Why can $y=0$ be considered the asymptote of $f(x)=fracsin xx$?

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Why can $y=0$ be considered the asymptote of $f(x)=fracsin xx$ when the two graphs don't get any closer when $x$ approaches infinity? Because isn't the asymptote something that a graph will touch and stay on once the graph reaches infinity? And yet, in $f(x)=fracsin xx$, the size of $x$ has no effect on the graph's proximity to the asymptote. It just keeps fluctuating around $y=0$! So that suggests that the graph won't touch and stay on the asymptote at infinity? And if it doesn't, how can $y=0$ still be considered the asymptote of $f(x)=fracsin xx$?




algebra-precalculus asymptotics




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    Here's a MathJax tutorial :)
    – Shaun
    Jul 23 at 14:33






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    "asymptote" simply means a line such that the distance between the line and the graph tends to zero. We don't care about fluctuations.
    – Wojowu
    Jul 23 at 14:34






  • 4




    "The size of x has no effect on the graph's proximity to the asymptote." Not so.
    – saulspatz
    Jul 23 at 14:36






  • 1




    @Wojowu According to Wikipedia (en.wikipedia.org/wiki/Asymptote), "Some sources include the requirement [in the definition of "asymptote"] that the curve may not cross the line infinitely often, but this is unusual for modern authors." A literal reading of the word's etymology would also support this requirement, so Ethan Chen is more justified than he might think. (In the context of rational functions, of course, the extra stipulation makes no difference.)
    – Connor Harris
    Jul 23 at 14:38















up vote
3
down vote

favorite
2












Why can $y=0$ be considered the asymptote of $f(x)=fracsin xx$ when the two graphs don't get any closer when $x$ approaches infinity? Because isn't the asymptote something that a graph will touch and stay on once the graph reaches infinity? And yet, in $f(x)=fracsin xx$, the size of $x$ has no effect on the graph's proximity to the asymptote. It just keeps fluctuating around $y=0$! So that suggests that the graph won't touch and stay on the asymptote at infinity? And if it doesn't, how can $y=0$ still be considered the asymptote of $f(x)=fracsin xx$?




algebra-precalculus asymptotics




share|cite|improve this question

















  • 1




    Here's a MathJax tutorial :)
    – Shaun
    Jul 23 at 14:33






  • 1




    "asymptote" simply means a line such that the distance between the line and the graph tends to zero. We don't care about fluctuations.
    – Wojowu
    Jul 23 at 14:34






  • 4




    "The size of x has no effect on the graph's proximity to the asymptote." Not so.
    – saulspatz
    Jul 23 at 14:36






  • 1




    @Wojowu According to Wikipedia (en.wikipedia.org/wiki/Asymptote), "Some sources include the requirement [in the definition of "asymptote"] that the curve may not cross the line infinitely often, but this is unusual for modern authors." A literal reading of the word's etymology would also support this requirement, so Ethan Chen is more justified than he might think. (In the context of rational functions, of course, the extra stipulation makes no difference.)
    – Connor Harris
    Jul 23 at 14:38













up vote
3
down vote

favorite
2









up vote
3
down vote

favorite
2






2





Why can $y=0$ be considered the asymptote of $f(x)=fracsin xx$ when the two graphs don't get any closer when $x$ approaches infinity? Because isn't the asymptote something that a graph will touch and stay on once the graph reaches infinity? And yet, in $f(x)=fracsin xx$, the size of $x$ has no effect on the graph's proximity to the asymptote. It just keeps fluctuating around $y=0$! So that suggests that the graph won't touch and stay on the asymptote at infinity? And if it doesn't, how can $y=0$ still be considered the asymptote of $f(x)=fracsin xx$?




algebra-precalculus asymptotics




share|cite|improve this question













Why can $y=0$ be considered the asymptote of $f(x)=fracsin xx$ when the two graphs don't get any closer when $x$ approaches infinity? Because isn't the asymptote something that a graph will touch and stay on once the graph reaches infinity? And yet, in $f(x)=fracsin xx$, the size of $x$ has no effect on the graph's proximity to the asymptote. It just keeps fluctuating around $y=0$! So that suggests that the graph won't touch and stay on the asymptote at infinity? And if it doesn't, how can $y=0$ still be considered the asymptote of $f(x)=fracsin xx$?





algebra-precalculus asymptotics





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edited Jul 23 at 17:11









peterh

2,14731631




2,14731631









asked Jul 23 at 14:31









Ethan Chan

603322




603322







  • 1




    Here's a MathJax tutorial :)
    – Shaun
    Jul 23 at 14:33






  • 1




    "asymptote" simply means a line such that the distance between the line and the graph tends to zero. We don't care about fluctuations.
    – Wojowu
    Jul 23 at 14:34






  • 4




    "The size of x has no effect on the graph's proximity to the asymptote." Not so.
    – saulspatz
    Jul 23 at 14:36






  • 1




    @Wojowu According to Wikipedia (en.wikipedia.org/wiki/Asymptote), "Some sources include the requirement [in the definition of "asymptote"] that the curve may not cross the line infinitely often, but this is unusual for modern authors." A literal reading of the word's etymology would also support this requirement, so Ethan Chen is more justified than he might think. (In the context of rational functions, of course, the extra stipulation makes no difference.)
    – Connor Harris
    Jul 23 at 14:38













  • 1




    Here's a MathJax tutorial :)
    – Shaun
    Jul 23 at 14:33






  • 1




    "asymptote" simply means a line such that the distance between the line and the graph tends to zero. We don't care about fluctuations.
    – Wojowu
    Jul 23 at 14:34






  • 4




    "The size of x has no effect on the graph's proximity to the asymptote." Not so.
    – saulspatz
    Jul 23 at 14:36






  • 1




    @Wojowu According to Wikipedia (en.wikipedia.org/wiki/Asymptote), "Some sources include the requirement [in the definition of "asymptote"] that the curve may not cross the line infinitely often, but this is unusual for modern authors." A literal reading of the word's etymology would also support this requirement, so Ethan Chen is more justified than he might think. (In the context of rational functions, of course, the extra stipulation makes no difference.)
    – Connor Harris
    Jul 23 at 14:38








1




1




Here's a MathJax tutorial :)
– Shaun
Jul 23 at 14:33




Here's a MathJax tutorial :)
– Shaun
Jul 23 at 14:33




1




1




"asymptote" simply means a line such that the distance between the line and the graph tends to zero. We don't care about fluctuations.
– Wojowu
Jul 23 at 14:34




"asymptote" simply means a line such that the distance between the line and the graph tends to zero. We don't care about fluctuations.
– Wojowu
Jul 23 at 14:34




4




4




"The size of x has no effect on the graph's proximity to the asymptote." Not so.
– saulspatz
Jul 23 at 14:36




"The size of x has no effect on the graph's proximity to the asymptote." Not so.
– saulspatz
Jul 23 at 14:36




1




1




@Wojowu According to Wikipedia (en.wikipedia.org/wiki/Asymptote), "Some sources include the requirement [in the definition of "asymptote"] that the curve may not cross the line infinitely often, but this is unusual for modern authors." A literal reading of the word's etymology would also support this requirement, so Ethan Chen is more justified than he might think. (In the context of rational functions, of course, the extra stipulation makes no difference.)
– Connor Harris
Jul 23 at 14:38





@Wojowu According to Wikipedia (en.wikipedia.org/wiki/Asymptote), "Some sources include the requirement [in the definition of "asymptote"] that the curve may not cross the line infinitely often, but this is unusual for modern authors." A literal reading of the word's etymology would also support this requirement, so Ethan Chen is more justified than he might think. (In the context of rational functions, of course, the extra stipulation makes no difference.)
– Connor Harris
Jul 23 at 14:38











4 Answers
4






active

oldest

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up vote
7
down vote













There are several misconceptions in your post.



$$fracsin xx$$ does get closer and closer to $0$ when $x$ increases. In fact,



$$left|fracsin xxright|lefrac1x$$ and the RHS clearly tends to $0$.



"... isn't the asymptote something that a graph will touch and stay on ...": no, quite often an asymptote doesn't touch the curve. It just needs to come arbitrarily close. In the case at hand, it crosses it infinitely often, but this does not matter.






share|cite|improve this answer






























    up vote
    4
    down vote













    I have already written to you about this. Look at the graph of your function:



    enter image description here



    Don't you agree that, as we move to the right, the graph gets closer to $0$? To be more precise, if we fix a number $varepsilon>0$, then, if $M$ is large enough, the graph of the restriction of $f$ to $[M,+infty)$ is between the lines $y=varepsilon$ and $y=-varepsilon$. That's what it means that the line $y=0$ is an asymptote of the graph of $f$.






    share|cite|improve this answer




























      up vote
      2
      down vote













      The definition of an horizontal asymptote is that
      $$limlimits_x to infty f(x) = c.$$



      Which is exactly the case here with $c=0$.



      In no way the definition of an asymptote implies that the function is supposed to stay on the same side of the asymptote for $x$ large enough.






      share|cite|improve this answer























      • So when x gets to infinity $sin(x)/x$ becomes 0? How so? Can you please explain this in the terms of someone who hasn't learnt calculus yet? Thanks.
        – Ethan Chan
        Jul 23 at 14:39






      • 2




        Using simple words: $sin x$ is bounded by $1$. If you divide $1$ by a large number you get a number close to $0$.
        – mathcounterexamples.net
        Jul 23 at 14:40











      • Ah right, I forgot about that. But how do we know sinx will output 1 when x=infinity?
        – Ethan Chan
        Jul 23 at 14:47






      • 1




        It will not output $1$... It will output something that is less or equal to $1$ which is even better!
        – mathcounterexamples.net
        Jul 23 at 14:52

















      up vote
      1
      down vote














      f(x)=sinx/x, the size of x has no effect on the graph's proximity to the asymptote.




      Yes it does:
      $$textmax_x>x_0left|fracsin(x)x-0right|leq frac1x_0$$






      share|cite|improve this answer





















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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        7
        down vote













        There are several misconceptions in your post.



        $$fracsin xx$$ does get closer and closer to $0$ when $x$ increases. In fact,



        $$left|fracsin xxright|lefrac1x$$ and the RHS clearly tends to $0$.



        "... isn't the asymptote something that a graph will touch and stay on ...": no, quite often an asymptote doesn't touch the curve. It just needs to come arbitrarily close. In the case at hand, it crosses it infinitely often, but this does not matter.






        share|cite|improve this answer



























          up vote
          7
          down vote













          There are several misconceptions in your post.



          $$fracsin xx$$ does get closer and closer to $0$ when $x$ increases. In fact,



          $$left|fracsin xxright|lefrac1x$$ and the RHS clearly tends to $0$.



          "... isn't the asymptote something that a graph will touch and stay on ...": no, quite often an asymptote doesn't touch the curve. It just needs to come arbitrarily close. In the case at hand, it crosses it infinitely often, but this does not matter.






          share|cite|improve this answer

























            up vote
            7
            down vote










            up vote
            7
            down vote









            There are several misconceptions in your post.



            $$fracsin xx$$ does get closer and closer to $0$ when $x$ increases. In fact,



            $$left|fracsin xxright|lefrac1x$$ and the RHS clearly tends to $0$.



            "... isn't the asymptote something that a graph will touch and stay on ...": no, quite often an asymptote doesn't touch the curve. It just needs to come arbitrarily close. In the case at hand, it crosses it infinitely often, but this does not matter.






            share|cite|improve this answer















            There are several misconceptions in your post.



            $$fracsin xx$$ does get closer and closer to $0$ when $x$ increases. In fact,



            $$left|fracsin xxright|lefrac1x$$ and the RHS clearly tends to $0$.



            "... isn't the asymptote something that a graph will touch and stay on ...": no, quite often an asymptote doesn't touch the curve. It just needs to come arbitrarily close. In the case at hand, it crosses it infinitely often, but this does not matter.







            share|cite|improve this answer















            share|cite|improve this answer



            share|cite|improve this answer








            edited Jul 23 at 15:52


























            answered Jul 23 at 14:48









            Yves Daoust

            111k665203




            111k665203




















                up vote
                4
                down vote













                I have already written to you about this. Look at the graph of your function:



                enter image description here



                Don't you agree that, as we move to the right, the graph gets closer to $0$? To be more precise, if we fix a number $varepsilon>0$, then, if $M$ is large enough, the graph of the restriction of $f$ to $[M,+infty)$ is between the lines $y=varepsilon$ and $y=-varepsilon$. That's what it means that the line $y=0$ is an asymptote of the graph of $f$.






                share|cite|improve this answer

























                  up vote
                  4
                  down vote













                  I have already written to you about this. Look at the graph of your function:



                  enter image description here



                  Don't you agree that, as we move to the right, the graph gets closer to $0$? To be more precise, if we fix a number $varepsilon>0$, then, if $M$ is large enough, the graph of the restriction of $f$ to $[M,+infty)$ is between the lines $y=varepsilon$ and $y=-varepsilon$. That's what it means that the line $y=0$ is an asymptote of the graph of $f$.






                  share|cite|improve this answer























                    up vote
                    4
                    down vote










                    up vote
                    4
                    down vote









                    I have already written to you about this. Look at the graph of your function:



                    enter image description here



                    Don't you agree that, as we move to the right, the graph gets closer to $0$? To be more precise, if we fix a number $varepsilon>0$, then, if $M$ is large enough, the graph of the restriction of $f$ to $[M,+infty)$ is between the lines $y=varepsilon$ and $y=-varepsilon$. That's what it means that the line $y=0$ is an asymptote of the graph of $f$.






                    share|cite|improve this answer













                    I have already written to you about this. Look at the graph of your function:



                    enter image description here



                    Don't you agree that, as we move to the right, the graph gets closer to $0$? To be more precise, if we fix a number $varepsilon>0$, then, if $M$ is large enough, the graph of the restriction of $f$ to $[M,+infty)$ is between the lines $y=varepsilon$ and $y=-varepsilon$. That's what it means that the line $y=0$ is an asymptote of the graph of $f$.







                    share|cite|improve this answer













                    share|cite|improve this answer



                    share|cite|improve this answer











                    answered Jul 23 at 14:45









                    José Carlos Santos

                    113k1698176




                    113k1698176




















                        up vote
                        2
                        down vote













                        The definition of an horizontal asymptote is that
                        $$limlimits_x to infty f(x) = c.$$



                        Which is exactly the case here with $c=0$.



                        In no way the definition of an asymptote implies that the function is supposed to stay on the same side of the asymptote for $x$ large enough.






                        share|cite|improve this answer























                        • So when x gets to infinity $sin(x)/x$ becomes 0? How so? Can you please explain this in the terms of someone who hasn't learnt calculus yet? Thanks.
                          – Ethan Chan
                          Jul 23 at 14:39






                        • 2




                          Using simple words: $sin x$ is bounded by $1$. If you divide $1$ by a large number you get a number close to $0$.
                          – mathcounterexamples.net
                          Jul 23 at 14:40











                        • Ah right, I forgot about that. But how do we know sinx will output 1 when x=infinity?
                          – Ethan Chan
                          Jul 23 at 14:47






                        • 1




                          It will not output $1$... It will output something that is less or equal to $1$ which is even better!
                          – mathcounterexamples.net
                          Jul 23 at 14:52














                        up vote
                        2
                        down vote













                        The definition of an horizontal asymptote is that
                        $$limlimits_x to infty f(x) = c.$$



                        Which is exactly the case here with $c=0$.



                        In no way the definition of an asymptote implies that the function is supposed to stay on the same side of the asymptote for $x$ large enough.






                        share|cite|improve this answer























                        • So when x gets to infinity $sin(x)/x$ becomes 0? How so? Can you please explain this in the terms of someone who hasn't learnt calculus yet? Thanks.
                          – Ethan Chan
                          Jul 23 at 14:39






                        • 2




                          Using simple words: $sin x$ is bounded by $1$. If you divide $1$ by a large number you get a number close to $0$.
                          – mathcounterexamples.net
                          Jul 23 at 14:40











                        • Ah right, I forgot about that. But how do we know sinx will output 1 when x=infinity?
                          – Ethan Chan
                          Jul 23 at 14:47






                        • 1




                          It will not output $1$... It will output something that is less or equal to $1$ which is even better!
                          – mathcounterexamples.net
                          Jul 23 at 14:52












                        up vote
                        2
                        down vote










                        up vote
                        2
                        down vote









                        The definition of an horizontal asymptote is that
                        $$limlimits_x to infty f(x) = c.$$



                        Which is exactly the case here with $c=0$.



                        In no way the definition of an asymptote implies that the function is supposed to stay on the same side of the asymptote for $x$ large enough.






                        share|cite|improve this answer















                        The definition of an horizontal asymptote is that
                        $$limlimits_x to infty f(x) = c.$$



                        Which is exactly the case here with $c=0$.



                        In no way the definition of an asymptote implies that the function is supposed to stay on the same side of the asymptote for $x$ large enough.







                        share|cite|improve this answer















                        share|cite|improve this answer



                        share|cite|improve this answer








                        edited Jul 23 at 14:39


























                        answered Jul 23 at 14:35









                        mathcounterexamples.net

                        23.9k21653




                        23.9k21653











                        • So when x gets to infinity $sin(x)/x$ becomes 0? How so? Can you please explain this in the terms of someone who hasn't learnt calculus yet? Thanks.
                          – Ethan Chan
                          Jul 23 at 14:39






                        • 2




                          Using simple words: $sin x$ is bounded by $1$. If you divide $1$ by a large number you get a number close to $0$.
                          – mathcounterexamples.net
                          Jul 23 at 14:40











                        • Ah right, I forgot about that. But how do we know sinx will output 1 when x=infinity?
                          – Ethan Chan
                          Jul 23 at 14:47






                        • 1




                          It will not output $1$... It will output something that is less or equal to $1$ which is even better!
                          – mathcounterexamples.net
                          Jul 23 at 14:52
















                        • So when x gets to infinity $sin(x)/x$ becomes 0? How so? Can you please explain this in the terms of someone who hasn't learnt calculus yet? Thanks.
                          – Ethan Chan
                          Jul 23 at 14:39






                        • 2




                          Using simple words: $sin x$ is bounded by $1$. If you divide $1$ by a large number you get a number close to $0$.
                          – mathcounterexamples.net
                          Jul 23 at 14:40











                        • Ah right, I forgot about that. But how do we know sinx will output 1 when x=infinity?
                          – Ethan Chan
                          Jul 23 at 14:47






                        • 1




                          It will not output $1$... It will output something that is less or equal to $1$ which is even better!
                          – mathcounterexamples.net
                          Jul 23 at 14:52















                        So when x gets to infinity $sin(x)/x$ becomes 0? How so? Can you please explain this in the terms of someone who hasn't learnt calculus yet? Thanks.
                        – Ethan Chan
                        Jul 23 at 14:39




                        So when x gets to infinity $sin(x)/x$ becomes 0? How so? Can you please explain this in the terms of someone who hasn't learnt calculus yet? Thanks.
                        – Ethan Chan
                        Jul 23 at 14:39




                        2




                        2




                        Using simple words: $sin x$ is bounded by $1$. If you divide $1$ by a large number you get a number close to $0$.
                        – mathcounterexamples.net
                        Jul 23 at 14:40





                        Using simple words: $sin x$ is bounded by $1$. If you divide $1$ by a large number you get a number close to $0$.
                        – mathcounterexamples.net
                        Jul 23 at 14:40













                        Ah right, I forgot about that. But how do we know sinx will output 1 when x=infinity?
                        – Ethan Chan
                        Jul 23 at 14:47




                        Ah right, I forgot about that. But how do we know sinx will output 1 when x=infinity?
                        – Ethan Chan
                        Jul 23 at 14:47




                        1




                        1




                        It will not output $1$... It will output something that is less or equal to $1$ which is even better!
                        – mathcounterexamples.net
                        Jul 23 at 14:52




                        It will not output $1$... It will output something that is less or equal to $1$ which is even better!
                        – mathcounterexamples.net
                        Jul 23 at 14:52










                        up vote
                        1
                        down vote














                        f(x)=sinx/x, the size of x has no effect on the graph's proximity to the asymptote.




                        Yes it does:
                        $$textmax_x>x_0left|fracsin(x)x-0right|leq frac1x_0$$






                        share|cite|improve this answer

























                          up vote
                          1
                          down vote














                          f(x)=sinx/x, the size of x has no effect on the graph's proximity to the asymptote.




                          Yes it does:
                          $$textmax_x>x_0left|fracsin(x)x-0right|leq frac1x_0$$






                          share|cite|improve this answer























                            up vote
                            1
                            down vote










                            up vote
                            1
                            down vote










                            f(x)=sinx/x, the size of x has no effect on the graph's proximity to the asymptote.




                            Yes it does:
                            $$textmax_x>x_0left|fracsin(x)x-0right|leq frac1x_0$$






                            share|cite|improve this answer














                            f(x)=sinx/x, the size of x has no effect on the graph's proximity to the asymptote.




                            Yes it does:
                            $$textmax_x>x_0left|fracsin(x)x-0right|leq frac1x_0$$







                            share|cite|improve this answer













                            share|cite|improve this answer



                            share|cite|improve this answer











                            answered Jul 23 at 14:37









                            Wouter

                            5,50021434




                            5,50021434






















                                 

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                                Color the edges and diagonals of a regular polygon

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                                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...
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                                Relationship between determinant of matrix and determinant of adjoint?

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                                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/…...
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