Proof Verification: Differentiability implies continuity.

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Proof:



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



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



Using the Triangle inequality, we get



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



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



Here's where I start getting unsure whether my proof is correct or not. This inequality is true for $epsilon'$ = $fracepsilondelta' - |f'(a)|$ where $delta' < delta$ such that $fracepsilondelta' > |f'(a)|$



This gives us



$ |f(x)-f(a)| < epsilon$



The logic behind the proof is that if someone gives asks me to find an appropriate delta for $epsilon = h$, I'll find a delta for $epsilon' = h$, say that delta is equal to $n$. Next I solve for $epsilon'$ using $epsilon' = frachn - |f'(a)|$. If on solving I get a negative $epsilon'$, I decrease $delta'$ to a value $n_0$ such that I get a positive value for $epsilon'$ and then get the corresponding delta, $n_1$. The minimum of $n_1$ and $n_0$ would be the required delta.



Please let me know where exactly the proof starts to go wrong. While alternate proofs are appreciated, the main goal here is understanding why this proof is wrong.



EDIT: After thinking about the problem for a while and reading a few of the answers, I found a way to communicate the idea more effectively.



Continuing from $|f(x)-f(a)| < (epsilon' + |f'(a)|) delta$



Now we need to show that we can represent any positive real number, $epsilon$, by taking appropriate values of $delta$ and $epsilon'$. Fix $epsilon' = 1$. Now $delta$ is just $fracepsilon1+$. To complete our answer, let one value of $delta$ for $epsilon' = 1$ be $delta_0$. Our final answer would be $delta = min(delta_0,fracepsilon1+)$










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  • There are some errors.. the derivative must be calculate on a and not in x
    – Federico Fallucca
    Nov 19 at 14:53











  • Fixed, thanks. Is the general idea of the proof correct? Because I haven't really seen the argument I used at the end in any other proof, I'm not sure if it's correct or not.
    – Star Platinum ZA WARUDO
    Nov 19 at 15:05










  • No, I think it is wrong because you use the fixed constant epsilon and delta to get the result but I guess that you must use only the variable x.
    – Federico Fallucca
    Nov 19 at 15:11














up vote
3
down vote

favorite
1












Proof:



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



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



Using the Triangle inequality, we get



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



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



Here's where I start getting unsure whether my proof is correct or not. This inequality is true for $epsilon'$ = $fracepsilondelta' - |f'(a)|$ where $delta' < delta$ such that $fracepsilondelta' > |f'(a)|$



This gives us



$ |f(x)-f(a)| < epsilon$



The logic behind the proof is that if someone gives asks me to find an appropriate delta for $epsilon = h$, I'll find a delta for $epsilon' = h$, say that delta is equal to $n$. Next I solve for $epsilon'$ using $epsilon' = frachn - |f'(a)|$. If on solving I get a negative $epsilon'$, I decrease $delta'$ to a value $n_0$ such that I get a positive value for $epsilon'$ and then get the corresponding delta, $n_1$. The minimum of $n_1$ and $n_0$ would be the required delta.



Please let me know where exactly the proof starts to go wrong. While alternate proofs are appreciated, the main goal here is understanding why this proof is wrong.



EDIT: After thinking about the problem for a while and reading a few of the answers, I found a way to communicate the idea more effectively.



Continuing from $|f(x)-f(a)| < (epsilon' + |f'(a)|) delta$



Now we need to show that we can represent any positive real number, $epsilon$, by taking appropriate values of $delta$ and $epsilon'$. Fix $epsilon' = 1$. Now $delta$ is just $fracepsilon1+$. To complete our answer, let one value of $delta$ for $epsilon' = 1$ be $delta_0$. Our final answer would be $delta = min(delta_0,fracepsilon1+)$










share|cite|improve this question























  • There are some errors.. the derivative must be calculate on a and not in x
    – Federico Fallucca
    Nov 19 at 14:53











  • Fixed, thanks. Is the general idea of the proof correct? Because I haven't really seen the argument I used at the end in any other proof, I'm not sure if it's correct or not.
    – Star Platinum ZA WARUDO
    Nov 19 at 15:05










  • No, I think it is wrong because you use the fixed constant epsilon and delta to get the result but I guess that you must use only the variable x.
    – Federico Fallucca
    Nov 19 at 15:11












up vote
3
down vote

favorite
1









up vote
3
down vote

favorite
1






1





Proof:



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



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



Using the Triangle inequality, we get



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



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



Here's where I start getting unsure whether my proof is correct or not. This inequality is true for $epsilon'$ = $fracepsilondelta' - |f'(a)|$ where $delta' < delta$ such that $fracepsilondelta' > |f'(a)|$



This gives us



$ |f(x)-f(a)| < epsilon$



The logic behind the proof is that if someone gives asks me to find an appropriate delta for $epsilon = h$, I'll find a delta for $epsilon' = h$, say that delta is equal to $n$. Next I solve for $epsilon'$ using $epsilon' = frachn - |f'(a)|$. If on solving I get a negative $epsilon'$, I decrease $delta'$ to a value $n_0$ such that I get a positive value for $epsilon'$ and then get the corresponding delta, $n_1$. The minimum of $n_1$ and $n_0$ would be the required delta.



Please let me know where exactly the proof starts to go wrong. While alternate proofs are appreciated, the main goal here is understanding why this proof is wrong.



EDIT: After thinking about the problem for a while and reading a few of the answers, I found a way to communicate the idea more effectively.



Continuing from $|f(x)-f(a)| < (epsilon' + |f'(a)|) delta$



Now we need to show that we can represent any positive real number, $epsilon$, by taking appropriate values of $delta$ and $epsilon'$. Fix $epsilon' = 1$. Now $delta$ is just $fracepsilon1+$. To complete our answer, let one value of $delta$ for $epsilon' = 1$ be $delta_0$. Our final answer would be $delta = min(delta_0,fracepsilon1+)$










share|cite|improve this question















Proof:



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



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



Using the Triangle inequality, we get



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



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



Here's where I start getting unsure whether my proof is correct or not. This inequality is true for $epsilon'$ = $fracepsilondelta' - |f'(a)|$ where $delta' < delta$ such that $fracepsilondelta' > |f'(a)|$



This gives us



$ |f(x)-f(a)| < epsilon$



The logic behind the proof is that if someone gives asks me to find an appropriate delta for $epsilon = h$, I'll find a delta for $epsilon' = h$, say that delta is equal to $n$. Next I solve for $epsilon'$ using $epsilon' = frachn - |f'(a)|$. If on solving I get a negative $epsilon'$, I decrease $delta'$ to a value $n_0$ such that I get a positive value for $epsilon'$ and then get the corresponding delta, $n_1$. The minimum of $n_1$ and $n_0$ would be the required delta.



Please let me know where exactly the proof starts to go wrong. While alternate proofs are appreciated, the main goal here is understanding why this proof is wrong.



EDIT: After thinking about the problem for a while and reading a few of the answers, I found a way to communicate the idea more effectively.



Continuing from $|f(x)-f(a)| < (epsilon' + |f'(a)|) delta$



Now we need to show that we can represent any positive real number, $epsilon$, by taking appropriate values of $delta$ and $epsilon'$. Fix $epsilon' = 1$. Now $delta$ is just $fracepsilon1+$. To complete our answer, let one value of $delta$ for $epsilon' = 1$ be $delta_0$. Our final answer would be $delta = min(delta_0,fracepsilon1+)$







real-analysis epsilon-delta






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edited Nov 20 at 1:42









miracle173

7,28822247




7,28822247










asked Nov 19 at 14:50









Star Platinum ZA WARUDO

33212




33212











  • There are some errors.. the derivative must be calculate on a and not in x
    – Federico Fallucca
    Nov 19 at 14:53











  • Fixed, thanks. Is the general idea of the proof correct? Because I haven't really seen the argument I used at the end in any other proof, I'm not sure if it's correct or not.
    – Star Platinum ZA WARUDO
    Nov 19 at 15:05










  • No, I think it is wrong because you use the fixed constant epsilon and delta to get the result but I guess that you must use only the variable x.
    – Federico Fallucca
    Nov 19 at 15:11
















  • There are some errors.. the derivative must be calculate on a and not in x
    – Federico Fallucca
    Nov 19 at 14:53











  • Fixed, thanks. Is the general idea of the proof correct? Because I haven't really seen the argument I used at the end in any other proof, I'm not sure if it's correct or not.
    – Star Platinum ZA WARUDO
    Nov 19 at 15:05










  • No, I think it is wrong because you use the fixed constant epsilon and delta to get the result but I guess that you must use only the variable x.
    – Federico Fallucca
    Nov 19 at 15:11















There are some errors.. the derivative must be calculate on a and not in x
– Federico Fallucca
Nov 19 at 14:53





There are some errors.. the derivative must be calculate on a and not in x
– Federico Fallucca
Nov 19 at 14:53













Fixed, thanks. Is the general idea of the proof correct? Because I haven't really seen the argument I used at the end in any other proof, I'm not sure if it's correct or not.
– Star Platinum ZA WARUDO
Nov 19 at 15:05




Fixed, thanks. Is the general idea of the proof correct? Because I haven't really seen the argument I used at the end in any other proof, I'm not sure if it's correct or not.
– Star Platinum ZA WARUDO
Nov 19 at 15:05












No, I think it is wrong because you use the fixed constant epsilon and delta to get the result but I guess that you must use only the variable x.
– Federico Fallucca
Nov 19 at 15:11




No, I think it is wrong because you use the fixed constant epsilon and delta to get the result but I guess that you must use only the variable x.
– Federico Fallucca
Nov 19 at 15:11










6 Answers
6






active

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up vote
7
down vote













Or just see that
$$lim_hrightarrow 0f(x+h)-f(x)=lim_hrightarrow 0underbracefracf(x+h)-f(x)h_rightarrow f'(x)h=0.$$






share|cite|improve this answer
















  • 2




    Nice job latexing.
    – djechlin
    Nov 19 at 18:36

















up vote
2
down vote













We can simply use the equivalent definition of differentiability



$$f(a+h)=f(a)+f'(a)cdot h +o(h) implies lim_xto a f(x)=lim_hto 0 f(a+h)=f(a)$$






share|cite|improve this answer




















  • Thanks for the answer. I know my proof is long and a bit harder to read, but is it correct? If it isn't, where exactly does it go wrong?
    – Star Platinum ZA WARUDO
    Nov 19 at 15:03










  • @StarPlatinumZAWARUDO It seems really over complicated to me, the implication is avery simple ansd trivial fact. I'll try to look to it more carefully later. Bye
    – gimusi
    Nov 19 at 15:10

















up vote
1
down vote













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



so



$ |f(x)-f(a)| <( epsilon' + |f'(a)|)|x-a|$



then for $xto a$ you have that



$|f(x)-f(a)|leq 0$






share|cite|improve this answer



























    up vote
    1
    down vote













    The heart of your argument is basically the following. Give me some $epsilon$, and I have to find some $delta$ so that $|f(x)-f(a)|<epsilon$ when $|x-a|<delta$. To start with, I'll just take any $delta$, it doesn't matter. Now, within the range $[a-delta, a+delta]$, the function $xtofracf(x)-f(a)x-a$ is bounded, because it's convergent at $a$ and convergent functions are locally bounded - this is a theorem which you essentially spend the first couple of lines of your proof re-proving by way of the value $epsilon'$. So say $lvertfracf(x)-f(a)x-arvert<M$ on the interval $[a-delta, a+delta]$. Well then certainly $|f(x)-f(a)|<Mdelta$, and by choosing a sufficiently small $delta$ we can make that less than $epsilon$ (since the $M$ only gets smaller as $delta$ gets smaller).



    If you notice, in your own proof, the $epsilon'$ is a bit pointless. Since at the end you're just going to say "and now take $delta$ as small as is necessary to make this true", you may as well initially take $epsilon'=10^100$. All that matters is that $epsilon'$ is finite and that $delta$ can be made arbitrarily small, in other words, that the difference quotient is locally bounded. The exact value of the derivative at $a$ also doesn't matter.






    share|cite|improve this answer



























      up vote
      1
      down vote













      Not quite right. One of the issues is:




      I define $epsilon'$ as $frac epsilon delta $




      As it stands, this doesn't make sense, because you started the proof by taking an arbitrary $epsilon'$. Perhaps you mean something like: Since this inequality is true for any $epsilon '$, it is true in particular for $frac epsilon delta$ ...



      But this doesn't quite make sense either; What's the $delta$ on the right-hand side? This $delta $ may depend on the $epsilon '$ you initially choose.




      Try this alternative:



      For any $epsilon> 0$ there is a $delta' > 0$, such that whenever $0 < |x-a| < delta'$, one has $left|fracf(x)-f(a)x-aright| < |f'(a)| + epsilon$. In particular, there is a $delta_0>0$ such that $left|fracf(x)-f(a)x-aright| < |f'(a)| + 1$



      Now, choose $delta = min left(delta_0, frac 1 right) $



      Can you complete the proof from here?






      share|cite|improve this answer






















      • I've made a few edits to the ending paragraphs to make it a bit easier to understand what I'm trying to say.
        – Star Platinum ZA WARUDO
        Nov 19 at 15:44

















      up vote
      0
      down vote













      I try to put your arguments in the right order.



      Proof:



      Assume that $f$ is differentiable in $a$. So $f'(a)$ exists.



      To prove that $f$ is continous in $a$ choose an arbitrary $varepsilon>0.$ Now you can choose $delta_1$ such
      $$0<delta_1<fracvarepsilonf'(a), text if f'(a)ne 0$$
      $$delta_1=1, text if |f'(a)|=0.$$
      So we have
      $$0<|f'(a)|<fracvarepsilondelta_1$$
      and define
      $$varepsilon_2=fracvarepsilondelta_1-|f'(a)|>0$$



      Because f is differentiable in $a$ we can find a $delta_2$ such that
      $$|fracf(x)-f(a)x-a - f'(a)| < varepsilon_2,; forall x: 0<|x-a|<delta_2$$



      Using the triangle inequality, we get



      $$ |fracf(x)-f(a)x-a| < varepsilon_2 + |f'(a)|,; forall x: 0<|x-a|<min(delta_1,delta_2)$$



      So we set
      $$delta=min(delta_1,delta_2)$$
      and get
      $$| f(x)-f(a)| < (varepsilon_2 + |f'(a)|) delta_2<fracvarepsilondelta_1min(delta_1,delta_2)le varepsilon,; forall x: 0<|x-a|<delta$$






      share|cite|improve this answer






















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        6 Answers
        6






        active

        oldest

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        6 Answers
        6






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        7
        down vote













        Or just see that
        $$lim_hrightarrow 0f(x+h)-f(x)=lim_hrightarrow 0underbracefracf(x+h)-f(x)h_rightarrow f'(x)h=0.$$






        share|cite|improve this answer
















        • 2




          Nice job latexing.
          – djechlin
          Nov 19 at 18:36














        up vote
        7
        down vote













        Or just see that
        $$lim_hrightarrow 0f(x+h)-f(x)=lim_hrightarrow 0underbracefracf(x+h)-f(x)h_rightarrow f'(x)h=0.$$






        share|cite|improve this answer
















        • 2




          Nice job latexing.
          – djechlin
          Nov 19 at 18:36












        up vote
        7
        down vote










        up vote
        7
        down vote









        Or just see that
        $$lim_hrightarrow 0f(x+h)-f(x)=lim_hrightarrow 0underbracefracf(x+h)-f(x)h_rightarrow f'(x)h=0.$$






        share|cite|improve this answer












        Or just see that
        $$lim_hrightarrow 0f(x+h)-f(x)=lim_hrightarrow 0underbracefracf(x+h)-f(x)h_rightarrow f'(x)h=0.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 19 at 15:01









        Peter Melech

        2,464813




        2,464813







        • 2




          Nice job latexing.
          – djechlin
          Nov 19 at 18:36












        • 2




          Nice job latexing.
          – djechlin
          Nov 19 at 18:36







        2




        2




        Nice job latexing.
        – djechlin
        Nov 19 at 18:36




        Nice job latexing.
        – djechlin
        Nov 19 at 18:36










        up vote
        2
        down vote













        We can simply use the equivalent definition of differentiability



        $$f(a+h)=f(a)+f'(a)cdot h +o(h) implies lim_xto a f(x)=lim_hto 0 f(a+h)=f(a)$$






        share|cite|improve this answer




















        • Thanks for the answer. I know my proof is long and a bit harder to read, but is it correct? If it isn't, where exactly does it go wrong?
          – Star Platinum ZA WARUDO
          Nov 19 at 15:03










        • @StarPlatinumZAWARUDO It seems really over complicated to me, the implication is avery simple ansd trivial fact. I'll try to look to it more carefully later. Bye
          – gimusi
          Nov 19 at 15:10














        up vote
        2
        down vote













        We can simply use the equivalent definition of differentiability



        $$f(a+h)=f(a)+f'(a)cdot h +o(h) implies lim_xto a f(x)=lim_hto 0 f(a+h)=f(a)$$






        share|cite|improve this answer




















        • Thanks for the answer. I know my proof is long and a bit harder to read, but is it correct? If it isn't, where exactly does it go wrong?
          – Star Platinum ZA WARUDO
          Nov 19 at 15:03










        • @StarPlatinumZAWARUDO It seems really over complicated to me, the implication is avery simple ansd trivial fact. I'll try to look to it more carefully later. Bye
          – gimusi
          Nov 19 at 15:10












        up vote
        2
        down vote










        up vote
        2
        down vote









        We can simply use the equivalent definition of differentiability



        $$f(a+h)=f(a)+f'(a)cdot h +o(h) implies lim_xto a f(x)=lim_hto 0 f(a+h)=f(a)$$






        share|cite|improve this answer












        We can simply use the equivalent definition of differentiability



        $$f(a+h)=f(a)+f'(a)cdot h +o(h) implies lim_xto a f(x)=lim_hto 0 f(a+h)=f(a)$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 19 at 14:54









        gimusi

        86.6k74393




        86.6k74393











        • Thanks for the answer. I know my proof is long and a bit harder to read, but is it correct? If it isn't, where exactly does it go wrong?
          – Star Platinum ZA WARUDO
          Nov 19 at 15:03










        • @StarPlatinumZAWARUDO It seems really over complicated to me, the implication is avery simple ansd trivial fact. I'll try to look to it more carefully later. Bye
          – gimusi
          Nov 19 at 15:10
















        • Thanks for the answer. I know my proof is long and a bit harder to read, but is it correct? If it isn't, where exactly does it go wrong?
          – Star Platinum ZA WARUDO
          Nov 19 at 15:03










        • @StarPlatinumZAWARUDO It seems really over complicated to me, the implication is avery simple ansd trivial fact. I'll try to look to it more carefully later. Bye
          – gimusi
          Nov 19 at 15:10















        Thanks for the answer. I know my proof is long and a bit harder to read, but is it correct? If it isn't, where exactly does it go wrong?
        – Star Platinum ZA WARUDO
        Nov 19 at 15:03




        Thanks for the answer. I know my proof is long and a bit harder to read, but is it correct? If it isn't, where exactly does it go wrong?
        – Star Platinum ZA WARUDO
        Nov 19 at 15:03












        @StarPlatinumZAWARUDO It seems really over complicated to me, the implication is avery simple ansd trivial fact. I'll try to look to it more carefully later. Bye
        – gimusi
        Nov 19 at 15:10




        @StarPlatinumZAWARUDO It seems really over complicated to me, the implication is avery simple ansd trivial fact. I'll try to look to it more carefully later. Bye
        – gimusi
        Nov 19 at 15:10










        up vote
        1
        down vote













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



        so



        $ |f(x)-f(a)| <( epsilon' + |f'(a)|)|x-a|$



        then for $xto a$ you have that



        $|f(x)-f(a)|leq 0$






        share|cite|improve this answer
























          up vote
          1
          down vote













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



          so



          $ |f(x)-f(a)| <( epsilon' + |f'(a)|)|x-a|$



          then for $xto a$ you have that



          $|f(x)-f(a)|leq 0$






          share|cite|improve this answer






















            up vote
            1
            down vote










            up vote
            1
            down vote









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



            so



            $ |f(x)-f(a)| <( epsilon' + |f'(a)|)|x-a|$



            then for $xto a$ you have that



            $|f(x)-f(a)|leq 0$






            share|cite|improve this answer












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



            so



            $ |f(x)-f(a)| <( epsilon' + |f'(a)|)|x-a|$



            then for $xto a$ you have that



            $|f(x)-f(a)|leq 0$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 19 at 14:57









            Federico Fallucca

            1,62818




            1,62818




















                up vote
                1
                down vote













                The heart of your argument is basically the following. Give me some $epsilon$, and I have to find some $delta$ so that $|f(x)-f(a)|<epsilon$ when $|x-a|<delta$. To start with, I'll just take any $delta$, it doesn't matter. Now, within the range $[a-delta, a+delta]$, the function $xtofracf(x)-f(a)x-a$ is bounded, because it's convergent at $a$ and convergent functions are locally bounded - this is a theorem which you essentially spend the first couple of lines of your proof re-proving by way of the value $epsilon'$. So say $lvertfracf(x)-f(a)x-arvert<M$ on the interval $[a-delta, a+delta]$. Well then certainly $|f(x)-f(a)|<Mdelta$, and by choosing a sufficiently small $delta$ we can make that less than $epsilon$ (since the $M$ only gets smaller as $delta$ gets smaller).



                If you notice, in your own proof, the $epsilon'$ is a bit pointless. Since at the end you're just going to say "and now take $delta$ as small as is necessary to make this true", you may as well initially take $epsilon'=10^100$. All that matters is that $epsilon'$ is finite and that $delta$ can be made arbitrarily small, in other words, that the difference quotient is locally bounded. The exact value of the derivative at $a$ also doesn't matter.






                share|cite|improve this answer
























                  up vote
                  1
                  down vote













                  The heart of your argument is basically the following. Give me some $epsilon$, and I have to find some $delta$ so that $|f(x)-f(a)|<epsilon$ when $|x-a|<delta$. To start with, I'll just take any $delta$, it doesn't matter. Now, within the range $[a-delta, a+delta]$, the function $xtofracf(x)-f(a)x-a$ is bounded, because it's convergent at $a$ and convergent functions are locally bounded - this is a theorem which you essentially spend the first couple of lines of your proof re-proving by way of the value $epsilon'$. So say $lvertfracf(x)-f(a)x-arvert<M$ on the interval $[a-delta, a+delta]$. Well then certainly $|f(x)-f(a)|<Mdelta$, and by choosing a sufficiently small $delta$ we can make that less than $epsilon$ (since the $M$ only gets smaller as $delta$ gets smaller).



                  If you notice, in your own proof, the $epsilon'$ is a bit pointless. Since at the end you're just going to say "and now take $delta$ as small as is necessary to make this true", you may as well initially take $epsilon'=10^100$. All that matters is that $epsilon'$ is finite and that $delta$ can be made arbitrarily small, in other words, that the difference quotient is locally bounded. The exact value of the derivative at $a$ also doesn't matter.






                  share|cite|improve this answer






















                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    The heart of your argument is basically the following. Give me some $epsilon$, and I have to find some $delta$ so that $|f(x)-f(a)|<epsilon$ when $|x-a|<delta$. To start with, I'll just take any $delta$, it doesn't matter. Now, within the range $[a-delta, a+delta]$, the function $xtofracf(x)-f(a)x-a$ is bounded, because it's convergent at $a$ and convergent functions are locally bounded - this is a theorem which you essentially spend the first couple of lines of your proof re-proving by way of the value $epsilon'$. So say $lvertfracf(x)-f(a)x-arvert<M$ on the interval $[a-delta, a+delta]$. Well then certainly $|f(x)-f(a)|<Mdelta$, and by choosing a sufficiently small $delta$ we can make that less than $epsilon$ (since the $M$ only gets smaller as $delta$ gets smaller).



                    If you notice, in your own proof, the $epsilon'$ is a bit pointless. Since at the end you're just going to say "and now take $delta$ as small as is necessary to make this true", you may as well initially take $epsilon'=10^100$. All that matters is that $epsilon'$ is finite and that $delta$ can be made arbitrarily small, in other words, that the difference quotient is locally bounded. The exact value of the derivative at $a$ also doesn't matter.






                    share|cite|improve this answer












                    The heart of your argument is basically the following. Give me some $epsilon$, and I have to find some $delta$ so that $|f(x)-f(a)|<epsilon$ when $|x-a|<delta$. To start with, I'll just take any $delta$, it doesn't matter. Now, within the range $[a-delta, a+delta]$, the function $xtofracf(x)-f(a)x-a$ is bounded, because it's convergent at $a$ and convergent functions are locally bounded - this is a theorem which you essentially spend the first couple of lines of your proof re-proving by way of the value $epsilon'$. So say $lvertfracf(x)-f(a)x-arvert<M$ on the interval $[a-delta, a+delta]$. Well then certainly $|f(x)-f(a)|<Mdelta$, and by choosing a sufficiently small $delta$ we can make that less than $epsilon$ (since the $M$ only gets smaller as $delta$ gets smaller).



                    If you notice, in your own proof, the $epsilon'$ is a bit pointless. Since at the end you're just going to say "and now take $delta$ as small as is necessary to make this true", you may as well initially take $epsilon'=10^100$. All that matters is that $epsilon'$ is finite and that $delta$ can be made arbitrarily small, in other words, that the difference quotient is locally bounded. The exact value of the derivative at $a$ also doesn't matter.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 19 at 20:08









                    Jack M

                    18.3k33778




                    18.3k33778




















                        up vote
                        1
                        down vote













                        Not quite right. One of the issues is:




                        I define $epsilon'$ as $frac epsilon delta $




                        As it stands, this doesn't make sense, because you started the proof by taking an arbitrary $epsilon'$. Perhaps you mean something like: Since this inequality is true for any $epsilon '$, it is true in particular for $frac epsilon delta$ ...



                        But this doesn't quite make sense either; What's the $delta$ on the right-hand side? This $delta $ may depend on the $epsilon '$ you initially choose.




                        Try this alternative:



                        For any $epsilon> 0$ there is a $delta' > 0$, such that whenever $0 < |x-a| < delta'$, one has $left|fracf(x)-f(a)x-aright| < |f'(a)| + epsilon$. In particular, there is a $delta_0>0$ such that $left|fracf(x)-f(a)x-aright| < |f'(a)| + 1$



                        Now, choose $delta = min left(delta_0, frac 1 right) $



                        Can you complete the proof from here?






                        share|cite|improve this answer






















                        • I've made a few edits to the ending paragraphs to make it a bit easier to understand what I'm trying to say.
                          – Star Platinum ZA WARUDO
                          Nov 19 at 15:44














                        up vote
                        1
                        down vote













                        Not quite right. One of the issues is:




                        I define $epsilon'$ as $frac epsilon delta $




                        As it stands, this doesn't make sense, because you started the proof by taking an arbitrary $epsilon'$. Perhaps you mean something like: Since this inequality is true for any $epsilon '$, it is true in particular for $frac epsilon delta$ ...



                        But this doesn't quite make sense either; What's the $delta$ on the right-hand side? This $delta $ may depend on the $epsilon '$ you initially choose.




                        Try this alternative:



                        For any $epsilon> 0$ there is a $delta' > 0$, such that whenever $0 < |x-a| < delta'$, one has $left|fracf(x)-f(a)x-aright| < |f'(a)| + epsilon$. In particular, there is a $delta_0>0$ such that $left|fracf(x)-f(a)x-aright| < |f'(a)| + 1$



                        Now, choose $delta = min left(delta_0, frac 1 right) $



                        Can you complete the proof from here?






                        share|cite|improve this answer






















                        • I've made a few edits to the ending paragraphs to make it a bit easier to understand what I'm trying to say.
                          – Star Platinum ZA WARUDO
                          Nov 19 at 15:44












                        up vote
                        1
                        down vote










                        up vote
                        1
                        down vote









                        Not quite right. One of the issues is:




                        I define $epsilon'$ as $frac epsilon delta $




                        As it stands, this doesn't make sense, because you started the proof by taking an arbitrary $epsilon'$. Perhaps you mean something like: Since this inequality is true for any $epsilon '$, it is true in particular for $frac epsilon delta$ ...



                        But this doesn't quite make sense either; What's the $delta$ on the right-hand side? This $delta $ may depend on the $epsilon '$ you initially choose.




                        Try this alternative:



                        For any $epsilon> 0$ there is a $delta' > 0$, such that whenever $0 < |x-a| < delta'$, one has $left|fracf(x)-f(a)x-aright| < |f'(a)| + epsilon$. In particular, there is a $delta_0>0$ such that $left|fracf(x)-f(a)x-aright| < |f'(a)| + 1$



                        Now, choose $delta = min left(delta_0, frac 1 right) $



                        Can you complete the proof from here?






                        share|cite|improve this answer














                        Not quite right. One of the issues is:




                        I define $epsilon'$ as $frac epsilon delta $




                        As it stands, this doesn't make sense, because you started the proof by taking an arbitrary $epsilon'$. Perhaps you mean something like: Since this inequality is true for any $epsilon '$, it is true in particular for $frac epsilon delta$ ...



                        But this doesn't quite make sense either; What's the $delta$ on the right-hand side? This $delta $ may depend on the $epsilon '$ you initially choose.




                        Try this alternative:



                        For any $epsilon> 0$ there is a $delta' > 0$, such that whenever $0 < |x-a| < delta'$, one has $left|fracf(x)-f(a)x-aright| < |f'(a)| + epsilon$. In particular, there is a $delta_0>0$ such that $left|fracf(x)-f(a)x-aright| < |f'(a)| + 1$



                        Now, choose $delta = min left(delta_0, frac 1 right) $



                        Can you complete the proof from here?







                        share|cite|improve this answer














                        share|cite|improve this answer



                        share|cite|improve this answer








                        edited Nov 19 at 23:50









                        miracle173

                        7,28822247




                        7,28822247










                        answered Nov 19 at 15:18









                        Praneet Srivastava

                        762516




                        762516











                        • I've made a few edits to the ending paragraphs to make it a bit easier to understand what I'm trying to say.
                          – Star Platinum ZA WARUDO
                          Nov 19 at 15:44
















                        • I've made a few edits to the ending paragraphs to make it a bit easier to understand what I'm trying to say.
                          – Star Platinum ZA WARUDO
                          Nov 19 at 15:44















                        I've made a few edits to the ending paragraphs to make it a bit easier to understand what I'm trying to say.
                        – Star Platinum ZA WARUDO
                        Nov 19 at 15:44




                        I've made a few edits to the ending paragraphs to make it a bit easier to understand what I'm trying to say.
                        – Star Platinum ZA WARUDO
                        Nov 19 at 15:44










                        up vote
                        0
                        down vote













                        I try to put your arguments in the right order.



                        Proof:



                        Assume that $f$ is differentiable in $a$. So $f'(a)$ exists.



                        To prove that $f$ is continous in $a$ choose an arbitrary $varepsilon>0.$ Now you can choose $delta_1$ such
                        $$0<delta_1<fracvarepsilonf'(a), text if f'(a)ne 0$$
                        $$delta_1=1, text if |f'(a)|=0.$$
                        So we have
                        $$0<|f'(a)|<fracvarepsilondelta_1$$
                        and define
                        $$varepsilon_2=fracvarepsilondelta_1-|f'(a)|>0$$



                        Because f is differentiable in $a$ we can find a $delta_2$ such that
                        $$|fracf(x)-f(a)x-a - f'(a)| < varepsilon_2,; forall x: 0<|x-a|<delta_2$$



                        Using the triangle inequality, we get



                        $$ |fracf(x)-f(a)x-a| < varepsilon_2 + |f'(a)|,; forall x: 0<|x-a|<min(delta_1,delta_2)$$



                        So we set
                        $$delta=min(delta_1,delta_2)$$
                        and get
                        $$| f(x)-f(a)| < (varepsilon_2 + |f'(a)|) delta_2<fracvarepsilondelta_1min(delta_1,delta_2)le varepsilon,; forall x: 0<|x-a|<delta$$






                        share|cite|improve this answer


























                          up vote
                          0
                          down vote













                          I try to put your arguments in the right order.



                          Proof:



                          Assume that $f$ is differentiable in $a$. So $f'(a)$ exists.



                          To prove that $f$ is continous in $a$ choose an arbitrary $varepsilon>0.$ Now you can choose $delta_1$ such
                          $$0<delta_1<fracvarepsilonf'(a), text if f'(a)ne 0$$
                          $$delta_1=1, text if |f'(a)|=0.$$
                          So we have
                          $$0<|f'(a)|<fracvarepsilondelta_1$$
                          and define
                          $$varepsilon_2=fracvarepsilondelta_1-|f'(a)|>0$$



                          Because f is differentiable in $a$ we can find a $delta_2$ such that
                          $$|fracf(x)-f(a)x-a - f'(a)| < varepsilon_2,; forall x: 0<|x-a|<delta_2$$



                          Using the triangle inequality, we get



                          $$ |fracf(x)-f(a)x-a| < varepsilon_2 + |f'(a)|,; forall x: 0<|x-a|<min(delta_1,delta_2)$$



                          So we set
                          $$delta=min(delta_1,delta_2)$$
                          and get
                          $$| f(x)-f(a)| < (varepsilon_2 + |f'(a)|) delta_2<fracvarepsilondelta_1min(delta_1,delta_2)le varepsilon,; forall x: 0<|x-a|<delta$$






                          share|cite|improve this answer
























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            I try to put your arguments in the right order.



                            Proof:



                            Assume that $f$ is differentiable in $a$. So $f'(a)$ exists.



                            To prove that $f$ is continous in $a$ choose an arbitrary $varepsilon>0.$ Now you can choose $delta_1$ such
                            $$0<delta_1<fracvarepsilonf'(a), text if f'(a)ne 0$$
                            $$delta_1=1, text if |f'(a)|=0.$$
                            So we have
                            $$0<|f'(a)|<fracvarepsilondelta_1$$
                            and define
                            $$varepsilon_2=fracvarepsilondelta_1-|f'(a)|>0$$



                            Because f is differentiable in $a$ we can find a $delta_2$ such that
                            $$|fracf(x)-f(a)x-a - f'(a)| < varepsilon_2,; forall x: 0<|x-a|<delta_2$$



                            Using the triangle inequality, we get



                            $$ |fracf(x)-f(a)x-a| < varepsilon_2 + |f'(a)|,; forall x: 0<|x-a|<min(delta_1,delta_2)$$



                            So we set
                            $$delta=min(delta_1,delta_2)$$
                            and get
                            $$| f(x)-f(a)| < (varepsilon_2 + |f'(a)|) delta_2<fracvarepsilondelta_1min(delta_1,delta_2)le varepsilon,; forall x: 0<|x-a|<delta$$






                            share|cite|improve this answer














                            I try to put your arguments in the right order.



                            Proof:



                            Assume that $f$ is differentiable in $a$. So $f'(a)$ exists.



                            To prove that $f$ is continous in $a$ choose an arbitrary $varepsilon>0.$ Now you can choose $delta_1$ such
                            $$0<delta_1<fracvarepsilonf'(a), text if f'(a)ne 0$$
                            $$delta_1=1, text if |f'(a)|=0.$$
                            So we have
                            $$0<|f'(a)|<fracvarepsilondelta_1$$
                            and define
                            $$varepsilon_2=fracvarepsilondelta_1-|f'(a)|>0$$



                            Because f is differentiable in $a$ we can find a $delta_2$ such that
                            $$|fracf(x)-f(a)x-a - f'(a)| < varepsilon_2,; forall x: 0<|x-a|<delta_2$$



                            Using the triangle inequality, we get



                            $$ |fracf(x)-f(a)x-a| < varepsilon_2 + |f'(a)|,; forall x: 0<|x-a|<min(delta_1,delta_2)$$



                            So we set
                            $$delta=min(delta_1,delta_2)$$
                            and get
                            $$| f(x)-f(a)| < (varepsilon_2 + |f'(a)|) delta_2<fracvarepsilondelta_1min(delta_1,delta_2)le varepsilon,; forall x: 0<|x-a|<delta$$







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Nov 20 at 7:29

























                            answered Nov 20 at 1:40









                            miracle173

                            7,28822247




                            7,28822247



























                                 

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