Is $F(x)=frac{x}{x+1}$ for $xin [0,1)$, $F(1)=frac{1}{2}$ continuous?












0












$begingroup$


How is this function continuous? The book before me states that it's continuous.



$$
F(x)=begin{cases} dfrac{x}{x+1},&;;xin [0,1),\\ ;; dfrac{1}{2},&;;x=1end{cases}
$$



I might be wrong



Let $epsilon_0=dfrac{1}{12},$ $x=dfrac{1}{2}$ and $y=1$, then begin{align}left|Fleft(dfrac{1}{2}right)-Fleft(1right)right|&=left|Fleft(dfrac{1}{2}right)-Fleft(1right)right|\&=left|Fleft(dfrac{1}{2}right)-Fleft(1right)right|\&=left|dfrac{1}{3}-dfrac{1}{2}right|\&=dfrac{1}{6}>dfrac{1}{12}=epsilon_0end{align}
Then, it's not continuous. I know we can use Pasting lemma but $[0,1)cap 1=phi.$






continuity epsilon-delta







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  • $begingroup$
    Why is $frac{x}{x+1}$ continuous on $[0, 1]$? Because $x$ is continuous, and $x+1$ is continuous and non-zero. There's no real need to split this function in cases
    $endgroup$
    – Jakobian
    Dec 28 '18 at 15:45












  • $begingroup$
    You are claiming that the function is discontinuous... at what point ? By your reasoning, it would seem that the function $f(x)=x$ should also be discontinuous...
    $endgroup$
    – leonbloy
    Dec 28 '18 at 15:46












  • $begingroup$
    By your reasoning there are no continuous functions except constants. Look at the definition - it does not require that $|F(1/2)-F(1)|<epsilon$.
    $endgroup$
    – David C. Ullrich
    Dec 28 '18 at 15:49


















0












$begingroup$


How is this function continuous? The book before me states that it's continuous.



$$
F(x)=begin{cases} dfrac{x}{x+1},&;;xin [0,1),\\ ;; dfrac{1}{2},&;;x=1end{cases}
$$



I might be wrong



Let $epsilon_0=dfrac{1}{12},$ $x=dfrac{1}{2}$ and $y=1$, then begin{align}left|Fleft(dfrac{1}{2}right)-Fleft(1right)right|&=left|Fleft(dfrac{1}{2}right)-Fleft(1right)right|\&=left|Fleft(dfrac{1}{2}right)-Fleft(1right)right|\&=left|dfrac{1}{3}-dfrac{1}{2}right|\&=dfrac{1}{6}>dfrac{1}{12}=epsilon_0end{align}
Then, it's not continuous. I know we can use Pasting lemma but $[0,1)cap 1=phi.$






continuity epsilon-delta







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why is $frac{x}{x+1}$ continuous on $[0, 1]$? Because $x$ is continuous, and $x+1$ is continuous and non-zero. There's no real need to split this function in cases
    $endgroup$
    – Jakobian
    Dec 28 '18 at 15:45












  • $begingroup$
    You are claiming that the function is discontinuous... at what point ? By your reasoning, it would seem that the function $f(x)=x$ should also be discontinuous...
    $endgroup$
    – leonbloy
    Dec 28 '18 at 15:46












  • $begingroup$
    By your reasoning there are no continuous functions except constants. Look at the definition - it does not require that $|F(1/2)-F(1)|<epsilon$.
    $endgroup$
    – David C. Ullrich
    Dec 28 '18 at 15:49
















0












0








0


1



$begingroup$


How is this function continuous? The book before me states that it's continuous.



$$
F(x)=begin{cases} dfrac{x}{x+1},&;;xin [0,1),\\ ;; dfrac{1}{2},&;;x=1end{cases}
$$



I might be wrong



Let $epsilon_0=dfrac{1}{12},$ $x=dfrac{1}{2}$ and $y=1$, then begin{align}left|Fleft(dfrac{1}{2}right)-Fleft(1right)right|&=left|Fleft(dfrac{1}{2}right)-Fleft(1right)right|\&=left|Fleft(dfrac{1}{2}right)-Fleft(1right)right|\&=left|dfrac{1}{3}-dfrac{1}{2}right|\&=dfrac{1}{6}>dfrac{1}{12}=epsilon_0end{align}
Then, it's not continuous. I know we can use Pasting lemma but $[0,1)cap 1=phi.$






continuity epsilon-delta







share|cite|improve this question











$endgroup$




How is this function continuous? The book before me states that it's continuous.



$$
F(x)=begin{cases} dfrac{x}{x+1},&;;xin [0,1),\\ ;; dfrac{1}{2},&;;x=1end{cases}
$$



I might be wrong



Let $epsilon_0=dfrac{1}{12},$ $x=dfrac{1}{2}$ and $y=1$, then begin{align}left|Fleft(dfrac{1}{2}right)-Fleft(1right)right|&=left|Fleft(dfrac{1}{2}right)-Fleft(1right)right|\&=left|Fleft(dfrac{1}{2}right)-Fleft(1right)right|\&=left|dfrac{1}{3}-dfrac{1}{2}right|\&=dfrac{1}{6}>dfrac{1}{12}=epsilon_0end{align}
Then, it's not continuous. I know we can use Pasting lemma but $[0,1)cap 1=phi.$






continuity epsilon-delta




continuity epsilon-delta






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edited Dec 28 '18 at 16:42









Martin Sleziak

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asked Dec 28 '18 at 15:42









Omojola MichealOmojola Micheal

1,988324




1,988324












  • $begingroup$
    Why is $frac{x}{x+1}$ continuous on $[0, 1]$? Because $x$ is continuous, and $x+1$ is continuous and non-zero. There's no real need to split this function in cases
    $endgroup$
    – Jakobian
    Dec 28 '18 at 15:45












  • $begingroup$
    You are claiming that the function is discontinuous... at what point ? By your reasoning, it would seem that the function $f(x)=x$ should also be discontinuous...
    $endgroup$
    – leonbloy
    Dec 28 '18 at 15:46












  • $begingroup$
    By your reasoning there are no continuous functions except constants. Look at the definition - it does not require that $|F(1/2)-F(1)|<epsilon$.
    $endgroup$
    – David C. Ullrich
    Dec 28 '18 at 15:49




















  • $begingroup$
    Why is $frac{x}{x+1}$ continuous on $[0, 1]$? Because $x$ is continuous, and $x+1$ is continuous and non-zero. There's no real need to split this function in cases
    $endgroup$
    – Jakobian
    Dec 28 '18 at 15:45












  • $begingroup$
    You are claiming that the function is discontinuous... at what point ? By your reasoning, it would seem that the function $f(x)=x$ should also be discontinuous...
    $endgroup$
    – leonbloy
    Dec 28 '18 at 15:46












  • $begingroup$
    By your reasoning there are no continuous functions except constants. Look at the definition - it does not require that $|F(1/2)-F(1)|<epsilon$.
    $endgroup$
    – David C. Ullrich
    Dec 28 '18 at 15:49


















$begingroup$
Why is $frac{x}{x+1}$ continuous on $[0, 1]$? Because $x$ is continuous, and $x+1$ is continuous and non-zero. There's no real need to split this function in cases
$endgroup$
– Jakobian
Dec 28 '18 at 15:45






$begingroup$
Why is $frac{x}{x+1}$ continuous on $[0, 1]$? Because $x$ is continuous, and $x+1$ is continuous and non-zero. There's no real need to split this function in cases
$endgroup$
– Jakobian
Dec 28 '18 at 15:45














$begingroup$
You are claiming that the function is discontinuous... at what point ? By your reasoning, it would seem that the function $f(x)=x$ should also be discontinuous...
$endgroup$
– leonbloy
Dec 28 '18 at 15:46






$begingroup$
You are claiming that the function is discontinuous... at what point ? By your reasoning, it would seem that the function $f(x)=x$ should also be discontinuous...
$endgroup$
– leonbloy
Dec 28 '18 at 15:46














$begingroup$
By your reasoning there are no continuous functions except constants. Look at the definition - it does not require that $|F(1/2)-F(1)|<epsilon$.
$endgroup$
– David C. Ullrich
Dec 28 '18 at 15:49






$begingroup$
By your reasoning there are no continuous functions except constants. Look at the definition - it does not require that $|F(1/2)-F(1)|<epsilon$.
$endgroup$
– David C. Ullrich
Dec 28 '18 at 15:49












2 Answers
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$begingroup$

Define$$begin{array}{rccc}Gcolon&[0,1]&longrightarrow&mathbb R\&x&mapsto&dfrac x{x+1}.end{array}$$Then $F=G$ and I suppose that it is clear to you that $G$ is continuous. Therefore, $F$ is continuous too.



On the other hand, there is no way of determining whther a function $varphicolon[0,1]longrightarrowmathbb R$ is continuous at $1$ just by using the values of $varphi$ at $1$ and at $frac12$, which is what you tried to do.






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    $begingroup$

    I like to think about it as a game between you and an adversary. You are tying to catch your adversary off guard by demanding that they produce a $delta>0$ for any $epsilon>0$ such that $0<|x-a|<delta$ implies that $|f(x)-f(a)|<epsilon$.



    The definition of being continuous does not let you chose your $delta$; that role is reserved for your adversary.



    So when you chose $epsilon_0=1/12$, you do not then also get to pick the $x$ and $y$ values. That is for the adversary to decide.






    share|cite|improve this answer









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      2 Answers
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      2 Answers
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      $begingroup$

      Define$$begin{array}{rccc}Gcolon&[0,1]&longrightarrow&mathbb R\&x&mapsto&dfrac x{x+1}.end{array}$$Then $F=G$ and I suppose that it is clear to you that $G$ is continuous. Therefore, $F$ is continuous too.



      On the other hand, there is no way of determining whther a function $varphicolon[0,1]longrightarrowmathbb R$ is continuous at $1$ just by using the values of $varphi$ at $1$ and at $frac12$, which is what you tried to do.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Define$$begin{array}{rccc}Gcolon&[0,1]&longrightarrow&mathbb R\&x&mapsto&dfrac x{x+1}.end{array}$$Then $F=G$ and I suppose that it is clear to you that $G$ is continuous. Therefore, $F$ is continuous too.



        On the other hand, there is no way of determining whther a function $varphicolon[0,1]longrightarrowmathbb R$ is continuous at $1$ just by using the values of $varphi$ at $1$ and at $frac12$, which is what you tried to do.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Define$$begin{array}{rccc}Gcolon&[0,1]&longrightarrow&mathbb R\&x&mapsto&dfrac x{x+1}.end{array}$$Then $F=G$ and I suppose that it is clear to you that $G$ is continuous. Therefore, $F$ is continuous too.



          On the other hand, there is no way of determining whther a function $varphicolon[0,1]longrightarrowmathbb R$ is continuous at $1$ just by using the values of $varphi$ at $1$ and at $frac12$, which is what you tried to do.






          share|cite|improve this answer









          $endgroup$



          Define$$begin{array}{rccc}Gcolon&[0,1]&longrightarrow&mathbb R\&x&mapsto&dfrac x{x+1}.end{array}$$Then $F=G$ and I suppose that it is clear to you that $G$ is continuous. Therefore, $F$ is continuous too.



          On the other hand, there is no way of determining whther a function $varphicolon[0,1]longrightarrowmathbb R$ is continuous at $1$ just by using the values of $varphi$ at $1$ and at $frac12$, which is what you tried to do.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 28 '18 at 15:47









          José Carlos SantosJosé Carlos Santos

          168k23132236




          168k23132236























              5












              $begingroup$

              I like to think about it as a game between you and an adversary. You are tying to catch your adversary off guard by demanding that they produce a $delta>0$ for any $epsilon>0$ such that $0<|x-a|<delta$ implies that $|f(x)-f(a)|<epsilon$.



              The definition of being continuous does not let you chose your $delta$; that role is reserved for your adversary.



              So when you chose $epsilon_0=1/12$, you do not then also get to pick the $x$ and $y$ values. That is for the adversary to decide.






              share|cite|improve this answer









              $endgroup$


















                5












                $begingroup$

                I like to think about it as a game between you and an adversary. You are tying to catch your adversary off guard by demanding that they produce a $delta>0$ for any $epsilon>0$ such that $0<|x-a|<delta$ implies that $|f(x)-f(a)|<epsilon$.



                The definition of being continuous does not let you chose your $delta$; that role is reserved for your adversary.



                So when you chose $epsilon_0=1/12$, you do not then also get to pick the $x$ and $y$ values. That is for the adversary to decide.






                share|cite|improve this answer









                $endgroup$
















                  5












                  5








                  5





                  $begingroup$

                  I like to think about it as a game between you and an adversary. You are tying to catch your adversary off guard by demanding that they produce a $delta>0$ for any $epsilon>0$ such that $0<|x-a|<delta$ implies that $|f(x)-f(a)|<epsilon$.



                  The definition of being continuous does not let you chose your $delta$; that role is reserved for your adversary.



                  So when you chose $epsilon_0=1/12$, you do not then also get to pick the $x$ and $y$ values. That is for the adversary to decide.






                  share|cite|improve this answer









                  $endgroup$



                  I like to think about it as a game between you and an adversary. You are tying to catch your adversary off guard by demanding that they produce a $delta>0$ for any $epsilon>0$ such that $0<|x-a|<delta$ implies that $|f(x)-f(a)|<epsilon$.



                  The definition of being continuous does not let you chose your $delta$; that role is reserved for your adversary.



                  So when you chose $epsilon_0=1/12$, you do not then also get to pick the $x$ and $y$ values. That is for the adversary to decide.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 28 '18 at 15:46









                  DudeManDudeMan

                  511




                  511






























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