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...
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
asked Dec 9 '18 at 20:17
Wesley Strik
1,498422
|
show 2 more comments
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
asked Dec 9 '18 at 20:17
Wesley Strik
1,498422
1
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.
– Melody
Dec 9 '18 at 20:21
What is your question exactly? You started from asking if something is true and you ended up seeking a functional theorem
– Rebellos
Dec 9 '18 at 20:22
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".
– Wesley Strik
Dec 9 '18 at 20:25
that makes more sense, then I wouldn't need the multiplication by $-1$, might be an error then.
– Wesley Strik
Dec 9 '18 at 20:27
1
So your actual question is if there is a formal way to prove the following equality? $$limlimits_{xto a}g(x)=limlimits_{hto 0}g(a+h)$$
– Martin Rosenau
Dec 9 '18 at 20:34
|
show 2 more comments
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
asked Dec 9 '18 at 20:17
Wesley Strik
1,498422
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
asked Dec 9 '18 at 20:17
Wesley Strik
1,498422
asked Dec 9 '18 at 20:17
Wesley Strik
1,498422
edited Dec 9 '18 at 20:56
asked Dec 9 '18 at 20:17
Wesley Strik
1,498422
asked Dec 9 '18 at 20:17
Wesley Strik
1,498422
asked Dec 9 '18 at 20:17
Wesley Strik
1,498422
1,498422
1
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.
– Melody
Dec 9 '18 at 20:21
What is your question exactly? You started from asking if something is true and you ended up seeking a functional theorem
– Rebellos
Dec 9 '18 at 20:22
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".
– Wesley Strik
Dec 9 '18 at 20:25
that makes more sense, then I wouldn't need the multiplication by $-1$, might be an error then.
– Wesley Strik
Dec 9 '18 at 20:27
1
So your actual question is if there is a formal way to prove the following equality? $$limlimits_{xto a}g(x)=limlimits_{hto 0}g(a+h)$$
– Martin Rosenau
Dec 9 '18 at 20:34
|
show 2 more comments
1
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.
– Melody
Dec 9 '18 at 20:21
What is your question exactly? You started from asking if something is true and you ended up seeking a functional theorem
– Rebellos
Dec 9 '18 at 20:22
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".
– Wesley Strik
Dec 9 '18 at 20:25
that makes more sense, then I wouldn't need the multiplication by $-1$, might be an error then.
– Wesley Strik
Dec 9 '18 at 20:27
1
So your actual question is if there is a formal way to prove the following equality? $$limlimits_{xto a}g(x)=limlimits_{hto 0}g(a+h)$$
– Martin Rosenau
Dec 9 '18 at 20:34
1
1
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.
– Melody
Dec 9 '18 at 20:21
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.
– Melody
Dec 9 '18 at 20:21
What is your question exactly? You started from asking if something is true and you ended up seeking a functional theorem
– Rebellos
Dec 9 '18 at 20:22
What is your question exactly? You started from asking if something is true and you ended up seeking a functional theorem
– Rebellos
Dec 9 '18 at 20:22
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".
– Wesley Strik
Dec 9 '18 at 20:25
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".
– Wesley Strik
Dec 9 '18 at 20:25
that makes more sense, then I wouldn't need the multiplication by $-1$, might be an error then.
– Wesley Strik
Dec 9 '18 at 20:27
that makes more sense, then I wouldn't need the multiplication by $-1$, might be an error then.
– Wesley Strik
Dec 9 '18 at 20:27
1
1
So your actual question is if there is a formal way to prove the following equality? $$limlimits_{xto a}g(x)=limlimits_{hto 0}g(a+h)$$
– Martin Rosenau
Dec 9 '18 at 20:34
So your actual question is if there is a formal way to prove the following equality? $$limlimits_{xto a}g(x)=limlimits_{hto 0}g(a+h)$$
– Martin Rosenau
Dec 9 '18 at 20:34
|
show 2 more comments
1 Answer
1
active
oldest
votes
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 $epsilon$ ...
answered Dec 9 '18 at 21:07
Martin Rosenau
1,156139
This is exactly the kind of thing I was seeking, thank you.
– Wesley Strik
Dec 9 '18 at 21:18
add a comment |
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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 $epsilon$ ...
answered Dec 9 '18 at 21:07
Martin Rosenau
1,156139
This is exactly the kind of thing I was seeking, thank you.
– Wesley Strik
Dec 9 '18 at 21:18
add a comment |
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 $epsilon$ ...
answered Dec 9 '18 at 21:07
Martin Rosenau
1,156139
This is exactly the kind of thing I was seeking, thank you.
– Wesley Strik
Dec 9 '18 at 21:18
add a comment |
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 $epsilon$ ...
answered Dec 9 '18 at 21:07
Martin Rosenau
1,156139
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 $epsilon$ ...
answered Dec 9 '18 at 21:07
Martin Rosenau
1,156139
answered Dec 9 '18 at 21:07
Martin Rosenau
1,156139
answered Dec 9 '18 at 21:07
Martin Rosenau
1,156139
answered Dec 9 '18 at 21:07
Martin Rosenau
1,156139
1,156139
This is exactly the kind of thing I was seeking, thank you.
– Wesley Strik
Dec 9 '18 at 21:18
add a comment |
This is exactly the kind of thing I was seeking, thank you.
– Wesley Strik
Dec 9 '18 at 21:18
This is exactly the kind of thing I was seeking, thank you.
– Wesley Strik
Dec 9 '18 at 21:18
This is exactly the kind of thing I was seeking, thank you.
– Wesley Strik
Dec 9 '18 at 21:18
add a comment |
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1
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.
– Melody
Dec 9 '18 at 20:21
What is your question exactly? You started from asking if something is true and you ended up seeking a functional theorem
– Rebellos
Dec 9 '18 at 20:22
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".
– Wesley Strik
Dec 9 '18 at 20:25
that makes more sense, then I wouldn't need the multiplication by $-1$, might be an error then.
– Wesley Strik
Dec 9 '18 at 20:27
1
So your actual question is if there is a formal way to prove the following equality? $$limlimits_{xto a}g(x)=limlimits_{hto 0}g(a+h)$$
– Martin Rosenau
Dec 9 '18 at 20:34