Value of $Bigllfloor,lim_xto0fracsin xxBigrrfloor$
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
What is the value of $Bigllfloor,lim_xto 0fracsin xxBigrrfloor$? Is it $1$ or $0$?
I was told that the answer is $0$ as $sinx$ is less than $x$ as $xrightarrow0$. Is it correct or are limits exact values?
I know that $lim_xto0Bigllfloorfracsin xxBigrrfloor$ will be $0$ due to the above mentioned fact.
calculus limits
edited Sep 4 at 13:36
gimusi
72.8k73889
asked Sep 4 at 11:46
Harshit Joshi
17512
add a comment |Â
up vote
0
down vote
favorite
What is the value of $Bigllfloor,lim_xto 0fracsin xxBigrrfloor$? Is it $1$ or $0$?
I was told that the answer is $0$ as $sinx$ is less than $x$ as $xrightarrow0$. Is it correct or are limits exact values?
I know that $lim_xto0Bigllfloorfracsin xxBigrrfloor$ will be $0$ due to the above mentioned fact.
calculus limits
edited Sep 4 at 13:36
gimusi
72.8k73889
asked Sep 4 at 11:46
Harshit Joshi
17512
1
$$ lfloor 1 rfloor = 1 quadtextandquad lim_xto 0fracsin xx = 1 $$
– Sobi
Sep 4 at 11:51
1
Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
– Shaun
Sep 4 at 11:58
@Shaun Will keep that in mind next time. I thought titles like this make your question appear interesting.
– Harshit Joshi
Sep 4 at 12:01
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
What is the value of $Bigllfloor,lim_xto 0fracsin xxBigrrfloor$? Is it $1$ or $0$?
I was told that the answer is $0$ as $sinx$ is less than $x$ as $xrightarrow0$. Is it correct or are limits exact values?
I know that $lim_xto0Bigllfloorfracsin xxBigrrfloor$ will be $0$ due to the above mentioned fact.
calculus limits
edited Sep 4 at 13:36
gimusi
72.8k73889
asked Sep 4 at 11:46
Harshit Joshi
17512
What is the value of $Bigllfloor,lim_xto 0fracsin xxBigrrfloor$? Is it $1$ or $0$?
I was told that the answer is $0$ as $sinx$ is less than $x$ as $xrightarrow0$. Is it correct or are limits exact values?
I know that $lim_xto0Bigllfloorfracsin xxBigrrfloor$ will be $0$ due to the above mentioned fact.
calculus limits
calculus limits
edited Sep 4 at 13:36
gimusi
72.8k73889
asked Sep 4 at 11:46
Harshit Joshi
17512
edited Sep 4 at 13:36
gimusi
72.8k73889
asked Sep 4 at 11:46
Harshit Joshi
17512
edited Sep 4 at 13:36
gimusi
72.8k73889
edited Sep 4 at 13:36
gimusi
72.8k73889
edited Sep 4 at 13:36
gimusi
72.8k73889
72.8k73889
asked Sep 4 at 11:46
Harshit Joshi
17512
asked Sep 4 at 11:46
Harshit Joshi
17512
asked Sep 4 at 11:46
Harshit Joshi
17512
17512
1
$$ lfloor 1 rfloor = 1 quadtextandquad lim_xto 0fracsin xx = 1 $$
– Sobi
Sep 4 at 11:51
1
Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
– Shaun
Sep 4 at 11:58
@Shaun Will keep that in mind next time. I thought titles like this make your question appear interesting.
– Harshit Joshi
Sep 4 at 12:01
add a comment |Â
1
$$ lfloor 1 rfloor = 1 quadtextandquad lim_xto 0fracsin xx = 1 $$
– Sobi
Sep 4 at 11:51
1
Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
– Shaun
Sep 4 at 11:58
@Shaun Will keep that in mind next time. I thought titles like this make your question appear interesting.
– Harshit Joshi
Sep 4 at 12:01
1
1
$$ lfloor 1 rfloor = 1 quadtextandquad lim_xto 0fracsin xx = 1 $$
– Sobi
Sep 4 at 11:51
$$ lfloor 1 rfloor = 1 quadtextandquad lim_xto 0fracsin xx = 1 $$
– Sobi
Sep 4 at 11:51
1
1
Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
– Shaun
Sep 4 at 11:58
Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
– Shaun
Sep 4 at 11:58
@Shaun Will keep that in mind next time. I thought titles like this make your question appear interesting.
– Harshit Joshi
Sep 4 at 12:01
@Shaun Will keep that in mind next time. I thought titles like this make your question appear interesting.
– Harshit Joshi
Sep 4 at 12:01
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
6
down vote
accepted
We have
$$lim_xto 0left(fracsin xxright)=lim_xto 0fraccos x1=cos (0) = 1.$$
Since $lfloor 1rfloor=1$ it follows that $leftlfloorleft(lim_xto 0left(fracsin xxright)right)rightrfloor=1$.
edited Sep 4 at 13:39
Arnaud D.
14.9k52142
answered Sep 4 at 11:52
YukiJ
1,6792624
So limits are exact values and not values tending to a number, am i right?
– Harshit Joshi
Sep 4 at 11:59
@nicomezi So are limits not exact values?
– Harshit Joshi
Sep 4 at 12:02
If they exist, yes. However, it might be that a limit does not exist (for example $lim_x to infty x$ does not exist).
– YukiJ
Sep 4 at 12:02
I missread the question. The answer is correct.
– nicomezi
Sep 4 at 12:03
use to instead of -> for tending
– Deepesh Meena
Sep 4 at 12:30
 |Â
show 1 more comment
up vote
4
down vote
Since
- $lim_xto0(fracsin xx)=1 $
- and for $xne 0$ sufficiently small: $0<fracsin xx<1$
we have that
$$Bigllfloor,lim_xto0fracsin xxBigrrfloor=1$$
and
$$lim_xto0Bigllfloorfracsin xxBigrrfloor=0$$
answered Sep 4 at 13:10
gimusi
72.8k73889
Thanks, but the question is answered already.
– Harshit Joshi
Sep 4 at 13:11
@HarshitJoshi Yes I see that, I've only add since the second answer is wrong and my aim was to give a full picture for the difference between the two cases.
– gimusi
Sep 4 at 13:15
Why is it wrong?
– Harshit Joshi
Sep 4 at 13:17
@HarshitJoshi As you mentioned for any $xne 0$ we have $0<fracsin xx<1$ and therefore $$Bigllfloorfracsin xxBigrrfloor=0 implies lim_x->0Bigllfloorfracsin xxBigrrfloor=0$$
– gimusi
Sep 4 at 13:20
Still don't understand why is it wrong?
– Harshit Joshi
Sep 4 at 13:22
 |Â
show 4 more comments
up vote
2
down vote
By the Maclaurin's expansion of $sin(x)$, we have,
$$sin(x) =sum^infty_k=0frac(-1)^k(2k+1)!x^2k+1=x-fracx^33!+fracx^55!-cdots$$
As $xto0$ we compute left hand limit,
$$lim_xto0^-fracsin(x)x =lim_xto0^-left( 1-fracx^23!+fracx^45!-cdotsright) $$
$$lim_xto0^-fracsin(x)x = 1$$
Further for right hand limit,
$$lim_xto0^+fracsin(x)x =lim_xto0^+left( 1-fracx^23!+fracx^45!-cdotsright) $$
$$lim_xto0^+fracsin(x)x = 1$$
Now,
$$Bigllfloorlim_xto0^+fracsin(x)xBigrrfloor = Bigllfloorlim_xto0^-fracsin(x)xBigrrfloor =1 $$
answered Sep 4 at 12:21
Sahil Silare
668
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
We have
$$lim_xto 0left(fracsin xxright)=lim_xto 0fraccos x1=cos (0) = 1.$$
Since $lfloor 1rfloor=1$ it follows that $leftlfloorleft(lim_xto 0left(fracsin xxright)right)rightrfloor=1$.
edited Sep 4 at 13:39
Arnaud D.
14.9k52142
answered Sep 4 at 11:52
YukiJ
1,6792624
So limits are exact values and not values tending to a number, am i right?
– Harshit Joshi
Sep 4 at 11:59
@nicomezi So are limits not exact values?
– Harshit Joshi
Sep 4 at 12:02
If they exist, yes. However, it might be that a limit does not exist (for example $lim_x to infty x$ does not exist).
– YukiJ
Sep 4 at 12:02
I missread the question. The answer is correct.
– nicomezi
Sep 4 at 12:03
use to instead of -> for tending
– Deepesh Meena
Sep 4 at 12:30
 |Â
show 1 more comment
up vote
6
down vote
accepted
We have
$$lim_xto 0left(fracsin xxright)=lim_xto 0fraccos x1=cos (0) = 1.$$
Since $lfloor 1rfloor=1$ it follows that $leftlfloorleft(lim_xto 0left(fracsin xxright)right)rightrfloor=1$.
edited Sep 4 at 13:39
Arnaud D.
14.9k52142
answered Sep 4 at 11:52
YukiJ
1,6792624
So limits are exact values and not values tending to a number, am i right?
– Harshit Joshi
Sep 4 at 11:59
@nicomezi So are limits not exact values?
– Harshit Joshi
Sep 4 at 12:02
If they exist, yes. However, it might be that a limit does not exist (for example $lim_x to infty x$ does not exist).
– YukiJ
Sep 4 at 12:02
I missread the question. The answer is correct.
– nicomezi
Sep 4 at 12:03
use to instead of -> for tending
– Deepesh Meena
Sep 4 at 12:30
 |Â
show 1 more comment
up vote
6
down vote
accepted
up vote
6
down vote
accepted
We have
$$lim_xto 0left(fracsin xxright)=lim_xto 0fraccos x1=cos (0) = 1.$$
Since $lfloor 1rfloor=1$ it follows that $leftlfloorleft(lim_xto 0left(fracsin xxright)right)rightrfloor=1$.
edited Sep 4 at 13:39
Arnaud D.
14.9k52142
answered Sep 4 at 11:52
YukiJ
1,6792624
We have
$$lim_xto 0left(fracsin xxright)=lim_xto 0fraccos x1=cos (0) = 1.$$
Since $lfloor 1rfloor=1$ it follows that $leftlfloorleft(lim_xto 0left(fracsin xxright)right)rightrfloor=1$.
edited Sep 4 at 13:39
Arnaud D.
14.9k52142
answered Sep 4 at 11:52
YukiJ
1,6792624
edited Sep 4 at 13:39
Arnaud D.
14.9k52142
edited Sep 4 at 13:39
Arnaud D.
14.9k52142
edited Sep 4 at 13:39
Arnaud D.
14.9k52142
14.9k52142
answered Sep 4 at 11:52
YukiJ
1,6792624
answered Sep 4 at 11:52
YukiJ
1,6792624
answered Sep 4 at 11:52
YukiJ
1,6792624
1,6792624
So limits are exact values and not values tending to a number, am i right?
– Harshit Joshi
Sep 4 at 11:59
@nicomezi So are limits not exact values?
– Harshit Joshi
Sep 4 at 12:02
If they exist, yes. However, it might be that a limit does not exist (for example $lim_x to infty x$ does not exist).
– YukiJ
Sep 4 at 12:02
I missread the question. The answer is correct.
– nicomezi
Sep 4 at 12:03
use to instead of -> for tending
– Deepesh Meena
Sep 4 at 12:30
 |Â
show 1 more comment
So limits are exact values and not values tending to a number, am i right?
– Harshit Joshi
Sep 4 at 11:59
@nicomezi So are limits not exact values?
– Harshit Joshi
Sep 4 at 12:02
If they exist, yes. However, it might be that a limit does not exist (for example $lim_x to infty x$ does not exist).
– YukiJ
Sep 4 at 12:02
I missread the question. The answer is correct.
– nicomezi
Sep 4 at 12:03
use to instead of -> for tending
– Deepesh Meena
Sep 4 at 12:30
So limits are exact values and not values tending to a number, am i right?
– Harshit Joshi
Sep 4 at 11:59
So limits are exact values and not values tending to a number, am i right?
– Harshit Joshi
Sep 4 at 11:59
@nicomezi So are limits not exact values?
– Harshit Joshi
Sep 4 at 12:02
@nicomezi So are limits not exact values?
– Harshit Joshi
Sep 4 at 12:02
If they exist, yes. However, it might be that a limit does not exist (for example $lim_x to infty x$ does not exist).
– YukiJ
Sep 4 at 12:02
If they exist, yes. However, it might be that a limit does not exist (for example $lim_x to infty x$ does not exist).
– YukiJ
Sep 4 at 12:02
I missread the question. The answer is correct.
– nicomezi
Sep 4 at 12:03
I missread the question. The answer is correct.
– nicomezi
Sep 4 at 12:03
use to instead of -> for tending
– Deepesh Meena
Sep 4 at 12:30
use to instead of -> for tending
– Deepesh Meena
Sep 4 at 12:30
 |Â
show 1 more comment
up vote
4
down vote
Since
- $lim_xto0(fracsin xx)=1 $
- and for $xne 0$ sufficiently small: $0<fracsin xx<1$
we have that
$$Bigllfloor,lim_xto0fracsin xxBigrrfloor=1$$
and
$$lim_xto0Bigllfloorfracsin xxBigrrfloor=0$$
answered Sep 4 at 13:10
gimusi
72.8k73889
Thanks, but the question is answered already.
– Harshit Joshi
Sep 4 at 13:11
@HarshitJoshi Yes I see that, I've only add since the second answer is wrong and my aim was to give a full picture for the difference between the two cases.
– gimusi
Sep 4 at 13:15
Why is it wrong?
– Harshit Joshi
Sep 4 at 13:17
@HarshitJoshi As you mentioned for any $xne 0$ we have $0<fracsin xx<1$ and therefore $$Bigllfloorfracsin xxBigrrfloor=0 implies lim_x->0Bigllfloorfracsin xxBigrrfloor=0$$
– gimusi
Sep 4 at 13:20
Still don't understand why is it wrong?
– Harshit Joshi
Sep 4 at 13:22
 |Â
show 4 more comments
up vote
4
down vote
Since
- $lim_xto0(fracsin xx)=1 $
- and for $xne 0$ sufficiently small: $0<fracsin xx<1$
we have that
$$Bigllfloor,lim_xto0fracsin xxBigrrfloor=1$$
and
$$lim_xto0Bigllfloorfracsin xxBigrrfloor=0$$
answered Sep 4 at 13:10
gimusi
72.8k73889
Thanks, but the question is answered already.
– Harshit Joshi
Sep 4 at 13:11
@HarshitJoshi Yes I see that, I've only add since the second answer is wrong and my aim was to give a full picture for the difference between the two cases.
– gimusi
Sep 4 at 13:15
Why is it wrong?
– Harshit Joshi
Sep 4 at 13:17
@HarshitJoshi As you mentioned for any $xne 0$ we have $0<fracsin xx<1$ and therefore $$Bigllfloorfracsin xxBigrrfloor=0 implies lim_x->0Bigllfloorfracsin xxBigrrfloor=0$$
– gimusi
Sep 4 at 13:20
Still don't understand why is it wrong?
– Harshit Joshi
Sep 4 at 13:22
 |Â
show 4 more comments
up vote
4
down vote
up vote
4
down vote
Since
- $lim_xto0(fracsin xx)=1 $
- and for $xne 0$ sufficiently small: $0<fracsin xx<1$
we have that
$$Bigllfloor,lim_xto0fracsin xxBigrrfloor=1$$
and
$$lim_xto0Bigllfloorfracsin xxBigrrfloor=0$$
answered Sep 4 at 13:10
gimusi
72.8k73889
Since
- $lim_xto0(fracsin xx)=1 $
- and for $xne 0$ sufficiently small: $0<fracsin xx<1$
we have that
$$Bigllfloor,lim_xto0fracsin xxBigrrfloor=1$$
and
$$lim_xto0Bigllfloorfracsin xxBigrrfloor=0$$
answered Sep 4 at 13:10
gimusi
72.8k73889
edited Sep 4 at 13:29
answered Sep 4 at 13:10
gimusi
72.8k73889
answered Sep 4 at 13:10
gimusi
72.8k73889
answered Sep 4 at 13:10
gimusi
72.8k73889
72.8k73889
Thanks, but the question is answered already.
– Harshit Joshi
Sep 4 at 13:11
@HarshitJoshi Yes I see that, I've only add since the second answer is wrong and my aim was to give a full picture for the difference between the two cases.
– gimusi
Sep 4 at 13:15
Why is it wrong?
– Harshit Joshi
Sep 4 at 13:17
@HarshitJoshi As you mentioned for any $xne 0$ we have $0<fracsin xx<1$ and therefore $$Bigllfloorfracsin xxBigrrfloor=0 implies lim_x->0Bigllfloorfracsin xxBigrrfloor=0$$
– gimusi
Sep 4 at 13:20
Still don't understand why is it wrong?
– Harshit Joshi
Sep 4 at 13:22
 |Â
show 4 more comments
Thanks, but the question is answered already.
– Harshit Joshi
Sep 4 at 13:11
@HarshitJoshi Yes I see that, I've only add since the second answer is wrong and my aim was to give a full picture for the difference between the two cases.
– gimusi
Sep 4 at 13:15
Why is it wrong?
– Harshit Joshi
Sep 4 at 13:17
@HarshitJoshi As you mentioned for any $xne 0$ we have $0<fracsin xx<1$ and therefore $$Bigllfloorfracsin xxBigrrfloor=0 implies lim_x->0Bigllfloorfracsin xxBigrrfloor=0$$
– gimusi
Sep 4 at 13:20
Still don't understand why is it wrong?
– Harshit Joshi
Sep 4 at 13:22
Thanks, but the question is answered already.
– Harshit Joshi
Sep 4 at 13:11
Thanks, but the question is answered already.
– Harshit Joshi
Sep 4 at 13:11
@HarshitJoshi Yes I see that, I've only add since the second answer is wrong and my aim was to give a full picture for the difference between the two cases.
– gimusi
Sep 4 at 13:15
@HarshitJoshi Yes I see that, I've only add since the second answer is wrong and my aim was to give a full picture for the difference between the two cases.
– gimusi
Sep 4 at 13:15
Why is it wrong?
– Harshit Joshi
Sep 4 at 13:17
Why is it wrong?
– Harshit Joshi
Sep 4 at 13:17
@HarshitJoshi As you mentioned for any $xne 0$ we have $0<fracsin xx<1$ and therefore $$Bigllfloorfracsin xxBigrrfloor=0 implies lim_x->0Bigllfloorfracsin xxBigrrfloor=0$$
– gimusi
Sep 4 at 13:20
@HarshitJoshi As you mentioned for any $xne 0$ we have $0<fracsin xx<1$ and therefore $$Bigllfloorfracsin xxBigrrfloor=0 implies lim_x->0Bigllfloorfracsin xxBigrrfloor=0$$
– gimusi
Sep 4 at 13:20
Still don't understand why is it wrong?
– Harshit Joshi
Sep 4 at 13:22
Still don't understand why is it wrong?
– Harshit Joshi
Sep 4 at 13:22
 |Â
show 4 more comments
up vote
2
down vote
By the Maclaurin's expansion of $sin(x)$, we have,
$$sin(x) =sum^infty_k=0frac(-1)^k(2k+1)!x^2k+1=x-fracx^33!+fracx^55!-cdots$$
As $xto0$ we compute left hand limit,
$$lim_xto0^-fracsin(x)x =lim_xto0^-left( 1-fracx^23!+fracx^45!-cdotsright) $$
$$lim_xto0^-fracsin(x)x = 1$$
Further for right hand limit,
$$lim_xto0^+fracsin(x)x =lim_xto0^+left( 1-fracx^23!+fracx^45!-cdotsright) $$
$$lim_xto0^+fracsin(x)x = 1$$
Now,
$$Bigllfloorlim_xto0^+fracsin(x)xBigrrfloor = Bigllfloorlim_xto0^-fracsin(x)xBigrrfloor =1 $$
answered Sep 4 at 12:21
Sahil Silare
668
add a comment |Â
up vote
2
down vote
By the Maclaurin's expansion of $sin(x)$, we have,
$$sin(x) =sum^infty_k=0frac(-1)^k(2k+1)!x^2k+1=x-fracx^33!+fracx^55!-cdots$$
As $xto0$ we compute left hand limit,
$$lim_xto0^-fracsin(x)x =lim_xto0^-left( 1-fracx^23!+fracx^45!-cdotsright) $$
$$lim_xto0^-fracsin(x)x = 1$$
Further for right hand limit,
$$lim_xto0^+fracsin(x)x =lim_xto0^+left( 1-fracx^23!+fracx^45!-cdotsright) $$
$$lim_xto0^+fracsin(x)x = 1$$
Now,
$$Bigllfloorlim_xto0^+fracsin(x)xBigrrfloor = Bigllfloorlim_xto0^-fracsin(x)xBigrrfloor =1 $$
answered Sep 4 at 12:21
Sahil Silare
668
add a comment |Â
up vote
2
down vote
up vote
2
down vote
By the Maclaurin's expansion of $sin(x)$, we have,
$$sin(x) =sum^infty_k=0frac(-1)^k(2k+1)!x^2k+1=x-fracx^33!+fracx^55!-cdots$$
As $xto0$ we compute left hand limit,
$$lim_xto0^-fracsin(x)x =lim_xto0^-left( 1-fracx^23!+fracx^45!-cdotsright) $$
$$lim_xto0^-fracsin(x)x = 1$$
Further for right hand limit,
$$lim_xto0^+fracsin(x)x =lim_xto0^+left( 1-fracx^23!+fracx^45!-cdotsright) $$
$$lim_xto0^+fracsin(x)x = 1$$
Now,
$$Bigllfloorlim_xto0^+fracsin(x)xBigrrfloor = Bigllfloorlim_xto0^-fracsin(x)xBigrrfloor =1 $$
answered Sep 4 at 12:21
Sahil Silare
668
By the Maclaurin's expansion of $sin(x)$, we have,
$$sin(x) =sum^infty_k=0frac(-1)^k(2k+1)!x^2k+1=x-fracx^33!+fracx^55!-cdots$$
As $xto0$ we compute left hand limit,
$$lim_xto0^-fracsin(x)x =lim_xto0^-left( 1-fracx^23!+fracx^45!-cdotsright) $$
$$lim_xto0^-fracsin(x)x = 1$$
Further for right hand limit,
$$lim_xto0^+fracsin(x)x =lim_xto0^+left( 1-fracx^23!+fracx^45!-cdotsright) $$
$$lim_xto0^+fracsin(x)x = 1$$
Now,
$$Bigllfloorlim_xto0^+fracsin(x)xBigrrfloor = Bigllfloorlim_xto0^-fracsin(x)xBigrrfloor =1 $$
answered Sep 4 at 12:21
Sahil Silare
668
edited Sep 4 at 14:37
answered Sep 4 at 12:21
Sahil Silare
668
answered Sep 4 at 12:21
Sahil Silare
668
answered Sep 4 at 12:21
Sahil Silare
668
668
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2904947%2fvalue-of-bigl-lfloor-lim-x-to0-frac-sin-xx-bigr-rfloor%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
$$ lfloor 1 rfloor = 1 quadtextandquad lim_xto 0fracsin xx = 1 $$
– Sobi
Sep 4 at 11:51
1
Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
– Shaun
Sep 4 at 11:58
@Shaun Will keep that in mind next time. I thought titles like this make your question appear interesting.
– Harshit Joshi
Sep 4 at 12:01