Verification that $lim_{x to a}frac{f(x)-f(a)}{x-a}$ and $lim_{hto 0}frac{f(a+h) -f(a)}{h}$ are equivalent...












0












$begingroup$


I wanted to verify that for definition of the derivative it is true that:




$$ lim_{x to a}frac{f(x)-f(a)}{x-a}= lim_{hto 0}frac{f(a+h) -f(a)}{h}$$




If I denote $h=x-a$, we can let $xto a$, this means that $h to 0$. Also notice that $f(x)=f(a+h)$
$$ lim_{x to a}frac{f(x)-f(a)}{x-a}=lim_{h to 0}frac{f(a+h)-f(a)}{h}$$
As desired.



Is the main idea here really just to use a variable substitution?



I suppose there is some formal theorem I also use here that when $f(x) to y$ we can let $g(f(x)) to g(y)$ if the function $g$ is continuous.






real-analysis derivatives proof-verification







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    On the R.H.S. you want $$lim_{hto0}dfrac{f(a+h)-f(a)}{h}.$$ You can't have $x$ be an varying on one side, and have a definite value on the other.
    $endgroup$
    – Melody
    Dec 9 '18 at 20:21












  • $begingroup$
    What is your question exactly? You started from asking if something is true and you ended up seeking a functional theorem
    $endgroup$
    – Rebellos
    Dec 9 '18 at 20:22










  • $begingroup$
    I seek a better understanding of this alternative definition, the book just says it is an easy exercise to rewrite it, thus left as a check for the reader , but I don't feel this is very "proper".
    $endgroup$
    – Wesley Strik
    Dec 9 '18 at 20:25












  • $begingroup$
    that makes more sense, then I wouldn't need the multiplication by $-1$, might be an error then.
    $endgroup$
    – Wesley Strik
    Dec 9 '18 at 20:27






  • 1




    $begingroup$
    So your actual question is if there is a formal way to prove the following equality? $$limlimits_{xto a}g(x)=limlimits_{hto 0}g(a+h)$$
    $endgroup$
    – Martin Rosenau
    Dec 9 '18 at 20:34
















0












$begingroup$


I wanted to verify that for definition of the derivative it is true that:




$$ lim_{x to a}frac{f(x)-f(a)}{x-a}= lim_{hto 0}frac{f(a+h) -f(a)}{h}$$




If I denote $h=x-a$, we can let $xto a$, this means that $h to 0$. Also notice that $f(x)=f(a+h)$
$$ lim_{x to a}frac{f(x)-f(a)}{x-a}=lim_{h to 0}frac{f(a+h)-f(a)}{h}$$
As desired.



Is the main idea here really just to use a variable substitution?



I suppose there is some formal theorem I also use here that when $f(x) to y$ we can let $g(f(x)) to g(y)$ if the function $g$ is continuous.






real-analysis derivatives proof-verification







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    On the R.H.S. you want $$lim_{hto0}dfrac{f(a+h)-f(a)}{h}.$$ You can't have $x$ be an varying on one side, and have a definite value on the other.
    $endgroup$
    – Melody
    Dec 9 '18 at 20:21












  • $begingroup$
    What is your question exactly? You started from asking if something is true and you ended up seeking a functional theorem
    $endgroup$
    – Rebellos
    Dec 9 '18 at 20:22










  • $begingroup$
    I seek a better understanding of this alternative definition, the book just says it is an easy exercise to rewrite it, thus left as a check for the reader , but I don't feel this is very "proper".
    $endgroup$
    – Wesley Strik
    Dec 9 '18 at 20:25












  • $begingroup$
    that makes more sense, then I wouldn't need the multiplication by $-1$, might be an error then.
    $endgroup$
    – Wesley Strik
    Dec 9 '18 at 20:27






  • 1




    $begingroup$
    So your actual question is if there is a formal way to prove the following equality? $$limlimits_{xto a}g(x)=limlimits_{hto 0}g(a+h)$$
    $endgroup$
    – Martin Rosenau
    Dec 9 '18 at 20:34














0












0








0





$begingroup$


I wanted to verify that for definition of the derivative it is true that:




$$ lim_{x to a}frac{f(x)-f(a)}{x-a}= lim_{hto 0}frac{f(a+h) -f(a)}{h}$$




If I denote $h=x-a$, we can let $xto a$, this means that $h to 0$. Also notice that $f(x)=f(a+h)$
$$ lim_{x to a}frac{f(x)-f(a)}{x-a}=lim_{h to 0}frac{f(a+h)-f(a)}{h}$$
As desired.



Is the main idea here really just to use a variable substitution?



I suppose there is some formal theorem I also use here that when $f(x) to y$ we can let $g(f(x)) to g(y)$ if the function $g$ is continuous.






real-analysis derivatives proof-verification







share|cite|improve this question











$endgroup$




I wanted to verify that for definition of the derivative it is true that:




$$ lim_{x to a}frac{f(x)-f(a)}{x-a}= lim_{hto 0}frac{f(a+h) -f(a)}{h}$$




If I denote $h=x-a$, we can let $xto a$, this means that $h to 0$. Also notice that $f(x)=f(a+h)$
$$ lim_{x to a}frac{f(x)-f(a)}{x-a}=lim_{h to 0}frac{f(a+h)-f(a)}{h}$$
As desired.



Is the main idea here really just to use a variable substitution?



I suppose there is some formal theorem I also use here that when $f(x) to y$ we can let $g(f(x)) to g(y)$ if the function $g$ is continuous.






real-analysis derivatives proof-verification




real-analysis derivatives proof-verification






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 20:56







Wesley Strik

















asked Dec 9 '18 at 20:17









Wesley StrikWesley Strik

1,761423




1,761423








  • 1




    $begingroup$
    On the R.H.S. you want $$lim_{hto0}dfrac{f(a+h)-f(a)}{h}.$$ You can't have $x$ be an varying on one side, and have a definite value on the other.
    $endgroup$
    – Melody
    Dec 9 '18 at 20:21












  • $begingroup$
    What is your question exactly? You started from asking if something is true and you ended up seeking a functional theorem
    $endgroup$
    – Rebellos
    Dec 9 '18 at 20:22










  • $begingroup$
    I seek a better understanding of this alternative definition, the book just says it is an easy exercise to rewrite it, thus left as a check for the reader , but I don't feel this is very "proper".
    $endgroup$
    – Wesley Strik
    Dec 9 '18 at 20:25












  • $begingroup$
    that makes more sense, then I wouldn't need the multiplication by $-1$, might be an error then.
    $endgroup$
    – Wesley Strik
    Dec 9 '18 at 20:27






  • 1




    $begingroup$
    So your actual question is if there is a formal way to prove the following equality? $$limlimits_{xto a}g(x)=limlimits_{hto 0}g(a+h)$$
    $endgroup$
    – Martin Rosenau
    Dec 9 '18 at 20:34














  • 1




    $begingroup$
    On the R.H.S. you want $$lim_{hto0}dfrac{f(a+h)-f(a)}{h}.$$ You can't have $x$ be an varying on one side, and have a definite value on the other.
    $endgroup$
    – Melody
    Dec 9 '18 at 20:21












  • $begingroup$
    What is your question exactly? You started from asking if something is true and you ended up seeking a functional theorem
    $endgroup$
    – Rebellos
    Dec 9 '18 at 20:22










  • $begingroup$
    I seek a better understanding of this alternative definition, the book just says it is an easy exercise to rewrite it, thus left as a check for the reader , but I don't feel this is very "proper".
    $endgroup$
    – Wesley Strik
    Dec 9 '18 at 20:25












  • $begingroup$
    that makes more sense, then I wouldn't need the multiplication by $-1$, might be an error then.
    $endgroup$
    – Wesley Strik
    Dec 9 '18 at 20:27






  • 1




    $begingroup$
    So your actual question is if there is a formal way to prove the following equality? $$limlimits_{xto a}g(x)=limlimits_{hto 0}g(a+h)$$
    $endgroup$
    – Martin Rosenau
    Dec 9 '18 at 20:34








1




1




$begingroup$
On the R.H.S. you want $$lim_{hto0}dfrac{f(a+h)-f(a)}{h}.$$ You can't have $x$ be an varying on one side, and have a definite value on the other.
$endgroup$
– Melody
Dec 9 '18 at 20:21






$begingroup$
On the R.H.S. you want $$lim_{hto0}dfrac{f(a+h)-f(a)}{h}.$$ You can't have $x$ be an varying on one side, and have a definite value on the other.
$endgroup$
– Melody
Dec 9 '18 at 20:21














$begingroup$
What is your question exactly? You started from asking if something is true and you ended up seeking a functional theorem
$endgroup$
– Rebellos
Dec 9 '18 at 20:22




$begingroup$
What is your question exactly? You started from asking if something is true and you ended up seeking a functional theorem
$endgroup$
– Rebellos
Dec 9 '18 at 20:22












$begingroup$
I seek a better understanding of this alternative definition, the book just says it is an easy exercise to rewrite it, thus left as a check for the reader , but I don't feel this is very "proper".
$endgroup$
– Wesley Strik
Dec 9 '18 at 20:25






$begingroup$
I seek a better understanding of this alternative definition, the book just says it is an easy exercise to rewrite it, thus left as a check for the reader , but I don't feel this is very "proper".
$endgroup$
– Wesley Strik
Dec 9 '18 at 20:25














$begingroup$
that makes more sense, then I wouldn't need the multiplication by $-1$, might be an error then.
$endgroup$
– Wesley Strik
Dec 9 '18 at 20:27




$begingroup$
that makes more sense, then I wouldn't need the multiplication by $-1$, might be an error then.
$endgroup$
– Wesley Strik
Dec 9 '18 at 20:27




1




1




$begingroup$
So your actual question is if there is a formal way to prove the following equality? $$limlimits_{xto a}g(x)=limlimits_{hto 0}g(a+h)$$
$endgroup$
– Martin Rosenau
Dec 9 '18 at 20:34




$begingroup$
So your actual question is if there is a formal way to prove the following equality? $$limlimits_{xto a}g(x)=limlimits_{hto 0}g(a+h)$$
$endgroup$
– Martin Rosenau
Dec 9 '18 at 20:34










1 Answer
1






active

oldest

votes


















1












$begingroup$

If I understand the problem correctly, the question is if it is possible to prove the following equality:



$$limlimits_{xto a}g(x)=limlimits_{hto 0}g(a+h)$$



Because in this case you can use: $$g(x) = frac{f(x)-f(a)}{x-a}$$




I suppose there is some formal theorem I also use here ...




If you want to use the formal definition of the limit:



$$limlimits_{xto a}g(x)=y$$ means that for every $epsilon>0$ given we can find a value $delta>0$ so that the following inequation applies for all $win[-delta,delta]$: $$|g(underbrace{a+w}_{"xto a"})-y|<epsilon$$



However, because $a+w=a+(0+w)$ this also means that: $$|g(a+underbrace{(0+w)}_{"hto 0"})-y|=|g(underbrace{a+w}_{"xto a"})-y|<epsilon$$



This however means that: $$limlimits_{hto 0}g(a+h)=y$$ ... because we can also find a value $delta$ for each given value $epsilon$ ...






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is exactly the kind of thing I was seeking, thank you.
    $endgroup$
    – Wesley Strik
    Dec 9 '18 at 21:18











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%2f3032926%2fverification-that-lim-x-to-a-fracfx-fax-a-and-lim-h-to-0-frac%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









1












$begingroup$

If I understand the problem correctly, the question is if it is possible to prove the following equality:



$$limlimits_{xto a}g(x)=limlimits_{hto 0}g(a+h)$$



Because in this case you can use: $$g(x) = frac{f(x)-f(a)}{x-a}$$




I suppose there is some formal theorem I also use here ...




If you want to use the formal definition of the limit:



$$limlimits_{xto a}g(x)=y$$ means that for every $epsilon>0$ given we can find a value $delta>0$ so that the following inequation applies for all $win[-delta,delta]$: $$|g(underbrace{a+w}_{"xto a"})-y|<epsilon$$



However, because $a+w=a+(0+w)$ this also means that: $$|g(a+underbrace{(0+w)}_{"hto 0"})-y|=|g(underbrace{a+w}_{"xto a"})-y|<epsilon$$



This however means that: $$limlimits_{hto 0}g(a+h)=y$$ ... because we can also find a value $delta$ for each given value $epsilon$ ...






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is exactly the kind of thing I was seeking, thank you.
    $endgroup$
    – Wesley Strik
    Dec 9 '18 at 21:18
















1












$begingroup$

If I understand the problem correctly, the question is if it is possible to prove the following equality:



$$limlimits_{xto a}g(x)=limlimits_{hto 0}g(a+h)$$



Because in this case you can use: $$g(x) = frac{f(x)-f(a)}{x-a}$$




I suppose there is some formal theorem I also use here ...




If you want to use the formal definition of the limit:



$$limlimits_{xto a}g(x)=y$$ means that for every $epsilon>0$ given we can find a value $delta>0$ so that the following inequation applies for all $win[-delta,delta]$: $$|g(underbrace{a+w}_{"xto a"})-y|<epsilon$$



However, because $a+w=a+(0+w)$ this also means that: $$|g(a+underbrace{(0+w)}_{"hto 0"})-y|=|g(underbrace{a+w}_{"xto a"})-y|<epsilon$$



This however means that: $$limlimits_{hto 0}g(a+h)=y$$ ... because we can also find a value $delta$ for each given value $epsilon$ ...






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is exactly the kind of thing I was seeking, thank you.
    $endgroup$
    – Wesley Strik
    Dec 9 '18 at 21:18














1












1








1





$begingroup$

If I understand the problem correctly, the question is if it is possible to prove the following equality:



$$limlimits_{xto a}g(x)=limlimits_{hto 0}g(a+h)$$



Because in this case you can use: $$g(x) = frac{f(x)-f(a)}{x-a}$$




I suppose there is some formal theorem I also use here ...




If you want to use the formal definition of the limit:



$$limlimits_{xto a}g(x)=y$$ means that for every $epsilon>0$ given we can find a value $delta>0$ so that the following inequation applies for all $win[-delta,delta]$: $$|g(underbrace{a+w}_{"xto a"})-y|<epsilon$$



However, because $a+w=a+(0+w)$ this also means that: $$|g(a+underbrace{(0+w)}_{"hto 0"})-y|=|g(underbrace{a+w}_{"xto a"})-y|<epsilon$$



This however means that: $$limlimits_{hto 0}g(a+h)=y$$ ... because we can also find a value $delta$ for each given value $epsilon$ ...






share|cite|improve this answer









$endgroup$



If I understand the problem correctly, the question is if it is possible to prove the following equality:



$$limlimits_{xto a}g(x)=limlimits_{hto 0}g(a+h)$$



Because in this case you can use: $$g(x) = frac{f(x)-f(a)}{x-a}$$




I suppose there is some formal theorem I also use here ...




If you want to use the formal definition of the limit:



$$limlimits_{xto a}g(x)=y$$ means that for every $epsilon>0$ given we can find a value $delta>0$ so that the following inequation applies for all $win[-delta,delta]$: $$|g(underbrace{a+w}_{"xto a"})-y|<epsilon$$



However, because $a+w=a+(0+w)$ this also means that: $$|g(a+underbrace{(0+w)}_{"hto 0"})-y|=|g(underbrace{a+w}_{"xto a"})-y|<epsilon$$



This however means that: $$limlimits_{hto 0}g(a+h)=y$$ ... because we can also find a value $delta$ for each given value $epsilon$ ...







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 9 '18 at 21:07









Martin RosenauMartin Rosenau

1,156139




1,156139












  • $begingroup$
    This is exactly the kind of thing I was seeking, thank you.
    $endgroup$
    – Wesley Strik
    Dec 9 '18 at 21:18


















  • $begingroup$
    This is exactly the kind of thing I was seeking, thank you.
    $endgroup$
    – Wesley Strik
    Dec 9 '18 at 21:18
















$begingroup$
This is exactly the kind of thing I was seeking, thank you.
$endgroup$
– Wesley Strik
Dec 9 '18 at 21:18




$begingroup$
This is exactly the kind of thing I was seeking, thank you.
$endgroup$
– Wesley Strik
Dec 9 '18 at 21:18


















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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3032926%2fverification-that-lim-x-to-a-fracfx-fax-a-and-lim-h-to-0-frac%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

Quarter-circle Tiles

Image
up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
Read more

build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

Image
0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
Read more

Mont Emei

Image
Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
Read more