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Differentiability implies continuity doubt in proof
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Let $f: [a,b] to mathbbR$ be a map that is differentiable at a point $x in [a,b]$. Then $f$ is continuous at $x$.
I'm trying to be very rigorously here. For reference, here's the definition of differentiability I'm using:
Let $f$ be defined (and real-valued) on $[a,b]$. For any $x in
[a,b]$, form the quotient $$phi(t) = fracf(t) - f(x)t-xquad (a
< t <b, t neq x)$$ and define $$f'(x) = lim_t to x phi(t)$$
Proof:
It suffices to show that $lim_t to x, t in [a,b] f(t) = f(x)$
And, $$lim_t to x, t in [a,b] f(t) = lim_tto x, t in [a,b] [(f(t)-f(x))+f(x)]$$
$$= lim_t to x, t in (a,b), t neq x [(f(t)-f(x))+f(x)]$$
$$= lim_t to x, t in (a,b), t neq x left[fracf(t)-f(x)t-x(t-x) + f(x)right]$$
$$= underbracelim_t to x, t in (a,b), t neq x fracf(t)-f(x)t-x_in mathbbR, by assumptionunderbracelim_t to x, t in (a,b), t neq x(t-x)_= 0 + f(x) = f(x)$$
Is everything I did correct? Nobody ever writes down these limit subranges. Why is that?
real-analysis derivatives
asked Jul 29 at 19:48
Math_QED
6,34131344
add a comment |Â
up vote
0
down vote
favorite
Let $f: [a,b] to mathbbR$ be a map that is differentiable at a point $x in [a,b]$. Then $f$ is continuous at $x$.
I'm trying to be very rigorously here. For reference, here's the definition of differentiability I'm using:
Let $f$ be defined (and real-valued) on $[a,b]$. For any $x in
[a,b]$, form the quotient $$phi(t) = fracf(t) - f(x)t-xquad (a
< t <b, t neq x)$$ and define $$f'(x) = lim_t to x phi(t)$$
Proof:
It suffices to show that $lim_t to x, t in [a,b] f(t) = f(x)$
And, $$lim_t to x, t in [a,b] f(t) = lim_tto x, t in [a,b] [(f(t)-f(x))+f(x)]$$
$$= lim_t to x, t in (a,b), t neq x [(f(t)-f(x))+f(x)]$$
$$= lim_t to x, t in (a,b), t neq x left[fracf(t)-f(x)t-x(t-x) + f(x)right]$$
$$= underbracelim_t to x, t in (a,b), t neq x fracf(t)-f(x)t-x_in mathbbR, by assumptionunderbracelim_t to x, t in (a,b), t neq x(t-x)_= 0 + f(x) = f(x)$$
Is everything I did correct? Nobody ever writes down these limit subranges. Why is that?
real-analysis derivatives
asked Jul 29 at 19:48
Math_QED
6,34131344
2
Looks correct. There was a similar post yesterday: math.stackexchange.com/a/2865578/168433
– md2perpe
Jul 29 at 19:53
In the case at hand the subranges are "tacitly understood". During a limit $tto x$ the variable $t$ does not assume the value $x$ by definition, and values of $t$ outside the domain of $f$ are not taken into consideration anyway.
– Christian Blatter
Jul 30 at 7:49
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $f: [a,b] to mathbbR$ be a map that is differentiable at a point $x in [a,b]$. Then $f$ is continuous at $x$.
I'm trying to be very rigorously here. For reference, here's the definition of differentiability I'm using:
Let $f$ be defined (and real-valued) on $[a,b]$. For any $x in
[a,b]$, form the quotient $$phi(t) = fracf(t) - f(x)t-xquad (a
< t <b, t neq x)$$ and define $$f'(x) = lim_t to x phi(t)$$
Proof:
It suffices to show that $lim_t to x, t in [a,b] f(t) = f(x)$
And, $$lim_t to x, t in [a,b] f(t) = lim_tto x, t in [a,b] [(f(t)-f(x))+f(x)]$$
$$= lim_t to x, t in (a,b), t neq x [(f(t)-f(x))+f(x)]$$
$$= lim_t to x, t in (a,b), t neq x left[fracf(t)-f(x)t-x(t-x) + f(x)right]$$
$$= underbracelim_t to x, t in (a,b), t neq x fracf(t)-f(x)t-x_in mathbbR, by assumptionunderbracelim_t to x, t in (a,b), t neq x(t-x)_= 0 + f(x) = f(x)$$
Is everything I did correct? Nobody ever writes down these limit subranges. Why is that?
real-analysis derivatives
asked Jul 29 at 19:48
Math_QED
6,34131344
Let $f: [a,b] to mathbbR$ be a map that is differentiable at a point $x in [a,b]$. Then $f$ is continuous at $x$.
I'm trying to be very rigorously here. For reference, here's the definition of differentiability I'm using:
Let $f$ be defined (and real-valued) on $[a,b]$. For any $x in
[a,b]$, form the quotient $$phi(t) = fracf(t) - f(x)t-xquad (a
< t <b, t neq x)$$ and define $$f'(x) = lim_t to x phi(t)$$
Proof:
It suffices to show that $lim_t to x, t in [a,b] f(t) = f(x)$
And, $$lim_t to x, t in [a,b] f(t) = lim_tto x, t in [a,b] [(f(t)-f(x))+f(x)]$$
$$= lim_t to x, t in (a,b), t neq x [(f(t)-f(x))+f(x)]$$
$$= lim_t to x, t in (a,b), t neq x left[fracf(t)-f(x)t-x(t-x) + f(x)right]$$
$$= underbracelim_t to x, t in (a,b), t neq x fracf(t)-f(x)t-x_in mathbbR, by assumptionunderbracelim_t to x, t in (a,b), t neq x(t-x)_= 0 + f(x) = f(x)$$
Is everything I did correct? Nobody ever writes down these limit subranges. Why is that?
real-analysis derivatives
asked Jul 29 at 19:48
Math_QED
6,34131344
asked Jul 29 at 19:48
Math_QED
6,34131344
asked Jul 29 at 19:48
Math_QED
6,34131344
asked Jul 29 at 19:48
Math_QED
6,34131344
6,34131344
2
Looks correct. There was a similar post yesterday: math.stackexchange.com/a/2865578/168433
– md2perpe
Jul 29 at 19:53
In the case at hand the subranges are "tacitly understood". During a limit $tto x$ the variable $t$ does not assume the value $x$ by definition, and values of $t$ outside the domain of $f$ are not taken into consideration anyway.
– Christian Blatter
Jul 30 at 7:49
add a comment |Â
2
Looks correct. There was a similar post yesterday: math.stackexchange.com/a/2865578/168433
– md2perpe
Jul 29 at 19:53
In the case at hand the subranges are "tacitly understood". During a limit $tto x$ the variable $t$ does not assume the value $x$ by definition, and values of $t$ outside the domain of $f$ are not taken into consideration anyway.
– Christian Blatter
Jul 30 at 7:49
2
2
Looks correct. There was a similar post yesterday: math.stackexchange.com/a/2865578/168433
– md2perpe
Jul 29 at 19:53
Looks correct. There was a similar post yesterday: math.stackexchange.com/a/2865578/168433
– md2perpe
Jul 29 at 19:53
In the case at hand the subranges are "tacitly understood". During a limit $tto x$ the variable $t$ does not assume the value $x$ by definition, and values of $t$ outside the domain of $f$ are not taken into consideration anyway.
– Christian Blatter
Jul 30 at 7:49
In the case at hand the subranges are "tacitly understood". During a limit $tto x$ the variable $t$ does not assume the value $x$ by definition, and values of $t$ outside the domain of $f$ are not taken into consideration anyway.
– Christian Blatter
Jul 30 at 7:49
add a comment |Â
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Looks correct. There was a similar post yesterday: math.stackexchange.com/a/2865578/168433
– md2perpe
Jul 29 at 19:53
In the case at hand the subranges are "tacitly understood". During a limit $tto x$ the variable $t$ does not assume the value $x$ by definition, and values of $t$ outside the domain of $f$ are not taken into consideration anyway.
– Christian Blatter
Jul 30 at 7:49