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Question about behaviour of $f(x)=sin(x)/x$ at $x=0$
Clash Royale CLAN TAG#URR8PPP
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It's possible to show that the limit of $sin(h)/h$ as $h$ tends to $0$ is $1$, famously.
Now, someone said to me the other day, "That means that $sin(0)/0$ is 1", and I strongly disagreed. I said to them that a limit $lim_xrightarrow af(x)$ equals $f(a)$ only if $f(x)$ is continuous at $a$ and we hadn't established that yet.
However, to establish whether $sin(x)/x$ is continuous at $x=0$, one needs consider a sequence $f(a_0), f(a_1), f(a_2),...$, where $a_i$ is a sequence that tends to $0$, and see whether the limit of the sequence is the same as $f(0)$ - but that's exactly our problem! We need to know the function is continuous at $x=0$ to evaluate it there using our limit, but to know whether it is continuous at $x=0$, we need to know its evaluation there!
I apologise if my reasoning is highly fallacious, as I am only an amateur enthusiast, but what, if any, is the way out of this circular reasoning? Is our function continuous at $x=0$?
real-analysis functions
asked Jul 16 at 20:07
Isky Mathews
777214
 |Â
show 3 more comments
up vote
1
down vote
favorite
It's possible to show that the limit of $sin(h)/h$ as $h$ tends to $0$ is $1$, famously.
Now, someone said to me the other day, "That means that $sin(0)/0$ is 1", and I strongly disagreed. I said to them that a limit $lim_xrightarrow af(x)$ equals $f(a)$ only if $f(x)$ is continuous at $a$ and we hadn't established that yet.
However, to establish whether $sin(x)/x$ is continuous at $x=0$, one needs consider a sequence $f(a_0), f(a_1), f(a_2),...$, where $a_i$ is a sequence that tends to $0$, and see whether the limit of the sequence is the same as $f(0)$ - but that's exactly our problem! We need to know the function is continuous at $x=0$ to evaluate it there using our limit, but to know whether it is continuous at $x=0$, we need to know its evaluation there!
I apologise if my reasoning is highly fallacious, as I am only an amateur enthusiast, but what, if any, is the way out of this circular reasoning? Is our function continuous at $x=0$?
real-analysis functions
asked Jul 16 at 20:07
Isky Mathews
777214
If $f(x)=sin x/x$ then $f(0)$ is not defined.
– Lord Shark the Unknown
Jul 16 at 20:08
So, $f(x)$ is discontinuous at $x=0$?
– Isky Mathews
Jul 16 at 20:10
1
$f(x)$ can be extended to a continuous function $hatf(x)$ where $hatf(0) = 1$ and $f$ and $hatf$ agree away from $0$.
– Peter Kagey
Jul 16 at 20:10
Ok. Well, there we go - my initial thoughts were correct. It's just that this person is normally correct about a lot of things...
– Isky Mathews
Jul 16 at 20:11
@IskyMathews There is a typo - I presume you mean $f(a)$ in your definition for continuity, not $a$...
– Daniele1234
Jul 16 at 21:58
 |Â
show 3 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
It's possible to show that the limit of $sin(h)/h$ as $h$ tends to $0$ is $1$, famously.
Now, someone said to me the other day, "That means that $sin(0)/0$ is 1", and I strongly disagreed. I said to them that a limit $lim_xrightarrow af(x)$ equals $f(a)$ only if $f(x)$ is continuous at $a$ and we hadn't established that yet.
However, to establish whether $sin(x)/x$ is continuous at $x=0$, one needs consider a sequence $f(a_0), f(a_1), f(a_2),...$, where $a_i$ is a sequence that tends to $0$, and see whether the limit of the sequence is the same as $f(0)$ - but that's exactly our problem! We need to know the function is continuous at $x=0$ to evaluate it there using our limit, but to know whether it is continuous at $x=0$, we need to know its evaluation there!
I apologise if my reasoning is highly fallacious, as I am only an amateur enthusiast, but what, if any, is the way out of this circular reasoning? Is our function continuous at $x=0$?
real-analysis functions
asked Jul 16 at 20:07
Isky Mathews
777214
It's possible to show that the limit of $sin(h)/h$ as $h$ tends to $0$ is $1$, famously.
Now, someone said to me the other day, "That means that $sin(0)/0$ is 1", and I strongly disagreed. I said to them that a limit $lim_xrightarrow af(x)$ equals $f(a)$ only if $f(x)$ is continuous at $a$ and we hadn't established that yet.
However, to establish whether $sin(x)/x$ is continuous at $x=0$, one needs consider a sequence $f(a_0), f(a_1), f(a_2),...$, where $a_i$ is a sequence that tends to $0$, and see whether the limit of the sequence is the same as $f(0)$ - but that's exactly our problem! We need to know the function is continuous at $x=0$ to evaluate it there using our limit, but to know whether it is continuous at $x=0$, we need to know its evaluation there!
I apologise if my reasoning is highly fallacious, as I am only an amateur enthusiast, but what, if any, is the way out of this circular reasoning? Is our function continuous at $x=0$?
real-analysis functions
asked Jul 16 at 20:07
Isky Mathews
777214
edited Jul 17 at 5:14
asked Jul 16 at 20:07
Isky Mathews
777214
asked Jul 16 at 20:07
Isky Mathews
777214
asked Jul 16 at 20:07
Isky Mathews
777214
777214
If $f(x)=sin x/x$ then $f(0)$ is not defined.
– Lord Shark the Unknown
Jul 16 at 20:08
So, $f(x)$ is discontinuous at $x=0$?
– Isky Mathews
Jul 16 at 20:10
1
$f(x)$ can be extended to a continuous function $hatf(x)$ where $hatf(0) = 1$ and $f$ and $hatf$ agree away from $0$.
– Peter Kagey
Jul 16 at 20:10
Ok. Well, there we go - my initial thoughts were correct. It's just that this person is normally correct about a lot of things...
– Isky Mathews
Jul 16 at 20:11
@IskyMathews There is a typo - I presume you mean $f(a)$ in your definition for continuity, not $a$...
– Daniele1234
Jul 16 at 21:58
 |Â
show 3 more comments
If $f(x)=sin x/x$ then $f(0)$ is not defined.
– Lord Shark the Unknown
Jul 16 at 20:08
So, $f(x)$ is discontinuous at $x=0$?
– Isky Mathews
Jul 16 at 20:10
1
$f(x)$ can be extended to a continuous function $hatf(x)$ where $hatf(0) = 1$ and $f$ and $hatf$ agree away from $0$.
– Peter Kagey
Jul 16 at 20:10
Ok. Well, there we go - my initial thoughts were correct. It's just that this person is normally correct about a lot of things...
– Isky Mathews
Jul 16 at 20:11
@IskyMathews There is a typo - I presume you mean $f(a)$ in your definition for continuity, not $a$...
– Daniele1234
Jul 16 at 21:58
If $f(x)=sin x/x$ then $f(0)$ is not defined.
– Lord Shark the Unknown
Jul 16 at 20:08
If $f(x)=sin x/x$ then $f(0)$ is not defined.
– Lord Shark the Unknown
Jul 16 at 20:08
So, $f(x)$ is discontinuous at $x=0$?
– Isky Mathews
Jul 16 at 20:10
So, $f(x)$ is discontinuous at $x=0$?
– Isky Mathews
Jul 16 at 20:10
1
1
$f(x)$ can be extended to a continuous function $hatf(x)$ where $hatf(0) = 1$ and $f$ and $hatf$ agree away from $0$.
– Peter Kagey
Jul 16 at 20:10
$f(x)$ can be extended to a continuous function $hatf(x)$ where $hatf(0) = 1$ and $f$ and $hatf$ agree away from $0$.
– Peter Kagey
Jul 16 at 20:10
Ok. Well, there we go - my initial thoughts were correct. It's just that this person is normally correct about a lot of things...
– Isky Mathews
Jul 16 at 20:11
Ok. Well, there we go - my initial thoughts were correct. It's just that this person is normally correct about a lot of things...
– Isky Mathews
Jul 16 at 20:11
@IskyMathews There is a typo - I presume you mean $f(a)$ in your definition for continuity, not $a$...
– Daniele1234
Jul 16 at 21:58
@IskyMathews There is a typo - I presume you mean $f(a)$ in your definition for continuity, not $a$...
– Daniele1234
Jul 16 at 21:58
 |Â
show 3 more comments
4 Answers
4
active
oldest
votes
up vote
3
down vote
accepted
First of all, I presume there is a typo in line 3 which should read "$lim_xrightarrow af(x)$ equal to $f(a)$", not "$a$." Quite simply, $f(0)$ is undefined, and so this is what we call a removable discontinuity. Indeed, $lim_xrightarrow 0(sinx)/x$ does exist and, by L'Hopitals Rule is equal to $lim_xrightarrow 0cosx=1$. However, as $lim_xrightarrow 0f(x)$ exists but has a value different from $f(0)$ (in this case undefined), we have a removable discontinuity.
If you want a very interesting problem: Let $f$ be an arbitrary function from the $R$ to $R$ (reals). Consider $D_f$, the set of points of points at which f is discontinuous. Show that $D_f$ is a F-sigma set - that is, $D_f$ can be written as the countable union of closed sets (hint: first consider monotone functions, then generalise from there).
Speak spoon g
NOTE: functional limits require $a$ to be a limit point of the domain (as 0 is in this case), however, v. important subtle difference is that continuity requires that $a$ actually be contained in the domain, and of course in this case, 0 is not as $sin0/0$ is undefined.
answered Jul 16 at 22:11
Daniele1234
716215
@IskyMathews what u think?
– Daniele1234
Jul 17 at 12:54
add a comment |Â
up vote
7
down vote
It only makes sense to talk about continuity for points $x$ in the domain of the function, so asking if $sin x/x$ is continuous at $0$ is not well-posed. However, we can define a function
$$
f(x)=begincases
sin x/x&x neq 0\
1&x=0
endcases
$$
which is continuous everywhere.
answered Jul 16 at 20:11
Foobaz John
18.1k41245
add a comment |Â
up vote
0
down vote
To some extent, this someone was right (though I bet many people here will disagree).
Because writing $dfracsin 00$ can be taken informally for an indeterminate expression where the limit of $dfracsin xx$ is implied (otherwise one would have written $dfracsin 00^2$ or $dfraclog(sin0+1)0$ or...), and because the natural choice for the value of the function when it is indeterminate is the limit.
So even if the statement lacks rigor, it can be understood as "the value of $dfracsin(0)0$ should be $1$", and that makes the function continuous.
By the way the "official" cardinal sine function is so defined.
answered Jul 16 at 20:27
Yves Daoust
111k665204
In fact that $sinalpha approx alpha approx tanalpha$ as $alpha to 0$, is known since millennia (and later refined with the Taylor series).
– G Cab
Jul 16 at 20:48
add a comment |Â
up vote
0
down vote
Since $f$ is undefined at 0, it is not continuous at 0.
answered Jul 17 at 14:45
nonremovable
7431619
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
First of all, I presume there is a typo in line 3 which should read "$lim_xrightarrow af(x)$ equal to $f(a)$", not "$a$." Quite simply, $f(0)$ is undefined, and so this is what we call a removable discontinuity. Indeed, $lim_xrightarrow 0(sinx)/x$ does exist and, by L'Hopitals Rule is equal to $lim_xrightarrow 0cosx=1$. However, as $lim_xrightarrow 0f(x)$ exists but has a value different from $f(0)$ (in this case undefined), we have a removable discontinuity.
If you want a very interesting problem: Let $f$ be an arbitrary function from the $R$ to $R$ (reals). Consider $D_f$, the set of points of points at which f is discontinuous. Show that $D_f$ is a F-sigma set - that is, $D_f$ can be written as the countable union of closed sets (hint: first consider monotone functions, then generalise from there).
Speak spoon g
NOTE: functional limits require $a$ to be a limit point of the domain (as 0 is in this case), however, v. important subtle difference is that continuity requires that $a$ actually be contained in the domain, and of course in this case, 0 is not as $sin0/0$ is undefined.
answered Jul 16 at 22:11
Daniele1234
716215
@IskyMathews what u think?
– Daniele1234
Jul 17 at 12:54
add a comment |Â
up vote
3
down vote
accepted
First of all, I presume there is a typo in line 3 which should read "$lim_xrightarrow af(x)$ equal to $f(a)$", not "$a$." Quite simply, $f(0)$ is undefined, and so this is what we call a removable discontinuity. Indeed, $lim_xrightarrow 0(sinx)/x$ does exist and, by L'Hopitals Rule is equal to $lim_xrightarrow 0cosx=1$. However, as $lim_xrightarrow 0f(x)$ exists but has a value different from $f(0)$ (in this case undefined), we have a removable discontinuity.
If you want a very interesting problem: Let $f$ be an arbitrary function from the $R$ to $R$ (reals). Consider $D_f$, the set of points of points at which f is discontinuous. Show that $D_f$ is a F-sigma set - that is, $D_f$ can be written as the countable union of closed sets (hint: first consider monotone functions, then generalise from there).
Speak spoon g
NOTE: functional limits require $a$ to be a limit point of the domain (as 0 is in this case), however, v. important subtle difference is that continuity requires that $a$ actually be contained in the domain, and of course in this case, 0 is not as $sin0/0$ is undefined.
answered Jul 16 at 22:11
Daniele1234
716215
@IskyMathews what u think?
– Daniele1234
Jul 17 at 12:54
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
First of all, I presume there is a typo in line 3 which should read "$lim_xrightarrow af(x)$ equal to $f(a)$", not "$a$." Quite simply, $f(0)$ is undefined, and so this is what we call a removable discontinuity. Indeed, $lim_xrightarrow 0(sinx)/x$ does exist and, by L'Hopitals Rule is equal to $lim_xrightarrow 0cosx=1$. However, as $lim_xrightarrow 0f(x)$ exists but has a value different from $f(0)$ (in this case undefined), we have a removable discontinuity.
If you want a very interesting problem: Let $f$ be an arbitrary function from the $R$ to $R$ (reals). Consider $D_f$, the set of points of points at which f is discontinuous. Show that $D_f$ is a F-sigma set - that is, $D_f$ can be written as the countable union of closed sets (hint: first consider monotone functions, then generalise from there).
Speak spoon g
NOTE: functional limits require $a$ to be a limit point of the domain (as 0 is in this case), however, v. important subtle difference is that continuity requires that $a$ actually be contained in the domain, and of course in this case, 0 is not as $sin0/0$ is undefined.
answered Jul 16 at 22:11
Daniele1234
716215
First of all, I presume there is a typo in line 3 which should read "$lim_xrightarrow af(x)$ equal to $f(a)$", not "$a$." Quite simply, $f(0)$ is undefined, and so this is what we call a removable discontinuity. Indeed, $lim_xrightarrow 0(sinx)/x$ does exist and, by L'Hopitals Rule is equal to $lim_xrightarrow 0cosx=1$. However, as $lim_xrightarrow 0f(x)$ exists but has a value different from $f(0)$ (in this case undefined), we have a removable discontinuity.
If you want a very interesting problem: Let $f$ be an arbitrary function from the $R$ to $R$ (reals). Consider $D_f$, the set of points of points at which f is discontinuous. Show that $D_f$ is a F-sigma set - that is, $D_f$ can be written as the countable union of closed sets (hint: first consider monotone functions, then generalise from there).
Speak spoon g
NOTE: functional limits require $a$ to be a limit point of the domain (as 0 is in this case), however, v. important subtle difference is that continuity requires that $a$ actually be contained in the domain, and of course in this case, 0 is not as $sin0/0$ is undefined.
answered Jul 16 at 22:11
Daniele1234
716215
answered Jul 16 at 22:11
Daniele1234
716215
answered Jul 16 at 22:11
Daniele1234
716215
answered Jul 16 at 22:11
Daniele1234
716215
716215
@IskyMathews what u think?
– Daniele1234
Jul 17 at 12:54
add a comment |Â
@IskyMathews what u think?
– Daniele1234
Jul 17 at 12:54
@IskyMathews what u think?
– Daniele1234
Jul 17 at 12:54
@IskyMathews what u think?
– Daniele1234
Jul 17 at 12:54
add a comment |Â
up vote
7
down vote
It only makes sense to talk about continuity for points $x$ in the domain of the function, so asking if $sin x/x$ is continuous at $0$ is not well-posed. However, we can define a function
$$
f(x)=begincases
sin x/x&x neq 0\
1&x=0
endcases
$$
which is continuous everywhere.
answered Jul 16 at 20:11
Foobaz John
18.1k41245
add a comment |Â
up vote
7
down vote
It only makes sense to talk about continuity for points $x$ in the domain of the function, so asking if $sin x/x$ is continuous at $0$ is not well-posed. However, we can define a function
$$
f(x)=begincases
sin x/x&x neq 0\
1&x=0
endcases
$$
which is continuous everywhere.
answered Jul 16 at 20:11
Foobaz John
18.1k41245
add a comment |Â
up vote
7
down vote
up vote
7
down vote
It only makes sense to talk about continuity for points $x$ in the domain of the function, so asking if $sin x/x$ is continuous at $0$ is not well-posed. However, we can define a function
$$
f(x)=begincases
sin x/x&x neq 0\
1&x=0
endcases
$$
which is continuous everywhere.
answered Jul 16 at 20:11
Foobaz John
18.1k41245
It only makes sense to talk about continuity for points $x$ in the domain of the function, so asking if $sin x/x$ is continuous at $0$ is not well-posed. However, we can define a function
$$
f(x)=begincases
sin x/x&x neq 0\
1&x=0
endcases
$$
which is continuous everywhere.
answered Jul 16 at 20:11
Foobaz John
18.1k41245
answered Jul 16 at 20:11
Foobaz John
18.1k41245
answered Jul 16 at 20:11
Foobaz John
18.1k41245
answered Jul 16 at 20:11
Foobaz John
18.1k41245
18.1k41245
add a comment |Â
add a comment |Â
up vote
0
down vote
To some extent, this someone was right (though I bet many people here will disagree).
Because writing $dfracsin 00$ can be taken informally for an indeterminate expression where the limit of $dfracsin xx$ is implied (otherwise one would have written $dfracsin 00^2$ or $dfraclog(sin0+1)0$ or...), and because the natural choice for the value of the function when it is indeterminate is the limit.
So even if the statement lacks rigor, it can be understood as "the value of $dfracsin(0)0$ should be $1$", and that makes the function continuous.
By the way the "official" cardinal sine function is so defined.
answered Jul 16 at 20:27
Yves Daoust
111k665204
In fact that $sinalpha approx alpha approx tanalpha$ as $alpha to 0$, is known since millennia (and later refined with the Taylor series).
– G Cab
Jul 16 at 20:48
add a comment |Â
up vote
0
down vote
To some extent, this someone was right (though I bet many people here will disagree).
Because writing $dfracsin 00$ can be taken informally for an indeterminate expression where the limit of $dfracsin xx$ is implied (otherwise one would have written $dfracsin 00^2$ or $dfraclog(sin0+1)0$ or...), and because the natural choice for the value of the function when it is indeterminate is the limit.
So even if the statement lacks rigor, it can be understood as "the value of $dfracsin(0)0$ should be $1$", and that makes the function continuous.
By the way the "official" cardinal sine function is so defined.
answered Jul 16 at 20:27
Yves Daoust
111k665204
In fact that $sinalpha approx alpha approx tanalpha$ as $alpha to 0$, is known since millennia (and later refined with the Taylor series).
– G Cab
Jul 16 at 20:48
add a comment |Â
up vote
0
down vote
up vote
0
down vote
To some extent, this someone was right (though I bet many people here will disagree).
Because writing $dfracsin 00$ can be taken informally for an indeterminate expression where the limit of $dfracsin xx$ is implied (otherwise one would have written $dfracsin 00^2$ or $dfraclog(sin0+1)0$ or...), and because the natural choice for the value of the function when it is indeterminate is the limit.
So even if the statement lacks rigor, it can be understood as "the value of $dfracsin(0)0$ should be $1$", and that makes the function continuous.
By the way the "official" cardinal sine function is so defined.
answered Jul 16 at 20:27
Yves Daoust
111k665204
To some extent, this someone was right (though I bet many people here will disagree).
Because writing $dfracsin 00$ can be taken informally for an indeterminate expression where the limit of $dfracsin xx$ is implied (otherwise one would have written $dfracsin 00^2$ or $dfraclog(sin0+1)0$ or...), and because the natural choice for the value of the function when it is indeterminate is the limit.
So even if the statement lacks rigor, it can be understood as "the value of $dfracsin(0)0$ should be $1$", and that makes the function continuous.
By the way the "official" cardinal sine function is so defined.
answered Jul 16 at 20:27
Yves Daoust
111k665204
answered Jul 16 at 20:27
Yves Daoust
111k665204
answered Jul 16 at 20:27
Yves Daoust
111k665204
answered Jul 16 at 20:27
Yves Daoust
111k665204
111k665204
In fact that $sinalpha approx alpha approx tanalpha$ as $alpha to 0$, is known since millennia (and later refined with the Taylor series).
– G Cab
Jul 16 at 20:48
add a comment |Â
In fact that $sinalpha approx alpha approx tanalpha$ as $alpha to 0$, is known since millennia (and later refined with the Taylor series).
– G Cab
Jul 16 at 20:48
In fact that $sinalpha approx alpha approx tanalpha$ as $alpha to 0$, is known since millennia (and later refined with the Taylor series).
– G Cab
Jul 16 at 20:48
In fact that $sinalpha approx alpha approx tanalpha$ as $alpha to 0$, is known since millennia (and later refined with the Taylor series).
– G Cab
Jul 16 at 20:48
add a comment |Â
up vote
0
down vote
Since $f$ is undefined at 0, it is not continuous at 0.
answered Jul 17 at 14:45
nonremovable
7431619
add a comment |Â
up vote
0
down vote
Since $f$ is undefined at 0, it is not continuous at 0.
answered Jul 17 at 14:45
nonremovable
7431619
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Since $f$ is undefined at 0, it is not continuous at 0.
answered Jul 17 at 14:45
nonremovable
7431619
Since $f$ is undefined at 0, it is not continuous at 0.
answered Jul 17 at 14:45
nonremovable
7431619
answered Jul 17 at 14:45
nonremovable
7431619
answered Jul 17 at 14:45
nonremovable
7431619
answered Jul 17 at 14:45
nonremovable
7431619
7431619
add a comment |Â
add a comment |Â
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If $f(x)=sin x/x$ then $f(0)$ is not defined.
– Lord Shark the Unknown
Jul 16 at 20:08
So, $f(x)$ is discontinuous at $x=0$?
– Isky Mathews
Jul 16 at 20:10
1
$f(x)$ can be extended to a continuous function $hatf(x)$ where $hatf(0) = 1$ and $f$ and $hatf$ agree away from $0$.
– Peter Kagey
Jul 16 at 20:10
Ok. Well, there we go - my initial thoughts were correct. It's just that this person is normally correct about a lot of things...
– Isky Mathews
Jul 16 at 20:11
@IskyMathews There is a typo - I presume you mean $f(a)$ in your definition for continuity, not $a$...
– Daniele1234
Jul 16 at 21:58