Proof Verification: Differentiability implies continuity.











up vote
3
down vote

favorite
1












Proof:



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



$implies |frac{f(x)-f(a)}{x-a} - f'(a)| < epsilon'$



Using the Triangle inequality, we get



$ |frac{f(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 $epsilon'$ = $frac{epsilon}{delta'} - |f'(a)|$ where $delta' < delta$ such that $frac{epsilon}{delta'} > |f'(a)|$



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 $epsilon' = frac{h}{n} - |f'(a)|$. 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 $frac{epsilon}{1+|f'(a)|}$. To complete our answer, let one value of $delta$ for $epsilon' = 1$ be $delta_0$. Our final answer would be $delta = min(delta_0,frac{epsilon}{1+|f'(a)|})$










share|cite|improve this question
























  • There are some errors.. the derivative must be calculate on a and not in x
    – Federico Fallucca
    12 hours ago












  • 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
    12 hours ago










  • 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
    12 hours ago















up vote
3
down vote

favorite
1












Proof:



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



$implies |frac{f(x)-f(a)}{x-a} - f'(a)| < epsilon'$



Using the Triangle inequality, we get



$ |frac{f(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 $epsilon'$ = $frac{epsilon}{delta'} - |f'(a)|$ where $delta' < delta$ such that $frac{epsilon}{delta'} > |f'(a)|$



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 $epsilon' = frac{h}{n} - |f'(a)|$. 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 $frac{epsilon}{1+|f'(a)|}$. To complete our answer, let one value of $delta$ for $epsilon' = 1$ be $delta_0$. Our final answer would be $delta = min(delta_0,frac{epsilon}{1+|f'(a)|})$










share|cite|improve this question
























  • There are some errors.. the derivative must be calculate on a and not in x
    – Federico Fallucca
    12 hours ago












  • 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
    12 hours ago










  • 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
    12 hours ago













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 $lim_{x to
a} frac{f(x)-f(a)}{x-a}$
exists and is equal to $f'(a)$. So, for any $epsilon' > 0$ there exists a $delta > 0$, such that $0<|x-a| < delta$



$implies |frac{f(x)-f(a)}{x-a} - f'(a)| < epsilon'$



Using the Triangle inequality, we get



$ |frac{f(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 $epsilon'$ = $frac{epsilon}{delta'} - |f'(a)|$ where $delta' < delta$ such that $frac{epsilon}{delta'} > |f'(a)|$



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 $epsilon' = frac{h}{n} - |f'(a)|$. 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 $frac{epsilon}{1+|f'(a)|}$. To complete our answer, let one value of $delta$ for $epsilon' = 1$ be $delta_0$. Our final answer would be $delta = min(delta_0,frac{epsilon}{1+|f'(a)|})$










share|cite|improve this question















Proof:



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



$implies |frac{f(x)-f(a)}{x-a} - f'(a)| < epsilon'$



Using the Triangle inequality, we get



$ |frac{f(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 $epsilon'$ = $frac{epsilon}{delta'} - |f'(a)|$ where $delta' < delta$ such that $frac{epsilon}{delta'} > |f'(a)|$



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 $epsilon' = frac{h}{n} - |f'(a)|$. 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 $frac{epsilon}{1+|f'(a)|}$. To complete our answer, let one value of $delta$ for $epsilon' = 1$ be $delta_0$. Our final answer would be $delta = min(delta_0,frac{epsilon}{1+|f'(a)|})$







real-analysis epsilon-delta






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 1 hour ago









miracle173

7,28822247




7,28822247










asked 12 hours ago









Star Platinum ZA WARUDO

33212




33212












  • There are some errors.. the derivative must be calculate on a and not in x
    – Federico Fallucca
    12 hours ago












  • 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
    12 hours ago










  • 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
    12 hours ago


















  • There are some errors.. the derivative must be calculate on a and not in x
    – Federico Fallucca
    12 hours ago












  • 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
    12 hours ago










  • 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
    12 hours ago
















There are some errors.. the derivative must be calculate on a and not in x
– Federico Fallucca
12 hours ago






There are some errors.. the derivative must be calculate on a and not in x
– Federico Fallucca
12 hours ago














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
12 hours ago




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
12 hours ago












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
12 hours ago




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
12 hours ago










6 Answers
6






active

oldest

votes

















up vote
7
down vote













Or just see that
$$lim_{hrightarrow 0}f(x+h)-f(x)=lim_{hrightarrow 0}underbrace{frac{f(x+h)-f(x)}{h}}_{rightarrow f'(x)}h=0.$$






share|cite|improve this answer

















  • 2




    Nice job latexing.
    – djechlin
    8 hours ago


















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
    12 hours ago










  • @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
    12 hours ago


















up vote
1
down vote













$ |frac{f(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 $xtofrac{f(x)-f(a)}{x-a}$ 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 $lvertfrac{f(x)-f(a)}{x-a}rvert<M$ 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 $left|frac{f(x)-f(a)}{x-a}right| < |f'(a)| + epsilon$. In particular, there is a $delta_0>0$ such that $left|frac{f(x)-f(a)}{x-a}right| < |f'(a)| + 1$



      Now, choose $delta = min left(delta_0, frac 1 {|f'(a)+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
        11 hours ago


















      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
      $$0<delta_1<frac{varepsilon}{|f'(a)|}, text{ if } f'(a)ne 0$$
      $$delta_1=1, text{ if } |f'(a)|=0.$$
      So we have
      $$0<|f'(a)|<frac{varepsilon}{delta_1}$$
      and define
      $$varepsilon_2=frac{varepsilon}{delta_1}-|f'(a)|>0$$



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



      Using the triangle inequality, we get



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



      $$implies| f(x)-f(a)| < (varepsilon_2 + |f'(a)|) delta_2<frac{varepsilon}{delta_1}min(delta_1,delta_2)le varepsilon$$






      share|cite|improve this answer





















        Your Answer





        StackExchange.ifUsing("editor", function () {
        return StackExchange.using("mathjaxEditing", function () {
        StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
        StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
        });
        });
        }, "mathjax-editing");

        StackExchange.ready(function() {
        var channelOptions = {
        tags: "".split(" "),
        id: "69"
        };
        initTagRenderer("".split(" "), "".split(" "), channelOptions);

        StackExchange.using("externalEditor", function() {
        // Have to fire editor after snippets, if snippets enabled
        if (StackExchange.settings.snippets.snippetsEnabled) {
        StackExchange.using("snippets", function() {
        createEditor();
        });
        }
        else {
        createEditor();
        }
        });

        function createEditor() {
        StackExchange.prepareEditor({
        heartbeatType: 'answer',
        convertImagesToLinks: true,
        noModals: true,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: 10,
        bindNavPrevention: true,
        postfix: "",
        imageUploader: {
        brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
        contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
        allowUrls: true
        },
        noCode: true, onDemand: true,
        discardSelector: ".discard-answer"
        ,immediatelyShowMarkdownHelp:true
        });


        }
        });














         

        draft saved


        draft discarded


















        StackExchange.ready(
        function () {
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3005021%2fproof-verification-differentiability-implies-continuity%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        6 Answers
        6






        active

        oldest

        votes








        6 Answers
        6






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        7
        down vote













        Or just see that
        $$lim_{hrightarrow 0}f(x+h)-f(x)=lim_{hrightarrow 0}underbrace{frac{f(x+h)-f(x)}{h}}_{rightarrow f'(x)}h=0.$$






        share|cite|improve this answer

















        • 2




          Nice job latexing.
          – djechlin
          8 hours ago















        up vote
        7
        down vote













        Or just see that
        $$lim_{hrightarrow 0}f(x+h)-f(x)=lim_{hrightarrow 0}underbrace{frac{f(x+h)-f(x)}{h}}_{rightarrow f'(x)}h=0.$$






        share|cite|improve this answer

















        • 2




          Nice job latexing.
          – djechlin
          8 hours ago













        up vote
        7
        down vote










        up vote
        7
        down vote









        Or just see that
        $$lim_{hrightarrow 0}f(x+h)-f(x)=lim_{hrightarrow 0}underbrace{frac{f(x+h)-f(x)}{h}}_{rightarrow f'(x)}h=0.$$






        share|cite|improve this answer












        Or just see that
        $$lim_{hrightarrow 0}f(x+h)-f(x)=lim_{hrightarrow 0}underbrace{frac{f(x+h)-f(x)}{h}}_{rightarrow f'(x)}h=0.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 12 hours ago









        Peter Melech

        2,444813




        2,444813








        • 2




          Nice job latexing.
          – djechlin
          8 hours ago














        • 2




          Nice job latexing.
          – djechlin
          8 hours ago








        2




        2




        Nice job latexing.
        – djechlin
        8 hours ago




        Nice job latexing.
        – djechlin
        8 hours ago










        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
          12 hours ago










        • @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
          12 hours ago















        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
          12 hours ago










        • @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
          12 hours ago













        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 12 hours ago









        gimusi

        85.5k74293




        85.5k74293












        • 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
          12 hours ago










        • @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
          12 hours ago


















        • 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
          12 hours ago










        • @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
          12 hours ago
















        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
        12 hours ago




        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
        12 hours ago












        @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
        12 hours ago




        @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
        12 hours ago










        up vote
        1
        down vote













        $ |frac{f(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













          $ |frac{f(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









            $ |frac{f(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












            $ |frac{f(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 12 hours ago









            Federico Fallucca

            1,63018




            1,63018






















                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 $xtofrac{f(x)-f(a)}{x-a}$ 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 $lvertfrac{f(x)-f(a)}{x-a}rvert<M$ 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 $xtofrac{f(x)-f(a)}{x-a}$ 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 $lvertfrac{f(x)-f(a)}{x-a}rvert<M$ 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 $xtofrac{f(x)-f(a)}{x-a}$ 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 $lvertfrac{f(x)-f(a)}{x-a}rvert<M$ 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 $xtofrac{f(x)-f(a)}{x-a}$ 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 $lvertfrac{f(x)-f(a)}{x-a}rvert<M$ 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 7 hours ago









                    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 $left|frac{f(x)-f(a)}{x-a}right| < |f'(a)| + epsilon$. In particular, there is a $delta_0>0$ such that $left|frac{f(x)-f(a)}{x-a}right| < |f'(a)| + 1$



                        Now, choose $delta = min left(delta_0, frac 1 {|f'(a)+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
                          11 hours ago















                        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 $left|frac{f(x)-f(a)}{x-a}right| < |f'(a)| + epsilon$. In particular, there is a $delta_0>0$ such that $left|frac{f(x)-f(a)}{x-a}right| < |f'(a)| + 1$



                        Now, choose $delta = min left(delta_0, frac 1 {|f'(a)+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
                          11 hours ago













                        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 $left|frac{f(x)-f(a)}{x-a}right| < |f'(a)| + epsilon$. In particular, there is a $delta_0>0$ such that $left|frac{f(x)-f(a)}{x-a}right| < |f'(a)| + 1$



                        Now, choose $delta = min left(delta_0, frac 1 {|f'(a)+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 $left|frac{f(x)-f(a)}{x-a}right| < |f'(a)| + epsilon$. In particular, there is a $delta_0>0$ such that $left|frac{f(x)-f(a)}{x-a}right| < |f'(a)| + 1$



                        Now, choose $delta = min left(delta_0, frac 1 {|f'(a)+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 3 hours ago









                        miracle173

                        7,28822247




                        7,28822247










                        answered 11 hours ago









                        Praneet Srivastava

                        750516




                        750516












                        • 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
                          11 hours ago


















                        • 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
                          11 hours ago
















                        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
                        11 hours ago




                        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
                        11 hours ago










                        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
                        $$0<delta_1<frac{varepsilon}{|f'(a)|}, text{ if } f'(a)ne 0$$
                        $$delta_1=1, text{ if } |f'(a)|=0.$$
                        So we have
                        $$0<|f'(a)|<frac{varepsilon}{delta_1}$$
                        and define
                        $$varepsilon_2=frac{varepsilon}{delta_1}-|f'(a)|>0$$



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



                        Using the triangle inequality, we get



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



                        $$implies| f(x)-f(a)| < (varepsilon_2 + |f'(a)|) delta_2<frac{varepsilon}{delta_1}min(delta_1,delta_2)le varepsilon$$






                        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
                          $$0<delta_1<frac{varepsilon}{|f'(a)|}, text{ if } f'(a)ne 0$$
                          $$delta_1=1, text{ if } |f'(a)|=0.$$
                          So we have
                          $$0<|f'(a)|<frac{varepsilon}{delta_1}$$
                          and define
                          $$varepsilon_2=frac{varepsilon}{delta_1}-|f'(a)|>0$$



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



                          Using the triangle inequality, we get



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



                          $$implies| f(x)-f(a)| < (varepsilon_2 + |f'(a)|) delta_2<frac{varepsilon}{delta_1}min(delta_1,delta_2)le varepsilon$$






                          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
                            $$0<delta_1<frac{varepsilon}{|f'(a)|}, text{ if } f'(a)ne 0$$
                            $$delta_1=1, text{ if } |f'(a)|=0.$$
                            So we have
                            $$0<|f'(a)|<frac{varepsilon}{delta_1}$$
                            and define
                            $$varepsilon_2=frac{varepsilon}{delta_1}-|f'(a)|>0$$



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



                            Using the triangle inequality, we get



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



                            $$implies| f(x)-f(a)| < (varepsilon_2 + |f'(a)|) delta_2<frac{varepsilon}{delta_1}min(delta_1,delta_2)le varepsilon$$






                            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
                            $$0<delta_1<frac{varepsilon}{|f'(a)|}, text{ if } f'(a)ne 0$$
                            $$delta_1=1, text{ if } |f'(a)|=0.$$
                            So we have
                            $$0<|f'(a)|<frac{varepsilon}{delta_1}$$
                            and define
                            $$varepsilon_2=frac{varepsilon}{delta_1}-|f'(a)|>0$$



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



                            Using the triangle inequality, we get



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



                            $$implies| f(x)-f(a)| < (varepsilon_2 + |f'(a)|) delta_2<frac{varepsilon}{delta_1}min(delta_1,delta_2)le varepsilon$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 1 hour ago









                            miracle173

                            7,28822247




                            7,28822247






























                                 

                                draft saved


                                draft discarded



















































                                 


                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function () {
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3005021%2fproof-verification-differentiability-implies-continuity%23new-answer', 'question_page');
                                }
                                );

                                Post as a guest















                                Required, but never shown





















































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown

































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown







                                Popular posts from this blog

                                404 Error Contact Form 7 ajax form submitting

                                How to know if a Active Directory user can login interactively

                                TypeError: fit_transform() missing 1 required positional argument: 'X'