Value of $lim_xto pi/2lfloor(x-fracpi2)/cos xrfloor$
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$$lim_xto fracpi2leftlfloorfracx-fracpi2cos xright rfloor$$
After I took the Right Hand Limit by using
$$lim_hto0 leftlfloorfracfracpi2+h-fracpi2cos(fracpi2+h)rightrfloor$$
I got $$lim_hto0 leftlfloor frach-sin hrightrfloor$$
From here onwards I am confused whether my answer is $-1$ or $-2$.
calculus
edited Aug 20 at 8:45
Henning Makholm
229k16294525
asked Aug 20 at 8:23
Samar Imam Zaidi
1,180316
add a comment |Â
up vote
4
down vote
favorite
$$lim_xto fracpi2leftlfloorfracx-fracpi2cos xright rfloor$$
After I took the Right Hand Limit by using
$$lim_hto0 leftlfloorfracfracpi2+h-fracpi2cos(fracpi2+h)rightrfloor$$
I got $$lim_hto0 leftlfloor frach-sin hrightrfloor$$
From here onwards I am confused whether my answer is $-1$ or $-2$.
calculus
edited Aug 20 at 8:45
Henning Makholm
229k16294525
asked Aug 20 at 8:23
Samar Imam Zaidi
1,180316
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
$$lim_xto fracpi2leftlfloorfracx-fracpi2cos xright rfloor$$
After I took the Right Hand Limit by using
$$lim_hto0 leftlfloorfracfracpi2+h-fracpi2cos(fracpi2+h)rightrfloor$$
I got $$lim_hto0 leftlfloor frach-sin hrightrfloor$$
From here onwards I am confused whether my answer is $-1$ or $-2$.
calculus
edited Aug 20 at 8:45
Henning Makholm
229k16294525
asked Aug 20 at 8:23
Samar Imam Zaidi
1,180316
$$lim_xto fracpi2leftlfloorfracx-fracpi2cos xright rfloor$$
After I took the Right Hand Limit by using
$$lim_hto0 leftlfloorfracfracpi2+h-fracpi2cos(fracpi2+h)rightrfloor$$
I got $$lim_hto0 leftlfloor frach-sin hrightrfloor$$
From here onwards I am confused whether my answer is $-1$ or $-2$.
calculus
edited Aug 20 at 8:45
Henning Makholm
229k16294525
asked Aug 20 at 8:23
Samar Imam Zaidi
1,180316
edited Aug 20 at 8:45
Henning Makholm
229k16294525
edited Aug 20 at 8:45
Henning Makholm
229k16294525
edited Aug 20 at 8:45
Henning Makholm
229k16294525
229k16294525
asked Aug 20 at 8:23
Samar Imam Zaidi
1,180316
asked Aug 20 at 8:23
Samar Imam Zaidi
1,180316
asked Aug 20 at 8:23
Samar Imam Zaidi
1,180316
1,180316
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
For $pi>h>0$
$$ sin(h)< h $$
For $-pi<h<0$
$$sin(h)> h $$
From this, for $hneq 0, -pi<h<pi$
$$ frachsin(h) > 1 implies -frachsin(h)< -1$$
As $$lim_hto 0frachsin(h) = 1 $$
For $h$ close enough to $0$, $-frachsin(h)$ is close to $-1$, so that
$$-2leq-frachsin(h)< -1 implies leftlfloor frachsin(h) rightrfloor = -2 $$
More formally, from definition of a limit, there exists $delta>0$ for which
$$ 0<|h|<delta implies -2leq-frachsin(h) $$
And the limit is $-2$, we can take $mindelta, pi$ for this limit.
answered Aug 20 at 8:38
Rumpelstiltskin
1,604415
I have one doubt about the equality sign of $-frachsin(h)< -1$ , how it is $ge -2$ , it can be greater than -2
– Samar Imam Zaidi
Aug 20 at 9:22
Can we say that -$frachsinh$ $in $(-2,-1)
– Samar Imam Zaidi
Aug 20 at 9:24
For $0<|h|<delta$, where $delta$ is the one in the answer, we can say that $-frachsin(h)in[-2, -1)$. Yes, it can be greater, and we can choose, maybe smaller, $delta_0>0$, for which $-frachsin(h)in(-2, -1)$. This doesn't change anything in our proof though. In fact, for any $x<-1$, we can choose $delta_0>0$ such that $-frachsin(h)in(x, -1)$.
– Rumpelstiltskin
Aug 20 at 9:26
I have one doubt. For $h<0$, $sin(h)>h$, which implies $h/sin(h)<1$, which should imply $-h/sin(h)>-1$.But your sixth line disagrees with my reasoning . Please clarify .
– Alphanerd
Aug 20 at 14:10
@Alphanerd I'm sorry, I noticed a mistake. $sin(h)<0$ for $-pi<h<0$, so in the expression $sin(h)>h$, if we try to divide by $sin(h)$, we need to change signs.
– Rumpelstiltskin
Aug 20 at 14:30
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
For $pi>h>0$
$$ sin(h)< h $$
For $-pi<h<0$
$$sin(h)> h $$
From this, for $hneq 0, -pi<h<pi$
$$ frachsin(h) > 1 implies -frachsin(h)< -1$$
As $$lim_hto 0frachsin(h) = 1 $$
For $h$ close enough to $0$, $-frachsin(h)$ is close to $-1$, so that
$$-2leq-frachsin(h)< -1 implies leftlfloor frachsin(h) rightrfloor = -2 $$
More formally, from definition of a limit, there exists $delta>0$ for which
$$ 0<|h|<delta implies -2leq-frachsin(h) $$
And the limit is $-2$, we can take $mindelta, pi$ for this limit.
answered Aug 20 at 8:38
Rumpelstiltskin
1,604415
I have one doubt about the equality sign of $-frachsin(h)< -1$ , how it is $ge -2$ , it can be greater than -2
– Samar Imam Zaidi
Aug 20 at 9:22
Can we say that -$frachsinh$ $in $(-2,-1)
– Samar Imam Zaidi
Aug 20 at 9:24
For $0<|h|<delta$, where $delta$ is the one in the answer, we can say that $-frachsin(h)in[-2, -1)$. Yes, it can be greater, and we can choose, maybe smaller, $delta_0>0$, for which $-frachsin(h)in(-2, -1)$. This doesn't change anything in our proof though. In fact, for any $x<-1$, we can choose $delta_0>0$ such that $-frachsin(h)in(x, -1)$.
– Rumpelstiltskin
Aug 20 at 9:26
I have one doubt. For $h<0$, $sin(h)>h$, which implies $h/sin(h)<1$, which should imply $-h/sin(h)>-1$.But your sixth line disagrees with my reasoning . Please clarify .
– Alphanerd
Aug 20 at 14:10
@Alphanerd I'm sorry, I noticed a mistake. $sin(h)<0$ for $-pi<h<0$, so in the expression $sin(h)>h$, if we try to divide by $sin(h)$, we need to change signs.
– Rumpelstiltskin
Aug 20 at 14:30
add a comment |Â
up vote
3
down vote
accepted
For $pi>h>0$
$$ sin(h)< h $$
For $-pi<h<0$
$$sin(h)> h $$
From this, for $hneq 0, -pi<h<pi$
$$ frachsin(h) > 1 implies -frachsin(h)< -1$$
As $$lim_hto 0frachsin(h) = 1 $$
For $h$ close enough to $0$, $-frachsin(h)$ is close to $-1$, so that
$$-2leq-frachsin(h)< -1 implies leftlfloor frachsin(h) rightrfloor = -2 $$
More formally, from definition of a limit, there exists $delta>0$ for which
$$ 0<|h|<delta implies -2leq-frachsin(h) $$
And the limit is $-2$, we can take $mindelta, pi$ for this limit.
answered Aug 20 at 8:38
Rumpelstiltskin
1,604415
I have one doubt about the equality sign of $-frachsin(h)< -1$ , how it is $ge -2$ , it can be greater than -2
– Samar Imam Zaidi
Aug 20 at 9:22
Can we say that -$frachsinh$ $in $(-2,-1)
– Samar Imam Zaidi
Aug 20 at 9:24
For $0<|h|<delta$, where $delta$ is the one in the answer, we can say that $-frachsin(h)in[-2, -1)$. Yes, it can be greater, and we can choose, maybe smaller, $delta_0>0$, for which $-frachsin(h)in(-2, -1)$. This doesn't change anything in our proof though. In fact, for any $x<-1$, we can choose $delta_0>0$ such that $-frachsin(h)in(x, -1)$.
– Rumpelstiltskin
Aug 20 at 9:26
I have one doubt. For $h<0$, $sin(h)>h$, which implies $h/sin(h)<1$, which should imply $-h/sin(h)>-1$.But your sixth line disagrees with my reasoning . Please clarify .
– Alphanerd
Aug 20 at 14:10
@Alphanerd I'm sorry, I noticed a mistake. $sin(h)<0$ for $-pi<h<0$, so in the expression $sin(h)>h$, if we try to divide by $sin(h)$, we need to change signs.
– Rumpelstiltskin
Aug 20 at 14:30
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
For $pi>h>0$
$$ sin(h)< h $$
For $-pi<h<0$
$$sin(h)> h $$
From this, for $hneq 0, -pi<h<pi$
$$ frachsin(h) > 1 implies -frachsin(h)< -1$$
As $$lim_hto 0frachsin(h) = 1 $$
For $h$ close enough to $0$, $-frachsin(h)$ is close to $-1$, so that
$$-2leq-frachsin(h)< -1 implies leftlfloor frachsin(h) rightrfloor = -2 $$
More formally, from definition of a limit, there exists $delta>0$ for which
$$ 0<|h|<delta implies -2leq-frachsin(h) $$
And the limit is $-2$, we can take $mindelta, pi$ for this limit.
answered Aug 20 at 8:38
Rumpelstiltskin
1,604415
For $pi>h>0$
$$ sin(h)< h $$
For $-pi<h<0$
$$sin(h)> h $$
From this, for $hneq 0, -pi<h<pi$
$$ frachsin(h) > 1 implies -frachsin(h)< -1$$
As $$lim_hto 0frachsin(h) = 1 $$
For $h$ close enough to $0$, $-frachsin(h)$ is close to $-1$, so that
$$-2leq-frachsin(h)< -1 implies leftlfloor frachsin(h) rightrfloor = -2 $$
More formally, from definition of a limit, there exists $delta>0$ for which
$$ 0<|h|<delta implies -2leq-frachsin(h) $$
And the limit is $-2$, we can take $mindelta, pi$ for this limit.
answered Aug 20 at 8:38
Rumpelstiltskin
1,604415
edited Aug 20 at 14:29
answered Aug 20 at 8:38
Rumpelstiltskin
1,604415
answered Aug 20 at 8:38
Rumpelstiltskin
1,604415
answered Aug 20 at 8:38
Rumpelstiltskin
1,604415
1,604415
I have one doubt about the equality sign of $-frachsin(h)< -1$ , how it is $ge -2$ , it can be greater than -2
– Samar Imam Zaidi
Aug 20 at 9:22
Can we say that -$frachsinh$ $in $(-2,-1)
– Samar Imam Zaidi
Aug 20 at 9:24
For $0<|h|<delta$, where $delta$ is the one in the answer, we can say that $-frachsin(h)in[-2, -1)$. Yes, it can be greater, and we can choose, maybe smaller, $delta_0>0$, for which $-frachsin(h)in(-2, -1)$. This doesn't change anything in our proof though. In fact, for any $x<-1$, we can choose $delta_0>0$ such that $-frachsin(h)in(x, -1)$.
– Rumpelstiltskin
Aug 20 at 9:26
I have one doubt. For $h<0$, $sin(h)>h$, which implies $h/sin(h)<1$, which should imply $-h/sin(h)>-1$.But your sixth line disagrees with my reasoning . Please clarify .
– Alphanerd
Aug 20 at 14:10
@Alphanerd I'm sorry, I noticed a mistake. $sin(h)<0$ for $-pi<h<0$, so in the expression $sin(h)>h$, if we try to divide by $sin(h)$, we need to change signs.
– Rumpelstiltskin
Aug 20 at 14:30
add a comment |Â
I have one doubt about the equality sign of $-frachsin(h)< -1$ , how it is $ge -2$ , it can be greater than -2
– Samar Imam Zaidi
Aug 20 at 9:22
Can we say that -$frachsinh$ $in $(-2,-1)
– Samar Imam Zaidi
Aug 20 at 9:24
For $0<|h|<delta$, where $delta$ is the one in the answer, we can say that $-frachsin(h)in[-2, -1)$. Yes, it can be greater, and we can choose, maybe smaller, $delta_0>0$, for which $-frachsin(h)in(-2, -1)$. This doesn't change anything in our proof though. In fact, for any $x<-1$, we can choose $delta_0>0$ such that $-frachsin(h)in(x, -1)$.
– Rumpelstiltskin
Aug 20 at 9:26
I have one doubt. For $h<0$, $sin(h)>h$, which implies $h/sin(h)<1$, which should imply $-h/sin(h)>-1$.But your sixth line disagrees with my reasoning . Please clarify .
– Alphanerd
Aug 20 at 14:10
@Alphanerd I'm sorry, I noticed a mistake. $sin(h)<0$ for $-pi<h<0$, so in the expression $sin(h)>h$, if we try to divide by $sin(h)$, we need to change signs.
– Rumpelstiltskin
Aug 20 at 14:30
I have one doubt about the equality sign of $-frachsin(h)< -1$ , how it is $ge -2$ , it can be greater than -2
– Samar Imam Zaidi
Aug 20 at 9:22
I have one doubt about the equality sign of $-frachsin(h)< -1$ , how it is $ge -2$ , it can be greater than -2
– Samar Imam Zaidi
Aug 20 at 9:22
Can we say that -$frachsinh$ $in $(-2,-1)
– Samar Imam Zaidi
Aug 20 at 9:24
Can we say that -$frachsinh$ $in $(-2,-1)
– Samar Imam Zaidi
Aug 20 at 9:24
For $0<|h|<delta$, where $delta$ is the one in the answer, we can say that $-frachsin(h)in[-2, -1)$. Yes, it can be greater, and we can choose, maybe smaller, $delta_0>0$, for which $-frachsin(h)in(-2, -1)$. This doesn't change anything in our proof though. In fact, for any $x<-1$, we can choose $delta_0>0$ such that $-frachsin(h)in(x, -1)$.
– Rumpelstiltskin
Aug 20 at 9:26
For $0<|h|<delta$, where $delta$ is the one in the answer, we can say that $-frachsin(h)in[-2, -1)$. Yes, it can be greater, and we can choose, maybe smaller, $delta_0>0$, for which $-frachsin(h)in(-2, -1)$. This doesn't change anything in our proof though. In fact, for any $x<-1$, we can choose $delta_0>0$ such that $-frachsin(h)in(x, -1)$.
– Rumpelstiltskin
Aug 20 at 9:26
I have one doubt. For $h<0$, $sin(h)>h$, which implies $h/sin(h)<1$, which should imply $-h/sin(h)>-1$.But your sixth line disagrees with my reasoning . Please clarify .
– Alphanerd
Aug 20 at 14:10
I have one doubt. For $h<0$, $sin(h)>h$, which implies $h/sin(h)<1$, which should imply $-h/sin(h)>-1$.But your sixth line disagrees with my reasoning . Please clarify .
– Alphanerd
Aug 20 at 14:10
@Alphanerd I'm sorry, I noticed a mistake. $sin(h)<0$ for $-pi<h<0$, so in the expression $sin(h)>h$, if we try to divide by $sin(h)$, we need to change signs.
– Rumpelstiltskin
Aug 20 at 14:30
@Alphanerd I'm sorry, I noticed a mistake. $sin(h)<0$ for $-pi<h<0$, so in the expression $sin(h)>h$, if we try to divide by $sin(h)$, we need to change signs.
– Rumpelstiltskin
Aug 20 at 14:30
add a comment |Â
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