Proving that an Epsilon-Delta Proof is not true












0












$begingroup$


We write $lim_{x to a} f(x)=L$ if the following is true
$(forallepsilon>0)(existsdelta>0)(forall x)(0<|x-a|<deltarightarrow|f(x)-L|<epsilon)$



Let $f:Bbb{R}rightarrowBbb{R}$ be given by



$f(x)=
begin{cases}
0, & text{if $x<0$} \
1/2, & text{if $=0$} \
1, & text{if $x>0$}
end{cases}$



We will show that it is not the case that $lim_{x to 0} f(x)=1/2$



(a) Write the negation of $lim_{x to 0} f(x)=1/2$ using the epsilon-delta definition given above




  • I attempted to find the negation of this and this what I got after some calculations


$(existsepsilon>0)(foralldelta>0)(exists x)(0<|x-0|<deltaland|f(x)-1/2|geepsilon)$



(b) Prove the assertion that you found in part (a)
Hint: $epsilon=1/4$




  • This is where I am stuck. How would I prove the epsilon-delta expression that I found in part (a). Any kind of help would be appreciated.






proof-verification proof-explanation epsilon-delta







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your negated formula is not correct: you should have $|f(x)-1/2|geqepsilon$ in the final bit.
    $endgroup$
    – Leo163
    Nov 30 '18 at 12:20
















0












$begingroup$


We write $lim_{x to a} f(x)=L$ if the following is true
$(forallepsilon>0)(existsdelta>0)(forall x)(0<|x-a|<deltarightarrow|f(x)-L|<epsilon)$



Let $f:Bbb{R}rightarrowBbb{R}$ be given by



$f(x)=
begin{cases}
0, & text{if $x<0$} \
1/2, & text{if $=0$} \
1, & text{if $x>0$}
end{cases}$



We will show that it is not the case that $lim_{x to 0} f(x)=1/2$



(a) Write the negation of $lim_{x to 0} f(x)=1/2$ using the epsilon-delta definition given above




  • I attempted to find the negation of this and this what I got after some calculations


$(existsepsilon>0)(foralldelta>0)(exists x)(0<|x-0|<deltaland|f(x)-1/2|geepsilon)$



(b) Prove the assertion that you found in part (a)
Hint: $epsilon=1/4$




  • This is where I am stuck. How would I prove the epsilon-delta expression that I found in part (a). Any kind of help would be appreciated.






proof-verification proof-explanation epsilon-delta







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your negated formula is not correct: you should have $|f(x)-1/2|geqepsilon$ in the final bit.
    $endgroup$
    – Leo163
    Nov 30 '18 at 12:20














0












0








0





$begingroup$


We write $lim_{x to a} f(x)=L$ if the following is true
$(forallepsilon>0)(existsdelta>0)(forall x)(0<|x-a|<deltarightarrow|f(x)-L|<epsilon)$



Let $f:Bbb{R}rightarrowBbb{R}$ be given by



$f(x)=
begin{cases}
0, & text{if $x<0$} \
1/2, & text{if $=0$} \
1, & text{if $x>0$}
end{cases}$



We will show that it is not the case that $lim_{x to 0} f(x)=1/2$



(a) Write the negation of $lim_{x to 0} f(x)=1/2$ using the epsilon-delta definition given above




  • I attempted to find the negation of this and this what I got after some calculations


$(existsepsilon>0)(foralldelta>0)(exists x)(0<|x-0|<deltaland|f(x)-1/2|geepsilon)$



(b) Prove the assertion that you found in part (a)
Hint: $epsilon=1/4$




  • This is where I am stuck. How would I prove the epsilon-delta expression that I found in part (a). Any kind of help would be appreciated.






proof-verification proof-explanation epsilon-delta







share|cite|improve this question











$endgroup$




We write $lim_{x to a} f(x)=L$ if the following is true
$(forallepsilon>0)(existsdelta>0)(forall x)(0<|x-a|<deltarightarrow|f(x)-L|<epsilon)$



Let $f:Bbb{R}rightarrowBbb{R}$ be given by



$f(x)=
begin{cases}
0, & text{if $x<0$} \
1/2, & text{if $=0$} \
1, & text{if $x>0$}
end{cases}$



We will show that it is not the case that $lim_{x to 0} f(x)=1/2$



(a) Write the negation of $lim_{x to 0} f(x)=1/2$ using the epsilon-delta definition given above




  • I attempted to find the negation of this and this what I got after some calculations


$(existsepsilon>0)(foralldelta>0)(exists x)(0<|x-0|<deltaland|f(x)-1/2|geepsilon)$



(b) Prove the assertion that you found in part (a)
Hint: $epsilon=1/4$




  • This is where I am stuck. How would I prove the epsilon-delta expression that I found in part (a). Any kind of help would be appreciated.






proof-verification proof-explanation epsilon-delta




proof-verification proof-explanation epsilon-delta






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share|cite|improve this question













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edited Nov 30 '18 at 12:25







Viserom

















asked Nov 30 '18 at 12:14









ViseromViserom

123




123












  • $begingroup$
    Your negated formula is not correct: you should have $|f(x)-1/2|geqepsilon$ in the final bit.
    $endgroup$
    – Leo163
    Nov 30 '18 at 12:20


















  • $begingroup$
    Your negated formula is not correct: you should have $|f(x)-1/2|geqepsilon$ in the final bit.
    $endgroup$
    – Leo163
    Nov 30 '18 at 12:20
















$begingroup$
Your negated formula is not correct: you should have $|f(x)-1/2|geqepsilon$ in the final bit.
$endgroup$
– Leo163
Nov 30 '18 at 12:20




$begingroup$
Your negated formula is not correct: you should have $|f(x)-1/2|geqepsilon$ in the final bit.
$endgroup$
– Leo163
Nov 30 '18 at 12:20










2 Answers
2






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$begingroup$

$$(existsepsilon>0)(foralldelta>0)(exists x)(0<|x-0|<deltaland|f(x)-1/2|color{blue}geepsilon)$$



Follow the hint,



Let $epsilon = frac14$, then $forall delta >0$, let $x= frac{delta}2$, then $$f(x)-frac12=1-frac12 ge epsilon$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Given $delta>0$, we look for $x$ such that



    $$0<|x|<delta $$
    and
    $$|f(x)-frac 12|ge frac 14.$$



    $$iff$$



    $$f(x)ge frac 34 text{ or } f(x)le frac 14$$



    so we can take $x_0=-frac{delta}{2}$
    with $f(x_0)=0$.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
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      2 Answers
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      2












      $begingroup$

      $$(existsepsilon>0)(foralldelta>0)(exists x)(0<|x-0|<deltaland|f(x)-1/2|color{blue}geepsilon)$$



      Follow the hint,



      Let $epsilon = frac14$, then $forall delta >0$, let $x= frac{delta}2$, then $$f(x)-frac12=1-frac12 ge epsilon$$






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        $$(existsepsilon>0)(foralldelta>0)(exists x)(0<|x-0|<deltaland|f(x)-1/2|color{blue}geepsilon)$$



        Follow the hint,



        Let $epsilon = frac14$, then $forall delta >0$, let $x= frac{delta}2$, then $$f(x)-frac12=1-frac12 ge epsilon$$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          $$(existsepsilon>0)(foralldelta>0)(exists x)(0<|x-0|<deltaland|f(x)-1/2|color{blue}geepsilon)$$



          Follow the hint,



          Let $epsilon = frac14$, then $forall delta >0$, let $x= frac{delta}2$, then $$f(x)-frac12=1-frac12 ge epsilon$$






          share|cite|improve this answer









          $endgroup$



          $$(existsepsilon>0)(foralldelta>0)(exists x)(0<|x-0|<deltaland|f(x)-1/2|color{blue}geepsilon)$$



          Follow the hint,



          Let $epsilon = frac14$, then $forall delta >0$, let $x= frac{delta}2$, then $$f(x)-frac12=1-frac12 ge epsilon$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 30 '18 at 12:20









          Siong Thye GohSiong Thye Goh

          101k1466118




          101k1466118























              1












              $begingroup$

              Given $delta>0$, we look for $x$ such that



              $$0<|x|<delta $$
              and
              $$|f(x)-frac 12|ge frac 14.$$



              $$iff$$



              $$f(x)ge frac 34 text{ or } f(x)le frac 14$$



              so we can take $x_0=-frac{delta}{2}$
              with $f(x_0)=0$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Given $delta>0$, we look for $x$ such that



                $$0<|x|<delta $$
                and
                $$|f(x)-frac 12|ge frac 14.$$



                $$iff$$



                $$f(x)ge frac 34 text{ or } f(x)le frac 14$$



                so we can take $x_0=-frac{delta}{2}$
                with $f(x_0)=0$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Given $delta>0$, we look for $x$ such that



                  $$0<|x|<delta $$
                  and
                  $$|f(x)-frac 12|ge frac 14.$$



                  $$iff$$



                  $$f(x)ge frac 34 text{ or } f(x)le frac 14$$



                  so we can take $x_0=-frac{delta}{2}$
                  with $f(x_0)=0$.






                  share|cite|improve this answer









                  $endgroup$



                  Given $delta>0$, we look for $x$ such that



                  $$0<|x|<delta $$
                  and
                  $$|f(x)-frac 12|ge frac 14.$$



                  $$iff$$



                  $$f(x)ge frac 34 text{ or } f(x)le frac 14$$



                  so we can take $x_0=-frac{delta}{2}$
                  with $f(x_0)=0$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 30 '18 at 13:18









                  hamam_Abdallahhamam_Abdallah

                  38.1k21634




                  38.1k21634






























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