Derivative definition with double limit












1














The following is an exercise from Calculus by Spivak.




Suppose that $f:mathbb{R}tomathbb{R}$ is differentiable at some point $xinmathbb{R}$. Prove that
$$f'(x)=lim_{h,kto 0^+}frac{f(x+h)-f(x-k)}{h+k}$$




My attempt - I tried something with substitutions, but it doesn't feel rigorous enough:



Let $a=x+h$, and $b=x-k$.
Then we have
begin{align*}
lim_{h,kto 0^+}frac{f(x+h)-f(x-k)}{h+k} & =lim_{kto 0^+}Bigg[lim_{hto 0^+}frac{f(x+h)-f(x-k)}{h+k}Bigg]
\ & = lim_{bto x^-}Bigg[lim_{ato b}frac{f(a)-f(b)}{a-b}Bigg]
\ & = lim_{bto x^-}f'(b)
\ & = f'(x)
end{align*}



This just feels quite off, but I wasn't sure how else to approach it. Any help is appreciated.






real-analysis proof-verification proof-writing







share|cite|improve this question






















  • Your first step is incorrect. There are functions $f(x,y)$ which are separately continuous in $x$ and $y$, but not jointly continuous in $(x,y)$. Also your next step is wrong: you have $ato x^+$.
    – Federico
    Nov 20 at 18:46












  • Try reasoning with $f(y)-f(x)-f'(x)(y-x)=o(|y-x|)$; it might be simpler
    – Federico
    Nov 20 at 18:50
















1














The following is an exercise from Calculus by Spivak.




Suppose that $f:mathbb{R}tomathbb{R}$ is differentiable at some point $xinmathbb{R}$. Prove that
$$f'(x)=lim_{h,kto 0^+}frac{f(x+h)-f(x-k)}{h+k}$$




My attempt - I tried something with substitutions, but it doesn't feel rigorous enough:



Let $a=x+h$, and $b=x-k$.
Then we have
begin{align*}
lim_{h,kto 0^+}frac{f(x+h)-f(x-k)}{h+k} & =lim_{kto 0^+}Bigg[lim_{hto 0^+}frac{f(x+h)-f(x-k)}{h+k}Bigg]
\ & = lim_{bto x^-}Bigg[lim_{ato b}frac{f(a)-f(b)}{a-b}Bigg]
\ & = lim_{bto x^-}f'(b)
\ & = f'(x)
end{align*}



This just feels quite off, but I wasn't sure how else to approach it. Any help is appreciated.






real-analysis proof-verification proof-writing







share|cite|improve this question






















  • Your first step is incorrect. There are functions $f(x,y)$ which are separately continuous in $x$ and $y$, but not jointly continuous in $(x,y)$. Also your next step is wrong: you have $ato x^+$.
    – Federico
    Nov 20 at 18:46












  • Try reasoning with $f(y)-f(x)-f'(x)(y-x)=o(|y-x|)$; it might be simpler
    – Federico
    Nov 20 at 18:50














1












1








1







The following is an exercise from Calculus by Spivak.




Suppose that $f:mathbb{R}tomathbb{R}$ is differentiable at some point $xinmathbb{R}$. Prove that
$$f'(x)=lim_{h,kto 0^+}frac{f(x+h)-f(x-k)}{h+k}$$




My attempt - I tried something with substitutions, but it doesn't feel rigorous enough:



Let $a=x+h$, and $b=x-k$.
Then we have
begin{align*}
lim_{h,kto 0^+}frac{f(x+h)-f(x-k)}{h+k} & =lim_{kto 0^+}Bigg[lim_{hto 0^+}frac{f(x+h)-f(x-k)}{h+k}Bigg]
\ & = lim_{bto x^-}Bigg[lim_{ato b}frac{f(a)-f(b)}{a-b}Bigg]
\ & = lim_{bto x^-}f'(b)
\ & = f'(x)
end{align*}



This just feels quite off, but I wasn't sure how else to approach it. Any help is appreciated.






real-analysis proof-verification proof-writing







share|cite|improve this question













The following is an exercise from Calculus by Spivak.




Suppose that $f:mathbb{R}tomathbb{R}$ is differentiable at some point $xinmathbb{R}$. Prove that
$$f'(x)=lim_{h,kto 0^+}frac{f(x+h)-f(x-k)}{h+k}$$




My attempt - I tried something with substitutions, but it doesn't feel rigorous enough:



Let $a=x+h$, and $b=x-k$.
Then we have
begin{align*}
lim_{h,kto 0^+}frac{f(x+h)-f(x-k)}{h+k} & =lim_{kto 0^+}Bigg[lim_{hto 0^+}frac{f(x+h)-f(x-k)}{h+k}Bigg]
\ & = lim_{bto x^-}Bigg[lim_{ato b}frac{f(a)-f(b)}{a-b}Bigg]
\ & = lim_{bto x^-}f'(b)
\ & = f'(x)
end{align*}



This just feels quite off, but I wasn't sure how else to approach it. Any help is appreciated.






real-analysis proof-verification proof-writing




real-analysis proof-verification proof-writing






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 20 at 18:04









高田航

1,283318




1,283318












  • Your first step is incorrect. There are functions $f(x,y)$ which are separately continuous in $x$ and $y$, but not jointly continuous in $(x,y)$. Also your next step is wrong: you have $ato x^+$.
    – Federico
    Nov 20 at 18:46












  • Try reasoning with $f(y)-f(x)-f'(x)(y-x)=o(|y-x|)$; it might be simpler
    – Federico
    Nov 20 at 18:50


















  • Your first step is incorrect. There are functions $f(x,y)$ which are separately continuous in $x$ and $y$, but not jointly continuous in $(x,y)$. Also your next step is wrong: you have $ato x^+$.
    – Federico
    Nov 20 at 18:46












  • Try reasoning with $f(y)-f(x)-f'(x)(y-x)=o(|y-x|)$; it might be simpler
    – Federico
    Nov 20 at 18:50
















Your first step is incorrect. There are functions $f(x,y)$ which are separately continuous in $x$ and $y$, but not jointly continuous in $(x,y)$. Also your next step is wrong: you have $ato x^+$.
– Federico
Nov 20 at 18:46






Your first step is incorrect. There are functions $f(x,y)$ which are separately continuous in $x$ and $y$, but not jointly continuous in $(x,y)$. Also your next step is wrong: you have $ato x^+$.
– Federico
Nov 20 at 18:46














Try reasoning with $f(y)-f(x)-f'(x)(y-x)=o(|y-x|)$; it might be simpler
– Federico
Nov 20 at 18:50




Try reasoning with $f(y)-f(x)-f'(x)(y-x)=o(|y-x|)$; it might be simpler
– Federico
Nov 20 at 18:50










1 Answer
1






active

oldest

votes


















2














$$
begin{split}
frac{f(x+h)-f(x-k)}{h+k} &= frac{f(x+h)-f(x)+f(x)-f(x-k)}{h+k} \
&= frac{f'(x)h+f'(x)k+o(h)+o(k)}{h+k} xrightarrow[substack{hto0^+\hto0^+}]{} f'(x) .
end{split}
$$






share|cite|improve this answer























  • I'm not well versed in asymptotic notation, and I'm not too sure how you get to that last step.
    – 高田航
    Nov 21 at 7:47






  • 1




    $o(h)$ is just any function $f(h)$ with the property that $lim_{hto0^+} frac{|f(h)|}{h}=0$. But then also $lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h+k} leq lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h} = 0$.
    – Federico
    Nov 21 at 15:13













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',
autoActivateHeartbeat: false,
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%2f3006678%2fderivative-definition-with-double-limit%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














$$
begin{split}
frac{f(x+h)-f(x-k)}{h+k} &= frac{f(x+h)-f(x)+f(x)-f(x-k)}{h+k} \
&= frac{f'(x)h+f'(x)k+o(h)+o(k)}{h+k} xrightarrow[substack{hto0^+\hto0^+}]{} f'(x) .
end{split}
$$






share|cite|improve this answer























  • I'm not well versed in asymptotic notation, and I'm not too sure how you get to that last step.
    – 高田航
    Nov 21 at 7:47






  • 1




    $o(h)$ is just any function $f(h)$ with the property that $lim_{hto0^+} frac{|f(h)|}{h}=0$. But then also $lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h+k} leq lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h} = 0$.
    – Federico
    Nov 21 at 15:13


















2














$$
begin{split}
frac{f(x+h)-f(x-k)}{h+k} &= frac{f(x+h)-f(x)+f(x)-f(x-k)}{h+k} \
&= frac{f'(x)h+f'(x)k+o(h)+o(k)}{h+k} xrightarrow[substack{hto0^+\hto0^+}]{} f'(x) .
end{split}
$$






share|cite|improve this answer























  • I'm not well versed in asymptotic notation, and I'm not too sure how you get to that last step.
    – 高田航
    Nov 21 at 7:47






  • 1




    $o(h)$ is just any function $f(h)$ with the property that $lim_{hto0^+} frac{|f(h)|}{h}=0$. But then also $lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h+k} leq lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h} = 0$.
    – Federico
    Nov 21 at 15:13
















2












2








2






$$
begin{split}
frac{f(x+h)-f(x-k)}{h+k} &= frac{f(x+h)-f(x)+f(x)-f(x-k)}{h+k} \
&= frac{f'(x)h+f'(x)k+o(h)+o(k)}{h+k} xrightarrow[substack{hto0^+\hto0^+}]{} f'(x) .
end{split}
$$






share|cite|improve this answer














$$
begin{split}
frac{f(x+h)-f(x-k)}{h+k} &= frac{f(x+h)-f(x)+f(x)-f(x-k)}{h+k} \
&= frac{f'(x)h+f'(x)k+o(h)+o(k)}{h+k} xrightarrow[substack{hto0^+\hto0^+}]{} f'(x) .
end{split}
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 21 at 15:15

























answered Nov 20 at 18:54









Federico

4,649514




4,649514












  • I'm not well versed in asymptotic notation, and I'm not too sure how you get to that last step.
    – 高田航
    Nov 21 at 7:47






  • 1




    $o(h)$ is just any function $f(h)$ with the property that $lim_{hto0^+} frac{|f(h)|}{h}=0$. But then also $lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h+k} leq lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h} = 0$.
    – Federico
    Nov 21 at 15:13




















  • I'm not well versed in asymptotic notation, and I'm not too sure how you get to that last step.
    – 高田航
    Nov 21 at 7:47






  • 1




    $o(h)$ is just any function $f(h)$ with the property that $lim_{hto0^+} frac{|f(h)|}{h}=0$. But then also $lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h+k} leq lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h} = 0$.
    – Federico
    Nov 21 at 15:13


















I'm not well versed in asymptotic notation, and I'm not too sure how you get to that last step.
– 高田航
Nov 21 at 7:47




I'm not well versed in asymptotic notation, and I'm not too sure how you get to that last step.
– 高田航
Nov 21 at 7:47




1




1




$o(h)$ is just any function $f(h)$ with the property that $lim_{hto0^+} frac{|f(h)|}{h}=0$. But then also $lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h+k} leq lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h} = 0$.
– Federico
Nov 21 at 15:13






$o(h)$ is just any function $f(h)$ with the property that $lim_{hto0^+} frac{|f(h)|}{h}=0$. But then also $lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h+k} leq lim_{substack{hto0^+\kto0^+}} frac{|f(h)|}{h} = 0$.
– Federico
Nov 21 at 15:13




















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3006678%2fderivative-definition-with-double-limit%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

How to change which sound is reproduced for terminal bell?

Image
3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
Read more

Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

Image
up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
Read more

Can I use Tabulator js library in my java Spring + Thymeleaf project?

Image
0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
Read more