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











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1 Answer
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1 Answer
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active

oldest

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active

oldest

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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


















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