How to give an example of a $f$ differentiable in a deleted neighborhood of $x_0$ such that $lim_{xto...
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 Nov 19 '18 at 13:57
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Nov 19 '18 at 13:46
kaisa
1019
add a comment |
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 Nov 19 '18 at 13:57
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Nov 19 '18 at 13:46
kaisa
1019
Must be your function continuous?
– Dog_69
Nov 19 '18 at 13:50
@Dog_69 No, it can be any function we can dream up
– kaisa
Nov 19 '18 at 13:51
I was thinking about the Heaviside's function but I will say the absolute value $|x|$ around $x=0$.
– Dog_69
Nov 19 '18 at 13:58
add a comment |
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 Nov 19 '18 at 13:57
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Nov 19 '18 at 13:46
kaisa
1019
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 Nov 19 '18 at 13:57
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Nov 19 '18 at 13:46
kaisa
1019
edited Nov 19 '18 at 13:57
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Nov 19 '18 at 13:46
kaisa
1019
edited Nov 19 '18 at 13:57
GNUSupporter 8964民主女神 地下教會
12.8k72445
edited Nov 19 '18 at 13:57
GNUSupporter 8964民主女神 地下教會
12.8k72445
edited Nov 19 '18 at 13:57
GNUSupporter 8964民主女神 地下教會
12.8k72445
12.8k72445
asked Nov 19 '18 at 13:46
kaisa
1019
asked Nov 19 '18 at 13:46
kaisa
1019
asked Nov 19 '18 at 13:46
kaisa
1019
1019
Must be your function continuous?
– Dog_69
Nov 19 '18 at 13:50
@Dog_69 No, it can be any function we can dream up
– kaisa
Nov 19 '18 at 13:51
I was thinking about the Heaviside's function but I will say the absolute value $|x|$ around $x=0$.
– Dog_69
Nov 19 '18 at 13:58
add a comment |
Must be your function continuous?
– Dog_69
Nov 19 '18 at 13:50
@Dog_69 No, it can be any function we can dream up
– kaisa
Nov 19 '18 at 13:51
I was thinking about the Heaviside's function but I will say the absolute value $|x|$ around $x=0$.
– Dog_69
Nov 19 '18 at 13:58
Must be your function continuous?
– Dog_69
Nov 19 '18 at 13:50
Must be your function continuous?
– Dog_69
Nov 19 '18 at 13:50
@Dog_69 No, it can be any function we can dream up
– kaisa
Nov 19 '18 at 13:51
@Dog_69 No, it can be any function we can dream up
– kaisa
Nov 19 '18 at 13:51
I was thinking about the Heaviside's function but I will say the absolute value $|x|$ around $x=0$.
– Dog_69
Nov 19 '18 at 13:58
I was thinking about the Heaviside's function but I will say the absolute value $|x|$ around $x=0$.
– Dog_69
Nov 19 '18 at 13:58
add a comment |
5 Answers
5
active
oldest
votes
Classic example:
$$sqrt[3]{(x-x_0)^2}$$
answered Nov 19 '18 at 13:51
trancelocation
9,1151521
add a comment |
Take $f(x) = x sin (1/x)$ near $0$
answered Nov 19 '18 at 13:51
Richard Martin
1,61118
add a comment |
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 Nov 19 '18 at 13:53
GNUSupporter 8964民主女神 地下教會
12.8k72445
2
This is a slightly better example than mine, in fact, because $f$ is differentiable at zero as well.
– Richard Martin
Nov 19 '18 at 14:09
add a comment |
Does $f(x)=x^frac 12 $ count?
$f'(x)=frac 1{2x^frac 12}$ which is discontinuous at $x=0$
answered Nov 19 '18 at 14:00
SmarthBansal
36412
1
Yes, it does! But see my remark on $x^2 cos 1/x$
– Richard Martin
Nov 19 '18 at 14:10
add a comment |
$$ln'(x) = dfrac{1}{x}$$
If you are looking for that exact derivative.
edited Dec 4 '18 at 9:56
GNUSupporter 8964民主女神 地下教會
12.8k72445
answered Nov 19 '18 at 17:08
rustypaper
84
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
Classic example:
$$sqrt[3]{(x-x_0)^2}$$
answered Nov 19 '18 at 13:51
trancelocation
9,1151521
add a comment |
Classic example:
$$sqrt[3]{(x-x_0)^2}$$
answered Nov 19 '18 at 13:51
trancelocation
9,1151521
add a comment |
Classic example:
$$sqrt[3]{(x-x_0)^2}$$
answered Nov 19 '18 at 13:51
trancelocation
9,1151521
Classic example:
$$sqrt[3]{(x-x_0)^2}$$
answered Nov 19 '18 at 13:51
trancelocation
9,1151521
answered Nov 19 '18 at 13:51
trancelocation
9,1151521
answered Nov 19 '18 at 13:51
trancelocation
9,1151521
answered Nov 19 '18 at 13:51
trancelocation
9,1151521
9,1151521
add a comment |
add a comment |
Take $f(x) = x sin (1/x)$ near $0$
answered Nov 19 '18 at 13:51
Richard Martin
1,61118
add a comment |
Take $f(x) = x sin (1/x)$ near $0$
answered Nov 19 '18 at 13:51
Richard Martin
1,61118
add a comment |
Take $f(x) = x sin (1/x)$ near $0$
answered Nov 19 '18 at 13:51
Richard Martin
1,61118
Take $f(x) = x sin (1/x)$ near $0$
answered Nov 19 '18 at 13:51
Richard Martin
1,61118
answered Nov 19 '18 at 13:51
Richard Martin
1,61118
answered Nov 19 '18 at 13:51
Richard Martin
1,61118
answered Nov 19 '18 at 13:51
Richard Martin
1,61118
1,61118
add a comment |
add a comment |
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 Nov 19 '18 at 13:53
GNUSupporter 8964民主女神 地下教會
12.8k72445
2
This is a slightly better example than mine, in fact, because $f$ is differentiable at zero as well.
– Richard Martin
Nov 19 '18 at 14:09
add a comment |
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 Nov 19 '18 at 13:53
GNUSupporter 8964民主女神 地下教會
12.8k72445
2
This is a slightly better example than mine, in fact, because $f$ is differentiable at zero as well.
– Richard Martin
Nov 19 '18 at 14:09
add a comment |
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 Nov 19 '18 at 13:53
GNUSupporter 8964民主女神 地下教會
12.8k72445
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 Nov 19 '18 at 13:53
GNUSupporter 8964民主女神 地下教會
12.8k72445
answered Nov 19 '18 at 13:53
GNUSupporter 8964民主女神 地下教會
12.8k72445
answered Nov 19 '18 at 13:53
GNUSupporter 8964民主女神 地下教會
12.8k72445
answered Nov 19 '18 at 13:53
GNUSupporter 8964民主女神 地下教會
12.8k72445
12.8k72445
2
This is a slightly better example than mine, in fact, because $f$ is differentiable at zero as well.
– Richard Martin
Nov 19 '18 at 14:09
add a comment |
2
This is a slightly better example than mine, in fact, because $f$ is differentiable at zero as well.
– Richard Martin
Nov 19 '18 at 14:09
2
2
This is a slightly better example than mine, in fact, because $f$ is differentiable at zero as well.
– Richard Martin
Nov 19 '18 at 14:09
This is a slightly better example than mine, in fact, because $f$ is differentiable at zero as well.
– Richard Martin
Nov 19 '18 at 14:09
add a comment |
Does $f(x)=x^frac 12 $ count?
$f'(x)=frac 1{2x^frac 12}$ which is discontinuous at $x=0$
answered Nov 19 '18 at 14:00
SmarthBansal
36412
1
Yes, it does! But see my remark on $x^2 cos 1/x$
– Richard Martin
Nov 19 '18 at 14:10
add a comment |
Does $f(x)=x^frac 12 $ count?
$f'(x)=frac 1{2x^frac 12}$ which is discontinuous at $x=0$
answered Nov 19 '18 at 14:00
SmarthBansal
36412
1
Yes, it does! But see my remark on $x^2 cos 1/x$
– Richard Martin
Nov 19 '18 at 14:10
add a comment |
Does $f(x)=x^frac 12 $ count?
$f'(x)=frac 1{2x^frac 12}$ which is discontinuous at $x=0$
answered Nov 19 '18 at 14:00
SmarthBansal
36412
Does $f(x)=x^frac 12 $ count?
$f'(x)=frac 1{2x^frac 12}$ which is discontinuous at $x=0$
answered Nov 19 '18 at 14:00
SmarthBansal
36412
answered Nov 19 '18 at 14:00
SmarthBansal
36412
answered Nov 19 '18 at 14:00
SmarthBansal
36412
answered Nov 19 '18 at 14:00
SmarthBansal
36412
36412
1
Yes, it does! But see my remark on $x^2 cos 1/x$
– Richard Martin
Nov 19 '18 at 14:10
add a comment |
1
Yes, it does! But see my remark on $x^2 cos 1/x$
– Richard Martin
Nov 19 '18 at 14:10
1
1
Yes, it does! But see my remark on $x^2 cos 1/x$
– Richard Martin
Nov 19 '18 at 14:10
Yes, it does! But see my remark on $x^2 cos 1/x$
– Richard Martin
Nov 19 '18 at 14:10
add a comment |
$$ln'(x) = dfrac{1}{x}$$
If you are looking for that exact derivative.
edited Dec 4 '18 at 9:56
GNUSupporter 8964民主女神 地下教會
12.8k72445
answered Nov 19 '18 at 17:08
rustypaper
84
add a comment |
$$ln'(x) = dfrac{1}{x}$$
If you are looking for that exact derivative.
edited Dec 4 '18 at 9:56
GNUSupporter 8964民主女神 地下教會
12.8k72445
answered Nov 19 '18 at 17:08
rustypaper
84
add a comment |
$$ln'(x) = dfrac{1}{x}$$
If you are looking for that exact derivative.
edited Dec 4 '18 at 9:56
GNUSupporter 8964民主女神 地下教會
12.8k72445
answered Nov 19 '18 at 17:08
rustypaper
84
$$ln'(x) = dfrac{1}{x}$$
If you are looking for that exact derivative.
edited Dec 4 '18 at 9:56
GNUSupporter 8964民主女神 地下教會
12.8k72445
answered Nov 19 '18 at 17:08
rustypaper
84
edited Dec 4 '18 at 9:56
GNUSupporter 8964民主女神 地下教會
12.8k72445
edited Dec 4 '18 at 9:56
GNUSupporter 8964民主女神 地下教會
12.8k72445
edited Dec 4 '18 at 9:56
GNUSupporter 8964民主女神 地下教會
12.8k72445
12.8k72445
answered Nov 19 '18 at 17:08
rustypaper
84
answered Nov 19 '18 at 17:08
rustypaper
84
answered Nov 19 '18 at 17:08
rustypaper
84
84
add a comment |
add a comment |
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Must be your function continuous?
– Dog_69
Nov 19 '18 at 13:50
@Dog_69 No, it can be any function we can dream up
– kaisa
Nov 19 '18 at 13:51
I was thinking about the Heaviside's function but I will say the absolute value $|x|$ around $x=0$.
– Dog_69
Nov 19 '18 at 13:58