How's g(x) = sin(x)/x removal discontinuity? [on hold]

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
0
down vote

favorite












Because when we draw a graph there have no break point and breakthrough where we can notice that function jumped from his path or removed?? https://i.stack.imgur.com/XGkQ8.jpg




calculus limits continuity




share|cite|improve this question













put on hold as off-topic by Arnaud Mortier, amWhy, Strants, John Ma, José Carlos Santos Aug 3 at 19:19


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Strants, John Ma, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 2




    A graph is not a proof of anything. Not ever. A graph gives you a hint as to what to expect when you write an actual proof using formulas (limits, in this case).
    – Arnaud Mortier
    Aug 3 at 14:12











  • In theory, yes, but looking at a graph as a human is not enough. You have to prove that a theoretically ideal observer looking at a graph drawn with a theoretical infinitely thin curve cannot see a break point or anything. That's usually done with more algebra than geometry.
    – Arthur
    Aug 3 at 14:15











  • I can claim that you aren't showing me the graph of $fracsin xx$ just as easily as you are claiming to be showing it to me.
    – Saucy O'Path
    Aug 3 at 14:17











  • What is your definition of a removable discontinuity?
    – Strants
    Aug 3 at 14:19














up vote
0
down vote

favorite












Because when we draw a graph there have no break point and breakthrough where we can notice that function jumped from his path or removed?? https://i.stack.imgur.com/XGkQ8.jpg




calculus limits continuity




share|cite|improve this question













put on hold as off-topic by Arnaud Mortier, amWhy, Strants, John Ma, José Carlos Santos Aug 3 at 19:19


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Strants, John Ma, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 2




    A graph is not a proof of anything. Not ever. A graph gives you a hint as to what to expect when you write an actual proof using formulas (limits, in this case).
    – Arnaud Mortier
    Aug 3 at 14:12











  • In theory, yes, but looking at a graph as a human is not enough. You have to prove that a theoretically ideal observer looking at a graph drawn with a theoretical infinitely thin curve cannot see a break point or anything. That's usually done with more algebra than geometry.
    – Arthur
    Aug 3 at 14:15











  • I can claim that you aren't showing me the graph of $fracsin xx$ just as easily as you are claiming to be showing it to me.
    – Saucy O'Path
    Aug 3 at 14:17











  • What is your definition of a removable discontinuity?
    – Strants
    Aug 3 at 14:19












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Because when we draw a graph there have no break point and breakthrough where we can notice that function jumped from his path or removed?? https://i.stack.imgur.com/XGkQ8.jpg




calculus limits continuity




share|cite|improve this question













Because when we draw a graph there have no break point and breakthrough where we can notice that function jumped from his path or removed?? https://i.stack.imgur.com/XGkQ8.jpg





calculus limits continuity





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 3 at 14:26









Mefitico

48913




48913









asked Aug 3 at 14:08









MIKEY SINGH

103




103




put on hold as off-topic by Arnaud Mortier, amWhy, Strants, John Ma, José Carlos Santos Aug 3 at 19:19


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Strants, John Ma, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.




put on hold as off-topic by Arnaud Mortier, amWhy, Strants, John Ma, José Carlos Santos Aug 3 at 19:19


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Strants, John Ma, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 2




    A graph is not a proof of anything. Not ever. A graph gives you a hint as to what to expect when you write an actual proof using formulas (limits, in this case).
    – Arnaud Mortier
    Aug 3 at 14:12











  • In theory, yes, but looking at a graph as a human is not enough. You have to prove that a theoretically ideal observer looking at a graph drawn with a theoretical infinitely thin curve cannot see a break point or anything. That's usually done with more algebra than geometry.
    – Arthur
    Aug 3 at 14:15











  • I can claim that you aren't showing me the graph of $fracsin xx$ just as easily as you are claiming to be showing it to me.
    – Saucy O'Path
    Aug 3 at 14:17











  • What is your definition of a removable discontinuity?
    – Strants
    Aug 3 at 14:19












  • 2




    A graph is not a proof of anything. Not ever. A graph gives you a hint as to what to expect when you write an actual proof using formulas (limits, in this case).
    – Arnaud Mortier
    Aug 3 at 14:12











  • In theory, yes, but looking at a graph as a human is not enough. You have to prove that a theoretically ideal observer looking at a graph drawn with a theoretical infinitely thin curve cannot see a break point or anything. That's usually done with more algebra than geometry.
    – Arthur
    Aug 3 at 14:15











  • I can claim that you aren't showing me the graph of $fracsin xx$ just as easily as you are claiming to be showing it to me.
    – Saucy O'Path
    Aug 3 at 14:17











  • What is your definition of a removable discontinuity?
    – Strants
    Aug 3 at 14:19







2




2




A graph is not a proof of anything. Not ever. A graph gives you a hint as to what to expect when you write an actual proof using formulas (limits, in this case).
– Arnaud Mortier
Aug 3 at 14:12





A graph is not a proof of anything. Not ever. A graph gives you a hint as to what to expect when you write an actual proof using formulas (limits, in this case).
– Arnaud Mortier
Aug 3 at 14:12













In theory, yes, but looking at a graph as a human is not enough. You have to prove that a theoretically ideal observer looking at a graph drawn with a theoretical infinitely thin curve cannot see a break point or anything. That's usually done with more algebra than geometry.
– Arthur
Aug 3 at 14:15





In theory, yes, but looking at a graph as a human is not enough. You have to prove that a theoretically ideal observer looking at a graph drawn with a theoretical infinitely thin curve cannot see a break point or anything. That's usually done with more algebra than geometry.
– Arthur
Aug 3 at 14:15













I can claim that you aren't showing me the graph of $fracsin xx$ just as easily as you are claiming to be showing it to me.
– Saucy O'Path
Aug 3 at 14:17





I can claim that you aren't showing me the graph of $fracsin xx$ just as easily as you are claiming to be showing it to me.
– Saucy O'Path
Aug 3 at 14:17













What is your definition of a removable discontinuity?
– Strants
Aug 3 at 14:19




What is your definition of a removable discontinuity?
– Strants
Aug 3 at 14:19










2 Answers
2






active

oldest

votes

















up vote
0
down vote



accepted










Let $f(x)$ be a continuously differentiable function at $x=0$ with $f(0)=0$ and let
beginalign*
g(x) :=fracf(x)x text for xneq 0.
endalign*
The following formula is a direct consequence of l'Hopital's rule:
beginalign*
lim_xrightarrow 0 fracf(x)x=f'(0)
endalign*



Therefore you can continuously extend $g$ at $0$ by setting $g(0):=f'(0)$.



This is directly applicable to $f(x)=sin(x)$.






share|cite|improve this answer






























    up vote
    5
    down vote













    The function $f : mathbb Rsetminus 0to mathbb R$ defined by the formula $f(x) = fracsin xx$ is not discontinuous at $x=0$, because it is not even defined there. Functions cannot be continuous or discontinuous at points outside their domain!



    However you know from a geometric argument (or Taylor series) that
    $$lim_xto 0 fracsin xx = 1, $$
    so you may define a continuous extension $ g : mathbb R to mathbb R$ of your function,
    $$g(x) = begincases frac sin x x & xneq 0, \ 1 & x=0endcases $$
    so the best you can say is that there exists a continuous extension of $f$ that has the real numbers as its domain.



    This you can do whenever a function is not defined at a point but has finite limit at that point.



    Addendum. It is not common practice in real analysis, at least that I know of, but you may define a function to have a removable singularity at a point whenever it is not defined at that point, but is defined in its immediate vicinity and has a continuous extension defined at that point. (This is more of a complex-analytic concept, though.)



    In this sense you may say thay your function has a removable singularity at $x=0$ (not discontinuity).






    share|cite|improve this answer



















    • 1




      Actually, it depends on the definitions you are using. It's not a mistake to say that a function is discontinuous on a point on which it is not defined. After all, a function is continuous if it has a limit on that point which is equal to the value of the function. Hence in any other way the function is discontinuous at the point.
      – Mark
      Aug 3 at 14:27











    • @Mark See my addendum.
      – giobrach
      Aug 3 at 14:29






    • 1




      That's fine. But for me it's still a point at which the function is not continuous.
      – Mark
      Aug 3 at 14:40






    • 1




      If you really want to do things properly you shouldn't start with "The function $f(x)=fracsin xx$..." because $f(x)=fracsin xx$ is not a function.
      – Arnaud Mortier
      Aug 3 at 14:55






    • 1




      Thanks ! It helped a lot. giobrach
      – MIKEY SINGH
      Aug 3 at 14:58

















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    0
    down vote



    accepted










    Let $f(x)$ be a continuously differentiable function at $x=0$ with $f(0)=0$ and let
    beginalign*
    g(x) :=fracf(x)x text for xneq 0.
    endalign*
    The following formula is a direct consequence of l'Hopital's rule:
    beginalign*
    lim_xrightarrow 0 fracf(x)x=f'(0)
    endalign*



    Therefore you can continuously extend $g$ at $0$ by setting $g(0):=f'(0)$.



    This is directly applicable to $f(x)=sin(x)$.






    share|cite|improve this answer



























      up vote
      0
      down vote



      accepted










      Let $f(x)$ be a continuously differentiable function at $x=0$ with $f(0)=0$ and let
      beginalign*
      g(x) :=fracf(x)x text for xneq 0.
      endalign*
      The following formula is a direct consequence of l'Hopital's rule:
      beginalign*
      lim_xrightarrow 0 fracf(x)x=f'(0)
      endalign*



      Therefore you can continuously extend $g$ at $0$ by setting $g(0):=f'(0)$.



      This is directly applicable to $f(x)=sin(x)$.






      share|cite|improve this answer

























        up vote
        0
        down vote



        accepted







        up vote
        0
        down vote



        accepted






        Let $f(x)$ be a continuously differentiable function at $x=0$ with $f(0)=0$ and let
        beginalign*
        g(x) :=fracf(x)x text for xneq 0.
        endalign*
        The following formula is a direct consequence of l'Hopital's rule:
        beginalign*
        lim_xrightarrow 0 fracf(x)x=f'(0)
        endalign*



        Therefore you can continuously extend $g$ at $0$ by setting $g(0):=f'(0)$.



        This is directly applicable to $f(x)=sin(x)$.






        share|cite|improve this answer















        Let $f(x)$ be a continuously differentiable function at $x=0$ with $f(0)=0$ and let
        beginalign*
        g(x) :=fracf(x)x text for xneq 0.
        endalign*
        The following formula is a direct consequence of l'Hopital's rule:
        beginalign*
        lim_xrightarrow 0 fracf(x)x=f'(0)
        endalign*



        Therefore you can continuously extend $g$ at $0$ by setting $g(0):=f'(0)$.



        This is directly applicable to $f(x)=sin(x)$.







        share|cite|improve this answer















        share|cite|improve this answer



        share|cite|improve this answer








        edited Aug 3 at 14:45


























        answered Aug 3 at 14:38









        Tobias

        483314




        483314




















            up vote
            5
            down vote













            The function $f : mathbb Rsetminus 0to mathbb R$ defined by the formula $f(x) = fracsin xx$ is not discontinuous at $x=0$, because it is not even defined there. Functions cannot be continuous or discontinuous at points outside their domain!



            However you know from a geometric argument (or Taylor series) that
            $$lim_xto 0 fracsin xx = 1, $$
            so you may define a continuous extension $ g : mathbb R to mathbb R$ of your function,
            $$g(x) = begincases frac sin x x & xneq 0, \ 1 & x=0endcases $$
            so the best you can say is that there exists a continuous extension of $f$ that has the real numbers as its domain.



            This you can do whenever a function is not defined at a point but has finite limit at that point.



            Addendum. It is not common practice in real analysis, at least that I know of, but you may define a function to have a removable singularity at a point whenever it is not defined at that point, but is defined in its immediate vicinity and has a continuous extension defined at that point. (This is more of a complex-analytic concept, though.)



            In this sense you may say thay your function has a removable singularity at $x=0$ (not discontinuity).






            share|cite|improve this answer



















            • 1




              Actually, it depends on the definitions you are using. It's not a mistake to say that a function is discontinuous on a point on which it is not defined. After all, a function is continuous if it has a limit on that point which is equal to the value of the function. Hence in any other way the function is discontinuous at the point.
              – Mark
              Aug 3 at 14:27











            • @Mark See my addendum.
              – giobrach
              Aug 3 at 14:29






            • 1




              That's fine. But for me it's still a point at which the function is not continuous.
              – Mark
              Aug 3 at 14:40






            • 1




              If you really want to do things properly you shouldn't start with "The function $f(x)=fracsin xx$..." because $f(x)=fracsin xx$ is not a function.
              – Arnaud Mortier
              Aug 3 at 14:55






            • 1




              Thanks ! It helped a lot. giobrach
              – MIKEY SINGH
              Aug 3 at 14:58














            up vote
            5
            down vote













            The function $f : mathbb Rsetminus 0to mathbb R$ defined by the formula $f(x) = fracsin xx$ is not discontinuous at $x=0$, because it is not even defined there. Functions cannot be continuous or discontinuous at points outside their domain!



            However you know from a geometric argument (or Taylor series) that
            $$lim_xto 0 fracsin xx = 1, $$
            so you may define a continuous extension $ g : mathbb R to mathbb R$ of your function,
            $$g(x) = begincases frac sin x x & xneq 0, \ 1 & x=0endcases $$
            so the best you can say is that there exists a continuous extension of $f$ that has the real numbers as its domain.



            This you can do whenever a function is not defined at a point but has finite limit at that point.



            Addendum. It is not common practice in real analysis, at least that I know of, but you may define a function to have a removable singularity at a point whenever it is not defined at that point, but is defined in its immediate vicinity and has a continuous extension defined at that point. (This is more of a complex-analytic concept, though.)



            In this sense you may say thay your function has a removable singularity at $x=0$ (not discontinuity).






            share|cite|improve this answer



















            • 1




              Actually, it depends on the definitions you are using. It's not a mistake to say that a function is discontinuous on a point on which it is not defined. After all, a function is continuous if it has a limit on that point which is equal to the value of the function. Hence in any other way the function is discontinuous at the point.
              – Mark
              Aug 3 at 14:27











            • @Mark See my addendum.
              – giobrach
              Aug 3 at 14:29






            • 1




              That's fine. But for me it's still a point at which the function is not continuous.
              – Mark
              Aug 3 at 14:40






            • 1




              If you really want to do things properly you shouldn't start with "The function $f(x)=fracsin xx$..." because $f(x)=fracsin xx$ is not a function.
              – Arnaud Mortier
              Aug 3 at 14:55






            • 1




              Thanks ! It helped a lot. giobrach
              – MIKEY SINGH
              Aug 3 at 14:58












            up vote
            5
            down vote










            up vote
            5
            down vote









            The function $f : mathbb Rsetminus 0to mathbb R$ defined by the formula $f(x) = fracsin xx$ is not discontinuous at $x=0$, because it is not even defined there. Functions cannot be continuous or discontinuous at points outside their domain!



            However you know from a geometric argument (or Taylor series) that
            $$lim_xto 0 fracsin xx = 1, $$
            so you may define a continuous extension $ g : mathbb R to mathbb R$ of your function,
            $$g(x) = begincases frac sin x x & xneq 0, \ 1 & x=0endcases $$
            so the best you can say is that there exists a continuous extension of $f$ that has the real numbers as its domain.



            This you can do whenever a function is not defined at a point but has finite limit at that point.



            Addendum. It is not common practice in real analysis, at least that I know of, but you may define a function to have a removable singularity at a point whenever it is not defined at that point, but is defined in its immediate vicinity and has a continuous extension defined at that point. (This is more of a complex-analytic concept, though.)



            In this sense you may say thay your function has a removable singularity at $x=0$ (not discontinuity).






            share|cite|improve this answer















            The function $f : mathbb Rsetminus 0to mathbb R$ defined by the formula $f(x) = fracsin xx$ is not discontinuous at $x=0$, because it is not even defined there. Functions cannot be continuous or discontinuous at points outside their domain!



            However you know from a geometric argument (or Taylor series) that
            $$lim_xto 0 fracsin xx = 1, $$
            so you may define a continuous extension $ g : mathbb R to mathbb R$ of your function,
            $$g(x) = begincases frac sin x x & xneq 0, \ 1 & x=0endcases $$
            so the best you can say is that there exists a continuous extension of $f$ that has the real numbers as its domain.



            This you can do whenever a function is not defined at a point but has finite limit at that point.



            Addendum. It is not common practice in real analysis, at least that I know of, but you may define a function to have a removable singularity at a point whenever it is not defined at that point, but is defined in its immediate vicinity and has a continuous extension defined at that point. (This is more of a complex-analytic concept, though.)



            In this sense you may say thay your function has a removable singularity at $x=0$ (not discontinuity).







            share|cite|improve this answer















            share|cite|improve this answer



            share|cite|improve this answer








            edited Aug 3 at 14:58


























            answered Aug 3 at 14:20









            giobrach

            2,469418




            2,469418







            • 1




              Actually, it depends on the definitions you are using. It's not a mistake to say that a function is discontinuous on a point on which it is not defined. After all, a function is continuous if it has a limit on that point which is equal to the value of the function. Hence in any other way the function is discontinuous at the point.
              – Mark
              Aug 3 at 14:27











            • @Mark See my addendum.
              – giobrach
              Aug 3 at 14:29






            • 1




              That's fine. But for me it's still a point at which the function is not continuous.
              – Mark
              Aug 3 at 14:40






            • 1




              If you really want to do things properly you shouldn't start with "The function $f(x)=fracsin xx$..." because $f(x)=fracsin xx$ is not a function.
              – Arnaud Mortier
              Aug 3 at 14:55






            • 1




              Thanks ! It helped a lot. giobrach
              – MIKEY SINGH
              Aug 3 at 14:58












            • 1




              Actually, it depends on the definitions you are using. It's not a mistake to say that a function is discontinuous on a point on which it is not defined. After all, a function is continuous if it has a limit on that point which is equal to the value of the function. Hence in any other way the function is discontinuous at the point.
              – Mark
              Aug 3 at 14:27











            • @Mark See my addendum.
              – giobrach
              Aug 3 at 14:29






            • 1




              That's fine. But for me it's still a point at which the function is not continuous.
              – Mark
              Aug 3 at 14:40






            • 1




              If you really want to do things properly you shouldn't start with "The function $f(x)=fracsin xx$..." because $f(x)=fracsin xx$ is not a function.
              – Arnaud Mortier
              Aug 3 at 14:55






            • 1




              Thanks ! It helped a lot. giobrach
              – MIKEY SINGH
              Aug 3 at 14:58







            1




            1




            Actually, it depends on the definitions you are using. It's not a mistake to say that a function is discontinuous on a point on which it is not defined. After all, a function is continuous if it has a limit on that point which is equal to the value of the function. Hence in any other way the function is discontinuous at the point.
            – Mark
            Aug 3 at 14:27





            Actually, it depends on the definitions you are using. It's not a mistake to say that a function is discontinuous on a point on which it is not defined. After all, a function is continuous if it has a limit on that point which is equal to the value of the function. Hence in any other way the function is discontinuous at the point.
            – Mark
            Aug 3 at 14:27













            @Mark See my addendum.
            – giobrach
            Aug 3 at 14:29




            @Mark See my addendum.
            – giobrach
            Aug 3 at 14:29




            1




            1




            That's fine. But for me it's still a point at which the function is not continuous.
            – Mark
            Aug 3 at 14:40




            That's fine. But for me it's still a point at which the function is not continuous.
            – Mark
            Aug 3 at 14:40




            1




            1




            If you really want to do things properly you shouldn't start with "The function $f(x)=fracsin xx$..." because $f(x)=fracsin xx$ is not a function.
            – Arnaud Mortier
            Aug 3 at 14:55




            If you really want to do things properly you shouldn't start with "The function $f(x)=fracsin xx$..." because $f(x)=fracsin xx$ is not a function.
            – Arnaud Mortier
            Aug 3 at 14:55




            1




            1




            Thanks ! It helped a lot. giobrach
            – MIKEY SINGH
            Aug 3 at 14:58




            Thanks ! It helped a lot. giobrach
            – MIKEY SINGH
            Aug 3 at 14:58


            Comments

            Post a Comment

            Popular posts from this blog

            What is the equation of a 3D cone with generalised tilt?

            Image
            Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an...
            Read more

            Color the edges and diagonals of a regular polygon

            Image
            Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne...
            Read more

            Relationship between determinant of matrix and determinant of adjoint?

            Image
            Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/…...
            Read more