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How to give an example of a $f$ differentiable in a deleted neighborhood of $x_0$ such that $lim_{xto...
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How would I give a simple example of a function $f$ differentiable in a deleted neighborhood of $x_0$ such that $lim_{xto x_0}f^prime(x)$ does not exist? I cannot seem to think of an example.
A delete neighborhood is an open interval about $x_0$ which does not contain $x_0$. So, $(x_0-delta,x_0+delta)-{x_0}$ for some $delta>0$.
How would something be differentiable in a deleted neighborhood if at the point of the derivative, the limit does not exist. Presumably, the derivative ends up looking something like $lim_{xto x_0} dfrac{1}{x}$, if it does not exist.
calculus derivatives examples-counterexamples
edited yesterday
GNUSupporter 8964民主女神 地下教會
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asked yesterday
kaisa
425
add a comment |
up vote
1
down vote
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How would I give a simple example of a function $f$ differentiable in a deleted neighborhood of $x_0$ such that $lim_{xto x_0}f^prime(x)$ does not exist? I cannot seem to think of an example.
A delete neighborhood is an open interval about $x_0$ which does not contain $x_0$. So, $(x_0-delta,x_0+delta)-{x_0}$ for some $delta>0$.
How would something be differentiable in a deleted neighborhood if at the point of the derivative, the limit does not exist. Presumably, the derivative ends up looking something like $lim_{xto x_0} dfrac{1}{x}$, if it does not exist.
calculus derivatives examples-counterexamples
edited yesterday
GNUSupporter 8964民主女神 地下教會
12.3k72343
asked yesterday
kaisa
425
Must be your function continuous?
– Dog_69
yesterday
@Dog_69 No, it can be any function we can dream up
– kaisa
yesterday
I was thinking about the Heaviside's function but I will say the absolute value $|x|$ around $x=0$.
– Dog_69
yesterday
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
How would I give a simple example of a function $f$ differentiable in a deleted neighborhood of $x_0$ such that $lim_{xto x_0}f^prime(x)$ does not exist? I cannot seem to think of an example.
A delete neighborhood is an open interval about $x_0$ which does not contain $x_0$. So, $(x_0-delta,x_0+delta)-{x_0}$ for some $delta>0$.
How would something be differentiable in a deleted neighborhood if at the point of the derivative, the limit does not exist. Presumably, the derivative ends up looking something like $lim_{xto x_0} dfrac{1}{x}$, if it does not exist.
calculus derivatives examples-counterexamples
edited yesterday
GNUSupporter 8964民主女神 地下教會
12.3k72343
asked yesterday
kaisa
425
How would I give a simple example of a function $f$ differentiable in a deleted neighborhood of $x_0$ such that $lim_{xto x_0}f^prime(x)$ does not exist? I cannot seem to think of an example.
A delete neighborhood is an open interval about $x_0$ which does not contain $x_0$. So, $(x_0-delta,x_0+delta)-{x_0}$ for some $delta>0$.
How would something be differentiable in a deleted neighborhood if at the point of the derivative, the limit does not exist. Presumably, the derivative ends up looking something like $lim_{xto x_0} dfrac{1}{x}$, if it does not exist.
calculus derivatives examples-counterexamples
calculus derivatives examples-counterexamples
edited yesterday
GNUSupporter 8964民主女神 地下教會
12.3k72343
asked yesterday
kaisa
425
edited yesterday
GNUSupporter 8964民主女神 地下教會
12.3k72343
asked yesterday
kaisa
425
edited yesterday
GNUSupporter 8964民主女神 地下教會
12.3k72343
edited yesterday
GNUSupporter 8964民主女神 地下教會
12.3k72343
edited yesterday
GNUSupporter 8964民主女神 地下教會
12.3k72343
12.3k72343
asked yesterday
kaisa
425
asked yesterday
kaisa
425
asked yesterday
kaisa
425
425
Must be your function continuous?
– Dog_69
yesterday
@Dog_69 No, it can be any function we can dream up
– kaisa
yesterday
I was thinking about the Heaviside's function but I will say the absolute value $|x|$ around $x=0$.
– Dog_69
yesterday
add a comment |
Must be your function continuous?
– Dog_69
yesterday
@Dog_69 No, it can be any function we can dream up
– kaisa
yesterday
I was thinking about the Heaviside's function but I will say the absolute value $|x|$ around $x=0$.
– Dog_69
yesterday
Must be your function continuous?
– Dog_69
yesterday
Must be your function continuous?
– Dog_69
yesterday
@Dog_69 No, it can be any function we can dream up
– kaisa
yesterday
@Dog_69 No, it can be any function we can dream up
– kaisa
yesterday
I was thinking about the Heaviside's function but I will say the absolute value $|x|$ around $x=0$.
– Dog_69
yesterday
I was thinking about the Heaviside's function but I will say the absolute value $|x|$ around $x=0$.
– Dog_69
yesterday
add a comment |
5 Answers
5
active
oldest
votes
up vote
4
down vote
accepted
Classic example:
$$sqrt[3]{(x-x_0)^2}$$
answered yesterday
trancelocation
8,0561519
add a comment |
up vote
6
down vote
Take $f(x) = x sin (1/x)$ near $0$
answered yesterday
Richard Martin
1,3238
add a comment |
up vote
3
down vote
You may try $f(x)=x^2cos(1/x)$, so that $f'(x)=2xcos(1/x)-sin(1/x)$ has a point of discontinuity at $x=0$.
answered yesterday
GNUSupporter 8964民主女神 地下教會
12.3k72343
2
This is a slightly better example than mine, in fact, because $f$ is differentiable at zero as well.
– Richard Martin
yesterday
add a comment |
up vote
2
down vote
Does $f(x)=x^frac 12 $ count?
$f'(x)=frac 1{2x^frac 12}$ which is discontinuous at $x=0$
answered yesterday
SmarthBansal
36412
1
Yes, it does! But see my remark on $x^2 cos 1/x$
– Richard Martin
yesterday
add a comment |
up vote
0
down vote
$$ln'(x) = dfrac{1}{x}$$
If you are looking for that exact derivative.
answered yesterday
rustypaper
84
add a comment |
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Classic example:
$$sqrt[3]{(x-x_0)^2}$$
answered yesterday
trancelocation
8,0561519
add a comment |
up vote
4
down vote
accepted
Classic example:
$$sqrt[3]{(x-x_0)^2}$$
answered yesterday
trancelocation
8,0561519
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Classic example:
$$sqrt[3]{(x-x_0)^2}$$
answered yesterday
trancelocation
8,0561519
Classic example:
$$sqrt[3]{(x-x_0)^2}$$
answered yesterday
trancelocation
8,0561519
answered yesterday
trancelocation
8,0561519
answered yesterday
trancelocation
8,0561519
answered yesterday
trancelocation
8,0561519
8,0561519
add a comment |
add a comment |
up vote
6
down vote
Take $f(x) = x sin (1/x)$ near $0$
answered yesterday
Richard Martin
1,3238
add a comment |
up vote
6
down vote
Take $f(x) = x sin (1/x)$ near $0$
answered yesterday
Richard Martin
1,3238
add a comment |
up vote
6
down vote
up vote
6
down vote
Take $f(x) = x sin (1/x)$ near $0$
answered yesterday
Richard Martin
1,3238
Take $f(x) = x sin (1/x)$ near $0$
answered yesterday
Richard Martin
1,3238
answered yesterday
Richard Martin
1,3238
answered yesterday
Richard Martin
1,3238
answered yesterday
Richard Martin
1,3238
1,3238
add a comment |
add a comment |
up vote
3
down vote
You may try $f(x)=x^2cos(1/x)$, so that $f'(x)=2xcos(1/x)-sin(1/x)$ has a point of discontinuity at $x=0$.
answered yesterday
GNUSupporter 8964民主女神 地下教會
12.3k72343
2
This is a slightly better example than mine, in fact, because $f$ is differentiable at zero as well.
– Richard Martin
yesterday
add a comment |
up vote
3
down vote
You may try $f(x)=x^2cos(1/x)$, so that $f'(x)=2xcos(1/x)-sin(1/x)$ has a point of discontinuity at $x=0$.
answered yesterday
GNUSupporter 8964民主女神 地下教會
12.3k72343
2
This is a slightly better example than mine, in fact, because $f$ is differentiable at zero as well.
– Richard Martin
yesterday
add a comment |
up vote
3
down vote
up vote
3
down vote
You may try $f(x)=x^2cos(1/x)$, so that $f'(x)=2xcos(1/x)-sin(1/x)$ has a point of discontinuity at $x=0$.
answered yesterday
GNUSupporter 8964民主女神 地下教會
12.3k72343
You may try $f(x)=x^2cos(1/x)$, so that $f'(x)=2xcos(1/x)-sin(1/x)$ has a point of discontinuity at $x=0$.
answered yesterday
GNUSupporter 8964民主女神 地下教會
12.3k72343
answered yesterday
GNUSupporter 8964民主女神 地下教會
12.3k72343
answered yesterday
GNUSupporter 8964民主女神 地下教會
12.3k72343
answered yesterday
GNUSupporter 8964民主女神 地下教會
12.3k72343
12.3k72343
2
This is a slightly better example than mine, in fact, because $f$ is differentiable at zero as well.
– Richard Martin
yesterday
add a comment |
2
This is a slightly better example than mine, in fact, because $f$ is differentiable at zero as well.
– Richard Martin
yesterday
2
2
This is a slightly better example than mine, in fact, because $f$ is differentiable at zero as well.
– Richard Martin
yesterday
This is a slightly better example than mine, in fact, because $f$ is differentiable at zero as well.
– Richard Martin
yesterday
add a comment |
up vote
2
down vote
Does $f(x)=x^frac 12 $ count?
$f'(x)=frac 1{2x^frac 12}$ which is discontinuous at $x=0$
answered yesterday
SmarthBansal
36412
1
Yes, it does! But see my remark on $x^2 cos 1/x$
– Richard Martin
yesterday
add a comment |
up vote
2
down vote
Does $f(x)=x^frac 12 $ count?
$f'(x)=frac 1{2x^frac 12}$ which is discontinuous at $x=0$
answered yesterday
SmarthBansal
36412
1
Yes, it does! But see my remark on $x^2 cos 1/x$
– Richard Martin
yesterday
add a comment |
up vote
2
down vote
up vote
2
down vote
Does $f(x)=x^frac 12 $ count?
$f'(x)=frac 1{2x^frac 12}$ which is discontinuous at $x=0$
answered yesterday
SmarthBansal
36412
Does $f(x)=x^frac 12 $ count?
$f'(x)=frac 1{2x^frac 12}$ which is discontinuous at $x=0$
answered yesterday
SmarthBansal
36412
answered yesterday
SmarthBansal
36412
answered yesterday
SmarthBansal
36412
answered yesterday
SmarthBansal
36412
36412
1
Yes, it does! But see my remark on $x^2 cos 1/x$
– Richard Martin
yesterday
add a comment |
1
Yes, it does! But see my remark on $x^2 cos 1/x$
– Richard Martin
yesterday
1
1
Yes, it does! But see my remark on $x^2 cos 1/x$
– Richard Martin
yesterday
Yes, it does! But see my remark on $x^2 cos 1/x$
– Richard Martin
yesterday
add a comment |
up vote
0
down vote
$$ln'(x) = dfrac{1}{x}$$
If you are looking for that exact derivative.
answered yesterday
rustypaper
84
add a comment |
up vote
0
down vote
$$ln'(x) = dfrac{1}{x}$$
If you are looking for that exact derivative.
answered yesterday
rustypaper
84
add a comment |
up vote
0
down vote
up vote
0
down vote
$$ln'(x) = dfrac{1}{x}$$
If you are looking for that exact derivative.
answered yesterday
rustypaper
84
$$ln'(x) = dfrac{1}{x}$$
If you are looking for that exact derivative.
answered yesterday
rustypaper
84
answered yesterday
rustypaper
84
answered yesterday
rustypaper
84
answered yesterday
rustypaper
84
84
add a comment |
add a comment |
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Must be your function continuous?
– Dog_69
yesterday
@Dog_69 No, it can be any function we can dream up
– kaisa
yesterday
I was thinking about the Heaviside's function but I will say the absolute value $|x|$ around $x=0$.
– Dog_69
yesterday