Epsilon Delta Proof of Differentiability implies Continuity

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Can someone check my proof if a function $f:Rmapsto R$ is differentiable at all points, it is continuous throughout as well?



Proof:



Let $x$ be a real number, since $f(x)$ is differentiable, or $lim_h to
0 fracf(x+h)-f(x)h$ exists and is equal to $f'(x)$. So, for any $epsilon > 0$ there exists a $delta > 0$, such that $0<|h - 0| < delta$



$implies |fracf(x+h)-f(x)h - f'(x)| < epsilon$



$implies |fracf(x+h)-f(x)-f'(x).hh| < epsilon$



$implies |f(x+h)-f(x)-f'(x).h| < |h|epsilon$



$implies -|h|epsilon<f(x+h)-f(x)-f'(x).h < |h|epsilon$



$implies f'(x).h-|h|epsilon<f(x+h)-f(x) < f'(x).h+|h|epsilon$



Let $epsilon'$ be min($f'(x).h+|h|epsilon - f(x+h)+f(x), f'(x).h-|h|epsilon-f(x+h)+f(x) $), then we can say,



$implies -epsilon'<f(x+h)-f(x) < epsilon'$



$implies |f(x+h)-f(x)| < epsilon'$



$implies lim_h to 0f(x+h)$ exists and is equal to $f(x)$



$implies f$ is continuous at $x$



NOTE: This is not a homework question, I am trying to learn basic calculus on my own through the web.




calculus limits derivatives proof-verification continuity




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  • 1




    May I give you a simple proof?
    – md2perpe
    Jul 28 at 20:26










  • @md2perpe, It would be helpful, please post the link.
    – avocado
    Jul 28 at 20:41










  • Should be $hto 0$, not $hmapsto 0$.
    – Shalop
    Jul 28 at 20:53






  • 1




    Thanks @Shalop, edited
    – avocado
    Jul 28 at 20:56














up vote
2
down vote

favorite












Can someone check my proof if a function $f:Rmapsto R$ is differentiable at all points, it is continuous throughout as well?



Proof:



Let $x$ be a real number, since $f(x)$ is differentiable, or $lim_h to
0 fracf(x+h)-f(x)h$ exists and is equal to $f'(x)$. So, for any $epsilon > 0$ there exists a $delta > 0$, such that $0<|h - 0| < delta$



$implies |fracf(x+h)-f(x)h - f'(x)| < epsilon$



$implies |fracf(x+h)-f(x)-f'(x).hh| < epsilon$



$implies |f(x+h)-f(x)-f'(x).h| < |h|epsilon$



$implies -|h|epsilon<f(x+h)-f(x)-f'(x).h < |h|epsilon$



$implies f'(x).h-|h|epsilon<f(x+h)-f(x) < f'(x).h+|h|epsilon$



Let $epsilon'$ be min($f'(x).h+|h|epsilon - f(x+h)+f(x), f'(x).h-|h|epsilon-f(x+h)+f(x) $), then we can say,



$implies -epsilon'<f(x+h)-f(x) < epsilon'$



$implies |f(x+h)-f(x)| < epsilon'$



$implies lim_h to 0f(x+h)$ exists and is equal to $f(x)$



$implies f$ is continuous at $x$



NOTE: This is not a homework question, I am trying to learn basic calculus on my own through the web.




calculus limits derivatives proof-verification continuity




share|cite|improve this question

















  • 1




    May I give you a simple proof?
    – md2perpe
    Jul 28 at 20:26










  • @md2perpe, It would be helpful, please post the link.
    – avocado
    Jul 28 at 20:41










  • Should be $hto 0$, not $hmapsto 0$.
    – Shalop
    Jul 28 at 20:53






  • 1




    Thanks @Shalop, edited
    – avocado
    Jul 28 at 20:56












up vote
2
down vote

favorite









up vote
2
down vote

favorite











Can someone check my proof if a function $f:Rmapsto R$ is differentiable at all points, it is continuous throughout as well?



Proof:



Let $x$ be a real number, since $f(x)$ is differentiable, or $lim_h to
0 fracf(x+h)-f(x)h$ exists and is equal to $f'(x)$. So, for any $epsilon > 0$ there exists a $delta > 0$, such that $0<|h - 0| < delta$



$implies |fracf(x+h)-f(x)h - f'(x)| < epsilon$



$implies |fracf(x+h)-f(x)-f'(x).hh| < epsilon$



$implies |f(x+h)-f(x)-f'(x).h| < |h|epsilon$



$implies -|h|epsilon<f(x+h)-f(x)-f'(x).h < |h|epsilon$



$implies f'(x).h-|h|epsilon<f(x+h)-f(x) < f'(x).h+|h|epsilon$



Let $epsilon'$ be min($f'(x).h+|h|epsilon - f(x+h)+f(x), f'(x).h-|h|epsilon-f(x+h)+f(x) $), then we can say,



$implies -epsilon'<f(x+h)-f(x) < epsilon'$



$implies |f(x+h)-f(x)| < epsilon'$



$implies lim_h to 0f(x+h)$ exists and is equal to $f(x)$



$implies f$ is continuous at $x$



NOTE: This is not a homework question, I am trying to learn basic calculus on my own through the web.




calculus limits derivatives proof-verification continuity




share|cite|improve this question













Can someone check my proof if a function $f:Rmapsto R$ is differentiable at all points, it is continuous throughout as well?



Proof:



Let $x$ be a real number, since $f(x)$ is differentiable, or $lim_h to
0 fracf(x+h)-f(x)h$ exists and is equal to $f'(x)$. So, for any $epsilon > 0$ there exists a $delta > 0$, such that $0<|h - 0| < delta$



$implies |fracf(x+h)-f(x)h - f'(x)| < epsilon$



$implies |fracf(x+h)-f(x)-f'(x).hh| < epsilon$



$implies |f(x+h)-f(x)-f'(x).h| < |h|epsilon$



$implies -|h|epsilon<f(x+h)-f(x)-f'(x).h < |h|epsilon$



$implies f'(x).h-|h|epsilon<f(x+h)-f(x) < f'(x).h+|h|epsilon$



Let $epsilon'$ be min($f'(x).h+|h|epsilon - f(x+h)+f(x), f'(x).h-|h|epsilon-f(x+h)+f(x) $), then we can say,



$implies -epsilon'<f(x+h)-f(x) < epsilon'$



$implies |f(x+h)-f(x)| < epsilon'$



$implies lim_h to 0f(x+h)$ exists and is equal to $f(x)$



$implies f$ is continuous at $x$



NOTE: This is not a homework question, I am trying to learn basic calculus on my own through the web.





calculus limits derivatives proof-verification continuity





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edited Jul 28 at 20:56
























asked Jul 28 at 20:19









avocado

133




133







  • 1




    May I give you a simple proof?
    – md2perpe
    Jul 28 at 20:26










  • @md2perpe, It would be helpful, please post the link.
    – avocado
    Jul 28 at 20:41










  • Should be $hto 0$, not $hmapsto 0$.
    – Shalop
    Jul 28 at 20:53






  • 1




    Thanks @Shalop, edited
    – avocado
    Jul 28 at 20:56












  • 1




    May I give you a simple proof?
    – md2perpe
    Jul 28 at 20:26










  • @md2perpe, It would be helpful, please post the link.
    – avocado
    Jul 28 at 20:41










  • Should be $hto 0$, not $hmapsto 0$.
    – Shalop
    Jul 28 at 20:53






  • 1




    Thanks @Shalop, edited
    – avocado
    Jul 28 at 20:56







1




1




May I give you a simple proof?
– md2perpe
Jul 28 at 20:26




May I give you a simple proof?
– md2perpe
Jul 28 at 20:26












@md2perpe, It would be helpful, please post the link.
– avocado
Jul 28 at 20:41




@md2perpe, It would be helpful, please post the link.
– avocado
Jul 28 at 20:41












Should be $hto 0$, not $hmapsto 0$.
– Shalop
Jul 28 at 20:53




Should be $hto 0$, not $hmapsto 0$.
– Shalop
Jul 28 at 20:53




1




1




Thanks @Shalop, edited
– avocado
Jul 28 at 20:56




Thanks @Shalop, edited
– avocado
Jul 28 at 20:56










3 Answers
3






active

oldest

votes

















up vote
2
down vote



accepted










This is not correct. You cannot choose $varepsilon'$. What you are supposed to prove is that for every $varepsilon'>0$, there is a $delta>0$ such that$$|h|<deltaimpliesbigl|f(x+h)-f(x)bigr|<varepsilon'.$$






share|cite|improve this answer





















  • Thanks for the edit as well as letting me realise my mistake. I'll think about it more.
    – avocado
    Jul 28 at 20:39










  • I'm glad I could help.
    – José Carlos Santos
    Jul 28 at 20:49

















up vote
2
down vote













Simple proof without explicit use of $epsilon$-$delta$:



For any $x$ we have
$$
left| f(x+h) - f(x) right|
= left| fracf(x+h) - f(x)h h right|
= left| fracf(x+h) - f(x)h right| |h|
to |f'(x)| cdot 0
= 0
$$
as $h to 0.$ This implies that $f(x+h) to f(x)$ as $h to 0.$



Rewritten using $epsilon$-$delta$



Fix $x in mathbb R.$



Let $delta_0>0$ be such that
$$left| fracf(x+h) - f(x)h - f'(x) right| < 1$$
whenever $|h| < delta_0.$ Such $delta_0$ exists by definition of $f'(x)$ as a limit of $(f(x+h)-f(x))/h.$



Given $epsilon>0,$ let $delta_1 = epsilon/(1+|f'(x)|)$ and let $delta = min(delta_0, delta_1).$ Then
$$
left| f(x+h) - f(x) right|
= left| fracf(x+h) - f(x)h h right|
= left| fracf(x+h) - f(x)h right| |h|
= left| fracf(x+h) - f(x)h - f'(x) + f'(x) right| |h| \
= text triangle inequality \
leq left( left| fracf(x+h) - f(x)h - f'(x) right| + left| f'(x) right| right) |h| \
< left( 1 + left| f'(x) right| right) fracepsilonf'(x)
= epsilon
$$
whenever $|h| < delta.$ Thus $f(x+h) to f(x)$ as $h to 0,$ i.e. $f$ is continuous in $x.$ Since $x$ was taken arbitrary, $f$ is continuous on all of $mathbb R.$






share|cite|improve this answer





















  • Thanks a lot @md2perpe for taking out time to write this answer. I really like the simplicity of both the proofs.
    – avocado
    Jul 28 at 21:29






  • 1




    The $epsilon$-$delta$ proof is certainly a bit more complicated that using limit laws. Even so, it could have been simpler if I hadn't cared about ending with $epsilon$ but accepted $(epsilon + |f'(x)|) epsilon$ which obviously can be made arbitrarily small. I don't know which one is least confusing; choosing $delta$ in a complicated way or not ending with $epsilon.$
    – md2perpe
    Jul 28 at 21:51










  • I feel choosing $delta$ in the proof may look a little strange at the beginning, however since it leads to a form which is seen throughout the text books. I personally feel this is slightly better than ending with $(epsilon+|f'(x)|)epsilon$.
    – avocado
    Jul 29 at 5:28











  • I've debated with myself the same question. I feel like maybe the best way to go actually shows the process of learning and doing Math: initially just arrive at something that can be made arbitrarily small and declare "This is pretty good--it is an adequate proof. But if you really need to see it with a simple $varepsilon$ we can use the work we have already done and just re-jigger the choice of $delta$..." Basically show the process of making an initial proof that is correct in spirit and then ironing out the details.
    – Addem
    Jul 30 at 16:49











  • @Addem. I agree.
    – md2perpe
    Jul 30 at 17:42

















up vote
1
down vote













Your proof does not work. It seems you have a misunderstanding of the definition of continuity. Your mistake is that your $epsilon'$ depends on $h$. But this makes no sense (why? hint: continuity means that for every arbitrary margin of error $epsilon$ you can find a neighborhood of $h$ where your function will be within that margin of error)



Try to fix it now that you understand and check the accepted answer here:



How to prove differentiability implies continuity with $epsilon-delta$ definition?






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  • Thanks John11, I'll surely check out the other answers.
    – avocado
    Jul 28 at 20:40










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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










This is not correct. You cannot choose $varepsilon'$. What you are supposed to prove is that for every $varepsilon'>0$, there is a $delta>0$ such that$$|h|<deltaimpliesbigl|f(x+h)-f(x)bigr|<varepsilon'.$$






share|cite|improve this answer





















  • Thanks for the edit as well as letting me realise my mistake. I'll think about it more.
    – avocado
    Jul 28 at 20:39










  • I'm glad I could help.
    – José Carlos Santos
    Jul 28 at 20:49














up vote
2
down vote



accepted










This is not correct. You cannot choose $varepsilon'$. What you are supposed to prove is that for every $varepsilon'>0$, there is a $delta>0$ such that$$|h|<deltaimpliesbigl|f(x+h)-f(x)bigr|<varepsilon'.$$






share|cite|improve this answer





















  • Thanks for the edit as well as letting me realise my mistake. I'll think about it more.
    – avocado
    Jul 28 at 20:39










  • I'm glad I could help.
    – José Carlos Santos
    Jul 28 at 20:49












up vote
2
down vote



accepted







up vote
2
down vote



accepted






This is not correct. You cannot choose $varepsilon'$. What you are supposed to prove is that for every $varepsilon'>0$, there is a $delta>0$ such that$$|h|<deltaimpliesbigl|f(x+h)-f(x)bigr|<varepsilon'.$$






share|cite|improve this answer













This is not correct. You cannot choose $varepsilon'$. What you are supposed to prove is that for every $varepsilon'>0$, there is a $delta>0$ such that$$|h|<deltaimpliesbigl|f(x+h)-f(x)bigr|<varepsilon'.$$







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 28 at 20:25









José Carlos Santos

112k1696173




112k1696173











  • Thanks for the edit as well as letting me realise my mistake. I'll think about it more.
    – avocado
    Jul 28 at 20:39










  • I'm glad I could help.
    – José Carlos Santos
    Jul 28 at 20:49
















  • Thanks for the edit as well as letting me realise my mistake. I'll think about it more.
    – avocado
    Jul 28 at 20:39










  • I'm glad I could help.
    – José Carlos Santos
    Jul 28 at 20:49















Thanks for the edit as well as letting me realise my mistake. I'll think about it more.
– avocado
Jul 28 at 20:39




Thanks for the edit as well as letting me realise my mistake. I'll think about it more.
– avocado
Jul 28 at 20:39












I'm glad I could help.
– José Carlos Santos
Jul 28 at 20:49




I'm glad I could help.
– José Carlos Santos
Jul 28 at 20:49










up vote
2
down vote













Simple proof without explicit use of $epsilon$-$delta$:



For any $x$ we have
$$
left| f(x+h) - f(x) right|
= left| fracf(x+h) - f(x)h h right|
= left| fracf(x+h) - f(x)h right| |h|
to |f'(x)| cdot 0
= 0
$$
as $h to 0.$ This implies that $f(x+h) to f(x)$ as $h to 0.$



Rewritten using $epsilon$-$delta$



Fix $x in mathbb R.$



Let $delta_0>0$ be such that
$$left| fracf(x+h) - f(x)h - f'(x) right| < 1$$
whenever $|h| < delta_0.$ Such $delta_0$ exists by definition of $f'(x)$ as a limit of $(f(x+h)-f(x))/h.$



Given $epsilon>0,$ let $delta_1 = epsilon/(1+|f'(x)|)$ and let $delta = min(delta_0, delta_1).$ Then
$$
left| f(x+h) - f(x) right|
= left| fracf(x+h) - f(x)h h right|
= left| fracf(x+h) - f(x)h right| |h|
= left| fracf(x+h) - f(x)h - f'(x) + f'(x) right| |h| \
= text triangle inequality \
leq left( left| fracf(x+h) - f(x)h - f'(x) right| + left| f'(x) right| right) |h| \
< left( 1 + left| f'(x) right| right) fracepsilonf'(x)
= epsilon
$$
whenever $|h| < delta.$ Thus $f(x+h) to f(x)$ as $h to 0,$ i.e. $f$ is continuous in $x.$ Since $x$ was taken arbitrary, $f$ is continuous on all of $mathbb R.$






share|cite|improve this answer





















  • Thanks a lot @md2perpe for taking out time to write this answer. I really like the simplicity of both the proofs.
    – avocado
    Jul 28 at 21:29






  • 1




    The $epsilon$-$delta$ proof is certainly a bit more complicated that using limit laws. Even so, it could have been simpler if I hadn't cared about ending with $epsilon$ but accepted $(epsilon + |f'(x)|) epsilon$ which obviously can be made arbitrarily small. I don't know which one is least confusing; choosing $delta$ in a complicated way or not ending with $epsilon.$
    – md2perpe
    Jul 28 at 21:51










  • I feel choosing $delta$ in the proof may look a little strange at the beginning, however since it leads to a form which is seen throughout the text books. I personally feel this is slightly better than ending with $(epsilon+|f'(x)|)epsilon$.
    – avocado
    Jul 29 at 5:28











  • I've debated with myself the same question. I feel like maybe the best way to go actually shows the process of learning and doing Math: initially just arrive at something that can be made arbitrarily small and declare "This is pretty good--it is an adequate proof. But if you really need to see it with a simple $varepsilon$ we can use the work we have already done and just re-jigger the choice of $delta$..." Basically show the process of making an initial proof that is correct in spirit and then ironing out the details.
    – Addem
    Jul 30 at 16:49











  • @Addem. I agree.
    – md2perpe
    Jul 30 at 17:42














up vote
2
down vote













Simple proof without explicit use of $epsilon$-$delta$:



For any $x$ we have
$$
left| f(x+h) - f(x) right|
= left| fracf(x+h) - f(x)h h right|
= left| fracf(x+h) - f(x)h right| |h|
to |f'(x)| cdot 0
= 0
$$
as $h to 0.$ This implies that $f(x+h) to f(x)$ as $h to 0.$



Rewritten using $epsilon$-$delta$



Fix $x in mathbb R.$



Let $delta_0>0$ be such that
$$left| fracf(x+h) - f(x)h - f'(x) right| < 1$$
whenever $|h| < delta_0.$ Such $delta_0$ exists by definition of $f'(x)$ as a limit of $(f(x+h)-f(x))/h.$



Given $epsilon>0,$ let $delta_1 = epsilon/(1+|f'(x)|)$ and let $delta = min(delta_0, delta_1).$ Then
$$
left| f(x+h) - f(x) right|
= left| fracf(x+h) - f(x)h h right|
= left| fracf(x+h) - f(x)h right| |h|
= left| fracf(x+h) - f(x)h - f'(x) + f'(x) right| |h| \
= text triangle inequality \
leq left( left| fracf(x+h) - f(x)h - f'(x) right| + left| f'(x) right| right) |h| \
< left( 1 + left| f'(x) right| right) fracepsilonf'(x)
= epsilon
$$
whenever $|h| < delta.$ Thus $f(x+h) to f(x)$ as $h to 0,$ i.e. $f$ is continuous in $x.$ Since $x$ was taken arbitrary, $f$ is continuous on all of $mathbb R.$






share|cite|improve this answer





















  • Thanks a lot @md2perpe for taking out time to write this answer. I really like the simplicity of both the proofs.
    – avocado
    Jul 28 at 21:29






  • 1




    The $epsilon$-$delta$ proof is certainly a bit more complicated that using limit laws. Even so, it could have been simpler if I hadn't cared about ending with $epsilon$ but accepted $(epsilon + |f'(x)|) epsilon$ which obviously can be made arbitrarily small. I don't know which one is least confusing; choosing $delta$ in a complicated way or not ending with $epsilon.$
    – md2perpe
    Jul 28 at 21:51










  • I feel choosing $delta$ in the proof may look a little strange at the beginning, however since it leads to a form which is seen throughout the text books. I personally feel this is slightly better than ending with $(epsilon+|f'(x)|)epsilon$.
    – avocado
    Jul 29 at 5:28











  • I've debated with myself the same question. I feel like maybe the best way to go actually shows the process of learning and doing Math: initially just arrive at something that can be made arbitrarily small and declare "This is pretty good--it is an adequate proof. But if you really need to see it with a simple $varepsilon$ we can use the work we have already done and just re-jigger the choice of $delta$..." Basically show the process of making an initial proof that is correct in spirit and then ironing out the details.
    – Addem
    Jul 30 at 16:49











  • @Addem. I agree.
    – md2perpe
    Jul 30 at 17:42












up vote
2
down vote










up vote
2
down vote









Simple proof without explicit use of $epsilon$-$delta$:



For any $x$ we have
$$
left| f(x+h) - f(x) right|
= left| fracf(x+h) - f(x)h h right|
= left| fracf(x+h) - f(x)h right| |h|
to |f'(x)| cdot 0
= 0
$$
as $h to 0.$ This implies that $f(x+h) to f(x)$ as $h to 0.$



Rewritten using $epsilon$-$delta$



Fix $x in mathbb R.$



Let $delta_0>0$ be such that
$$left| fracf(x+h) - f(x)h - f'(x) right| < 1$$
whenever $|h| < delta_0.$ Such $delta_0$ exists by definition of $f'(x)$ as a limit of $(f(x+h)-f(x))/h.$



Given $epsilon>0,$ let $delta_1 = epsilon/(1+|f'(x)|)$ and let $delta = min(delta_0, delta_1).$ Then
$$
left| f(x+h) - f(x) right|
= left| fracf(x+h) - f(x)h h right|
= left| fracf(x+h) - f(x)h right| |h|
= left| fracf(x+h) - f(x)h - f'(x) + f'(x) right| |h| \
= text triangle inequality \
leq left( left| fracf(x+h) - f(x)h - f'(x) right| + left| f'(x) right| right) |h| \
< left( 1 + left| f'(x) right| right) fracepsilonf'(x)
= epsilon
$$
whenever $|h| < delta.$ Thus $f(x+h) to f(x)$ as $h to 0,$ i.e. $f$ is continuous in $x.$ Since $x$ was taken arbitrary, $f$ is continuous on all of $mathbb R.$






share|cite|improve this answer













Simple proof without explicit use of $epsilon$-$delta$:



For any $x$ we have
$$
left| f(x+h) - f(x) right|
= left| fracf(x+h) - f(x)h h right|
= left| fracf(x+h) - f(x)h right| |h|
to |f'(x)| cdot 0
= 0
$$
as $h to 0.$ This implies that $f(x+h) to f(x)$ as $h to 0.$



Rewritten using $epsilon$-$delta$



Fix $x in mathbb R.$



Let $delta_0>0$ be such that
$$left| fracf(x+h) - f(x)h - f'(x) right| < 1$$
whenever $|h| < delta_0.$ Such $delta_0$ exists by definition of $f'(x)$ as a limit of $(f(x+h)-f(x))/h.$



Given $epsilon>0,$ let $delta_1 = epsilon/(1+|f'(x)|)$ and let $delta = min(delta_0, delta_1).$ Then
$$
left| f(x+h) - f(x) right|
= left| fracf(x+h) - f(x)h h right|
= left| fracf(x+h) - f(x)h right| |h|
= left| fracf(x+h) - f(x)h - f'(x) + f'(x) right| |h| \
= text triangle inequality \
leq left( left| fracf(x+h) - f(x)h - f'(x) right| + left| f'(x) right| right) |h| \
< left( 1 + left| f'(x) right| right) fracepsilonf'(x)
= epsilon
$$
whenever $|h| < delta.$ Thus $f(x+h) to f(x)$ as $h to 0,$ i.e. $f$ is continuous in $x.$ Since $x$ was taken arbitrary, $f$ is continuous on all of $mathbb R.$







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 28 at 21:18









md2perpe

5,7801922




5,7801922











  • Thanks a lot @md2perpe for taking out time to write this answer. I really like the simplicity of both the proofs.
    – avocado
    Jul 28 at 21:29






  • 1




    The $epsilon$-$delta$ proof is certainly a bit more complicated that using limit laws. Even so, it could have been simpler if I hadn't cared about ending with $epsilon$ but accepted $(epsilon + |f'(x)|) epsilon$ which obviously can be made arbitrarily small. I don't know which one is least confusing; choosing $delta$ in a complicated way or not ending with $epsilon.$
    – md2perpe
    Jul 28 at 21:51










  • I feel choosing $delta$ in the proof may look a little strange at the beginning, however since it leads to a form which is seen throughout the text books. I personally feel this is slightly better than ending with $(epsilon+|f'(x)|)epsilon$.
    – avocado
    Jul 29 at 5:28











  • I've debated with myself the same question. I feel like maybe the best way to go actually shows the process of learning and doing Math: initially just arrive at something that can be made arbitrarily small and declare "This is pretty good--it is an adequate proof. But if you really need to see it with a simple $varepsilon$ we can use the work we have already done and just re-jigger the choice of $delta$..." Basically show the process of making an initial proof that is correct in spirit and then ironing out the details.
    – Addem
    Jul 30 at 16:49











  • @Addem. I agree.
    – md2perpe
    Jul 30 at 17:42
















  • Thanks a lot @md2perpe for taking out time to write this answer. I really like the simplicity of both the proofs.
    – avocado
    Jul 28 at 21:29






  • 1




    The $epsilon$-$delta$ proof is certainly a bit more complicated that using limit laws. Even so, it could have been simpler if I hadn't cared about ending with $epsilon$ but accepted $(epsilon + |f'(x)|) epsilon$ which obviously can be made arbitrarily small. I don't know which one is least confusing; choosing $delta$ in a complicated way or not ending with $epsilon.$
    – md2perpe
    Jul 28 at 21:51










  • I feel choosing $delta$ in the proof may look a little strange at the beginning, however since it leads to a form which is seen throughout the text books. I personally feel this is slightly better than ending with $(epsilon+|f'(x)|)epsilon$.
    – avocado
    Jul 29 at 5:28











  • I've debated with myself the same question. I feel like maybe the best way to go actually shows the process of learning and doing Math: initially just arrive at something that can be made arbitrarily small and declare "This is pretty good--it is an adequate proof. But if you really need to see it with a simple $varepsilon$ we can use the work we have already done and just re-jigger the choice of $delta$..." Basically show the process of making an initial proof that is correct in spirit and then ironing out the details.
    – Addem
    Jul 30 at 16:49











  • @Addem. I agree.
    – md2perpe
    Jul 30 at 17:42















Thanks a lot @md2perpe for taking out time to write this answer. I really like the simplicity of both the proofs.
– avocado
Jul 28 at 21:29




Thanks a lot @md2perpe for taking out time to write this answer. I really like the simplicity of both the proofs.
– avocado
Jul 28 at 21:29




1




1




The $epsilon$-$delta$ proof is certainly a bit more complicated that using limit laws. Even so, it could have been simpler if I hadn't cared about ending with $epsilon$ but accepted $(epsilon + |f'(x)|) epsilon$ which obviously can be made arbitrarily small. I don't know which one is least confusing; choosing $delta$ in a complicated way or not ending with $epsilon.$
– md2perpe
Jul 28 at 21:51




The $epsilon$-$delta$ proof is certainly a bit more complicated that using limit laws. Even so, it could have been simpler if I hadn't cared about ending with $epsilon$ but accepted $(epsilon + |f'(x)|) epsilon$ which obviously can be made arbitrarily small. I don't know which one is least confusing; choosing $delta$ in a complicated way or not ending with $epsilon.$
– md2perpe
Jul 28 at 21:51












I feel choosing $delta$ in the proof may look a little strange at the beginning, however since it leads to a form which is seen throughout the text books. I personally feel this is slightly better than ending with $(epsilon+|f'(x)|)epsilon$.
– avocado
Jul 29 at 5:28





I feel choosing $delta$ in the proof may look a little strange at the beginning, however since it leads to a form which is seen throughout the text books. I personally feel this is slightly better than ending with $(epsilon+|f'(x)|)epsilon$.
– avocado
Jul 29 at 5:28













I've debated with myself the same question. I feel like maybe the best way to go actually shows the process of learning and doing Math: initially just arrive at something that can be made arbitrarily small and declare "This is pretty good--it is an adequate proof. But if you really need to see it with a simple $varepsilon$ we can use the work we have already done and just re-jigger the choice of $delta$..." Basically show the process of making an initial proof that is correct in spirit and then ironing out the details.
– Addem
Jul 30 at 16:49





I've debated with myself the same question. I feel like maybe the best way to go actually shows the process of learning and doing Math: initially just arrive at something that can be made arbitrarily small and declare "This is pretty good--it is an adequate proof. But if you really need to see it with a simple $varepsilon$ we can use the work we have already done and just re-jigger the choice of $delta$..." Basically show the process of making an initial proof that is correct in spirit and then ironing out the details.
– Addem
Jul 30 at 16:49













@Addem. I agree.
– md2perpe
Jul 30 at 17:42




@Addem. I agree.
– md2perpe
Jul 30 at 17:42










up vote
1
down vote













Your proof does not work. It seems you have a misunderstanding of the definition of continuity. Your mistake is that your $epsilon'$ depends on $h$. But this makes no sense (why? hint: continuity means that for every arbitrary margin of error $epsilon$ you can find a neighborhood of $h$ where your function will be within that margin of error)



Try to fix it now that you understand and check the accepted answer here:



How to prove differentiability implies continuity with $epsilon-delta$ definition?






share|cite|improve this answer





















  • Thanks John11, I'll surely check out the other answers.
    – avocado
    Jul 28 at 20:40














up vote
1
down vote













Your proof does not work. It seems you have a misunderstanding of the definition of continuity. Your mistake is that your $epsilon'$ depends on $h$. But this makes no sense (why? hint: continuity means that for every arbitrary margin of error $epsilon$ you can find a neighborhood of $h$ where your function will be within that margin of error)



Try to fix it now that you understand and check the accepted answer here:



How to prove differentiability implies continuity with $epsilon-delta$ definition?






share|cite|improve this answer





















  • Thanks John11, I'll surely check out the other answers.
    – avocado
    Jul 28 at 20:40












up vote
1
down vote










up vote
1
down vote









Your proof does not work. It seems you have a misunderstanding of the definition of continuity. Your mistake is that your $epsilon'$ depends on $h$. But this makes no sense (why? hint: continuity means that for every arbitrary margin of error $epsilon$ you can find a neighborhood of $h$ where your function will be within that margin of error)



Try to fix it now that you understand and check the accepted answer here:



How to prove differentiability implies continuity with $epsilon-delta$ definition?






share|cite|improve this answer













Your proof does not work. It seems you have a misunderstanding of the definition of continuity. Your mistake is that your $epsilon'$ depends on $h$. But this makes no sense (why? hint: continuity means that for every arbitrary margin of error $epsilon$ you can find a neighborhood of $h$ where your function will be within that margin of error)



Try to fix it now that you understand and check the accepted answer here:



How to prove differentiability implies continuity with $epsilon-delta$ definition?







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 28 at 20:28









John11

7751720




7751720











  • Thanks John11, I'll surely check out the other answers.
    – avocado
    Jul 28 at 20:40
















  • Thanks John11, I'll surely check out the other answers.
    – avocado
    Jul 28 at 20:40















Thanks John11, I'll surely check out the other answers.
– avocado
Jul 28 at 20:40




Thanks John11, I'll surely check out the other answers.
– avocado
Jul 28 at 20:40












 

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