41719
$begingroup$
There's no limit. If $x>0$ is really small the term is zero. If $x<0$ is really small we get $-frac{1}{x}$ which goes to $-infty$.
$endgroup$
– Yanko
Jan 15 at 13:55
1
$begingroup$
For your edit: Well now it seems like the limit is $1$. You need to calculate the limit for $xrightarrow 0$ but $x>0$ and for $xrightarrow 0$ but $x<0$.
$endgroup$
– Yanko
Jan 15 at 14:01
$begingroup$
Are you sure you didn't mean least integer function?
$endgroup$
– Cuhrazatee
Jan 15 at 14:36
add a comment |
$begingroup$
There's no limit. If $x>0$ is really small the term is zero. If $x<0$ is really small we get $-frac{1}{x}$ which goes to $-infty$.
$endgroup$
– Yanko
Jan 15 at 13:55
1
$begingroup$
For your edit: Well now it seems like the limit is $1$. You need to calculate the limit for $xrightarrow 0$ but $x>0$ and for $xrightarrow 0$ but $x<0$.
$endgroup$
– Yanko
Jan 15 at 14:01
$begingroup$
Are you sure you didn't mean least integer function?
$endgroup$
– Cuhrazatee
Jan 15 at 14:36
$begingroup$
There's no limit. If $x>0$ is really small the term is zero. If $x<0$ is really small we get $-frac{1}{x}$ which goes to $-infty$.
$endgroup$
– Yanko
Jan 15 at 13:55
$begingroup$
There's no limit. If $x>0$ is really small the term is zero. If $x<0$ is really small we get $-frac{1}{x}$ which goes to $-infty$.
$endgroup$
– Yanko
Jan 15 at 13:55
1
1
$begingroup$
For your edit: Well now it seems like the limit is $1$. You need to calculate the limit for $xrightarrow 0$ but $x>0$ and for $xrightarrow 0$ but $x<0$.
$endgroup$
– Yanko
Jan 15 at 14:01
$begingroup$
For your edit: Well now it seems like the limit is $1$. You need to calculate the limit for $xrightarrow 0$ but $x>0$ and for $xrightarrow 0$ but $x<0$.
$endgroup$
– Yanko
Jan 15 at 14:01
$begingroup$
Are you sure you didn't mean least integer function?
$endgroup$
– Cuhrazatee
Jan 15 at 14:36
$begingroup$
Are you sure you didn't mean least integer function?
$endgroup$
– Cuhrazatee
Jan 15 at 14:36
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
if the GIF function contains a value that tends to infinity then the gif function can be removed.
hence [1/x] in the second expression will simply become 1/x.
now if the GIF contains some other expression then we must understand that the limit is applied on value given by GIF and not on the expression in the GIF.
Right Hand Limit And Left hand limit must be found(bcoz in negative inputs it gives further negative number ie [-0.2] = -1) and then solved.
Ex. $ lim_{x->0^+}[sinx/x] $
now since sine of a value is always less than the actual value hence sinx/x will be little less than 1 so GIF will return zero so
= lim_{x->0}0=0
answered Jan 15 at 14:45
Harsh WasnikHarsh Wasnik
365
$endgroup$
add a comment |
$begingroup$
Defining $u={1over x}$ we must obtain two limits$$lim_{uto infty}{lfloor urfloorover u}$$and $$lim_{uto -infty}{lfloor urfloorover u}$$for $uto infty$ we can write$${lfloor urfloorover lfloor urfloor+1}< {lfloor urfloorover u}le1$$and for $uto -infty$$$1le {lfloor urfloorover u}<{lfloor urfloorover lfloor urfloor+1}$$by Squeeze theorem, both tend to $1$ and so does $lim_{xto 0}xlfloor{1over x}rfloor$
answered Jan 15 at 15:10
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
if the GIF function contains a value that tends to infinity then the gif function can be removed.
hence [1/x] in the second expression will simply become 1/x.
now if the GIF contains some other expression then we must understand that the limit is applied on value given by GIF and not on the expression in the GIF.
Right Hand Limit And Left hand limit must be found(bcoz in negative inputs it gives further negative number ie [-0.2] = -1) and then solved.
Ex. $ lim_{x->0^+}[sinx/x] $
now since sine of a value is always less than the actual value hence sinx/x will be little less than 1 so GIF will return zero so
= lim_{x->0}0=0
answered Jan 15 at 14:45
Harsh WasnikHarsh Wasnik
365
$endgroup$
add a comment |
$begingroup$
if the GIF function contains a value that tends to infinity then the gif function can be removed.
hence [1/x] in the second expression will simply become 1/x.
now if the GIF contains some other expression then we must understand that the limit is applied on value given by GIF and not on the expression in the GIF.
Right Hand Limit And Left hand limit must be found(bcoz in negative inputs it gives further negative number ie [-0.2] = -1) and then solved.
Ex. $ lim_{x->0^+}[sinx/x] $
now since sine of a value is always less than the actual value hence sinx/x will be little less than 1 so GIF will return zero so
= lim_{x->0}0=0
answered Jan 15 at 14:45
Harsh WasnikHarsh Wasnik
365
$endgroup$
add a comment |
$begingroup$
if the GIF function contains a value that tends to infinity then the gif function can be removed.
hence [1/x] in the second expression will simply become 1/x.
now if the GIF contains some other expression then we must understand that the limit is applied on value given by GIF and not on the expression in the GIF.
Right Hand Limit And Left hand limit must be found(bcoz in negative inputs it gives further negative number ie [-0.2] = -1) and then solved.
Ex. $ lim_{x->0^+}[sinx/x] $
now since sine of a value is always less than the actual value hence sinx/x will be little less than 1 so GIF will return zero so
= lim_{x->0}0=0
answered Jan 15 at 14:45
Harsh WasnikHarsh Wasnik
365
$endgroup$
if the GIF function contains a value that tends to infinity then the gif function can be removed.
hence [1/x] in the second expression will simply become 1/x.
now if the GIF contains some other expression then we must understand that the limit is applied on value given by GIF and not on the expression in the GIF.
Right Hand Limit And Left hand limit must be found(bcoz in negative inputs it gives further negative number ie [-0.2] = -1) and then solved.
Ex. $ lim_{x->0^+}[sinx/x] $
now since sine of a value is always less than the actual value hence sinx/x will be little less than 1 so GIF will return zero so
= lim_{x->0}0=0
answered Jan 15 at 14:45
Harsh WasnikHarsh Wasnik
365
answered Jan 15 at 14:45
Harsh WasnikHarsh Wasnik
365
answered Jan 15 at 14:45
Harsh WasnikHarsh Wasnik
365
answered Jan 15 at 14:45
Harsh WasnikHarsh Wasnik
365
365
add a comment |
add a comment |
$begingroup$
Defining $u={1over x}$ we must obtain two limits$$lim_{uto infty}{lfloor urfloorover u}$$and $$lim_{uto -infty}{lfloor urfloorover u}$$for $uto infty$ we can write$${lfloor urfloorover lfloor urfloor+1}< {lfloor urfloorover u}le1$$and for $uto -infty$$$1le {lfloor urfloorover u}<{lfloor urfloorover lfloor urfloor+1}$$by Squeeze theorem, both tend to $1$ and so does $lim_{xto 0}xlfloor{1over x}rfloor$
answered Jan 15 at 15:10
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
$begingroup$
Defining $u={1over x}$ we must obtain two limits$$lim_{uto infty}{lfloor urfloorover u}$$and $$lim_{uto -infty}{lfloor urfloorover u}$$for $uto infty$ we can write$${lfloor urfloorover lfloor urfloor+1}< {lfloor urfloorover u}le1$$and for $uto -infty$$$1le {lfloor urfloorover u}<{lfloor urfloorover lfloor urfloor+1}$$by Squeeze theorem, both tend to $1$ and so does $lim_{xto 0}xlfloor{1over x}rfloor$
answered Jan 15 at 15:10
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
$begingroup$
Defining $u={1over x}$ we must obtain two limits$$lim_{uto infty}{lfloor urfloorover u}$$and $$lim_{uto -infty}{lfloor urfloorover u}$$for $uto infty$ we can write$${lfloor urfloorover lfloor urfloor+1}< {lfloor urfloorover u}le1$$and for $uto -infty$$$1le {lfloor urfloorover u}<{lfloor urfloorover lfloor urfloor+1}$$by Squeeze theorem, both tend to $1$ and so does $lim_{xto 0}xlfloor{1over x}rfloor$
answered Jan 15 at 15:10
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
Defining $u={1over x}$ we must obtain two limits$$lim_{uto infty}{lfloor urfloorover u}$$and $$lim_{uto -infty}{lfloor urfloorover u}$$for $uto infty$ we can write$${lfloor urfloorover lfloor urfloor+1}< {lfloor urfloorover u}le1$$and for $uto -infty$$$1le {lfloor urfloorover u}<{lfloor urfloorover lfloor urfloor+1}$$by Squeeze theorem, both tend to $1$ and so does $lim_{xto 0}xlfloor{1over x}rfloor$
answered Jan 15 at 15:10
Mostafa AyazMostafa Ayaz
18.1k31040
answered Jan 15 at 15:10
Mostafa AyazMostafa Ayaz
18.1k31040
answered Jan 15 at 15:10
Mostafa AyazMostafa Ayaz
18.1k31040
answered Jan 15 at 15:10
Mostafa AyazMostafa Ayaz
18.1k31040
18.1k31040
add a comment |
add a comment |
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$begingroup$
There's no limit. If $x>0$ is really small the term is zero. If $x<0$ is really small we get $-frac{1}{x}$ which goes to $-infty$.
$endgroup$
– Yanko
Jan 15 at 13:55
1
$begingroup$
For your edit: Well now it seems like the limit is $1$. You need to calculate the limit for $xrightarrow 0$ but $x>0$ and for $xrightarrow 0$ but $x<0$.
$endgroup$
– Yanko
Jan 15 at 14:01
$begingroup$
Are you sure you didn't mean least integer function?
$endgroup$
– Cuhrazatee
Jan 15 at 14:36