Differentiability of a function that is $x^2$ on rational and 0 on irrational. [closed]
$begingroup$
Let f = $begin{cases}x^2,& x inmathbb{Q}\
0,& xin mathbb{R}setminus mathbb{Q}end{cases}$
Is $f$ differentiable at $x = 0$?
calculus real-analysis derivatives
edited Dec 2 '18 at 1:44
qbert
22.1k32460
asked Dec 2 '18 at 1:27
HarshitHarshit
11
$endgroup$
closed as off-topic by Saad, KReiser, Brahadeesh, NCh, RRL Dec 3 '18 at 5:22
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, KReiser, Brahadeesh, NCh, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let f = $begin{cases}x^2,& x inmathbb{Q}\
0,& xin mathbb{R}setminus mathbb{Q}end{cases}$
Is $f$ differentiable at $x = 0$?
calculus real-analysis derivatives
edited Dec 2 '18 at 1:44
qbert
22.1k32460
asked Dec 2 '18 at 1:27
HarshitHarshit
11
$endgroup$
closed as off-topic by Saad, KReiser, Brahadeesh, NCh, RRL Dec 3 '18 at 5:22
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, KReiser, Brahadeesh, NCh, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let f = $begin{cases}x^2,& x inmathbb{Q}\
0,& xin mathbb{R}setminus mathbb{Q}end{cases}$
Is $f$ differentiable at $x = 0$?
calculus real-analysis derivatives
edited Dec 2 '18 at 1:44
qbert
22.1k32460
asked Dec 2 '18 at 1:27
HarshitHarshit
11
$endgroup$
Let f = $begin{cases}x^2,& x inmathbb{Q}\
0,& xin mathbb{R}setminus mathbb{Q}end{cases}$
Is $f$ differentiable at $x = 0$?
calculus real-analysis derivatives
calculus real-analysis derivatives
edited Dec 2 '18 at 1:44
qbert
22.1k32460
asked Dec 2 '18 at 1:27
HarshitHarshit
11
edited Dec 2 '18 at 1:44
qbert
22.1k32460
asked Dec 2 '18 at 1:27
HarshitHarshit
11
edited Dec 2 '18 at 1:44
qbert
22.1k32460
edited Dec 2 '18 at 1:44
qbert
22.1k32460
edited Dec 2 '18 at 1:44
qbert
22.1k32460
22.1k32460
asked Dec 2 '18 at 1:27
HarshitHarshit
11
asked Dec 2 '18 at 1:27
HarshitHarshit
11
asked Dec 2 '18 at 1:27
HarshitHarshit
11
11
closed as off-topic by Saad, KReiser, Brahadeesh, NCh, RRL Dec 3 '18 at 5:22
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, KReiser, Brahadeesh, NCh, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, KReiser, Brahadeesh, NCh, RRL Dec 3 '18 at 5:22
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, KReiser, Brahadeesh, NCh, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Just compute the limit: for $x_nto 0$ (obviously $x_nne 0$) a sequence,
$$
lim_{nto infty}f(x_n)/x_n=begin{cases}0,& x_nin mathbb{Q}\
x_n,&x_nin mathbb{R}setminus mathbb{Q} \
end{cases}
$$
In either case, the limit is $0$.
If you like your epsilons and deltas, fix $epsilon>0$. Then choose $delta=sqrt{epsilon}$. Then, if $0<|x|<delta$, we have
$$
|f(x)/x|<epsilon
$$
since again, the $f(x)/x$ is either $0$ or $x$. So the derivative exists and is $0$.
answered Dec 2 '18 at 1:32
qbertqbert
22.1k32460
$endgroup$
add a comment |
$begingroup$
It's very easy to show that if $g$ is a bounded function defined on an interval containing zero then $f(x):=x^2g(x)$ is differentiable at $x=0$ and $f'(0)=0$. Now choose
$$g(x) = begin{cases}1,& x inmathbb{Q};\
0,& xin mathbb{R}setminus mathbb{Q}.end{cases}$$
answered Dec 2 '18 at 15:02
Michael HoppeMichael Hoppe
10.8k31834
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Just compute the limit: for $x_nto 0$ (obviously $x_nne 0$) a sequence,
$$
lim_{nto infty}f(x_n)/x_n=begin{cases}0,& x_nin mathbb{Q}\
x_n,&x_nin mathbb{R}setminus mathbb{Q} \
end{cases}
$$
In either case, the limit is $0$.
If you like your epsilons and deltas, fix $epsilon>0$. Then choose $delta=sqrt{epsilon}$. Then, if $0<|x|<delta$, we have
$$
|f(x)/x|<epsilon
$$
since again, the $f(x)/x$ is either $0$ or $x$. So the derivative exists and is $0$.
answered Dec 2 '18 at 1:32
qbertqbert
22.1k32460
$endgroup$
add a comment |
$begingroup$
Just compute the limit: for $x_nto 0$ (obviously $x_nne 0$) a sequence,
$$
lim_{nto infty}f(x_n)/x_n=begin{cases}0,& x_nin mathbb{Q}\
x_n,&x_nin mathbb{R}setminus mathbb{Q} \
end{cases}
$$
In either case, the limit is $0$.
If you like your epsilons and deltas, fix $epsilon>0$. Then choose $delta=sqrt{epsilon}$. Then, if $0<|x|<delta$, we have
$$
|f(x)/x|<epsilon
$$
since again, the $f(x)/x$ is either $0$ or $x$. So the derivative exists and is $0$.
answered Dec 2 '18 at 1:32
qbertqbert
22.1k32460
$endgroup$
add a comment |
$begingroup$
Just compute the limit: for $x_nto 0$ (obviously $x_nne 0$) a sequence,
$$
lim_{nto infty}f(x_n)/x_n=begin{cases}0,& x_nin mathbb{Q}\
x_n,&x_nin mathbb{R}setminus mathbb{Q} \
end{cases}
$$
In either case, the limit is $0$.
If you like your epsilons and deltas, fix $epsilon>0$. Then choose $delta=sqrt{epsilon}$. Then, if $0<|x|<delta$, we have
$$
|f(x)/x|<epsilon
$$
since again, the $f(x)/x$ is either $0$ or $x$. So the derivative exists and is $0$.
answered Dec 2 '18 at 1:32
qbertqbert
22.1k32460
$endgroup$
Just compute the limit: for $x_nto 0$ (obviously $x_nne 0$) a sequence,
$$
lim_{nto infty}f(x_n)/x_n=begin{cases}0,& x_nin mathbb{Q}\
x_n,&x_nin mathbb{R}setminus mathbb{Q} \
end{cases}
$$
In either case, the limit is $0$.
If you like your epsilons and deltas, fix $epsilon>0$. Then choose $delta=sqrt{epsilon}$. Then, if $0<|x|<delta$, we have
$$
|f(x)/x|<epsilon
$$
since again, the $f(x)/x$ is either $0$ or $x$. So the derivative exists and is $0$.
answered Dec 2 '18 at 1:32
qbertqbert
22.1k32460
answered Dec 2 '18 at 1:32
qbertqbert
22.1k32460
answered Dec 2 '18 at 1:32
qbertqbert
22.1k32460
answered Dec 2 '18 at 1:32
qbertqbert
22.1k32460
22.1k32460
add a comment |
add a comment |
$begingroup$
It's very easy to show that if $g$ is a bounded function defined on an interval containing zero then $f(x):=x^2g(x)$ is differentiable at $x=0$ and $f'(0)=0$. Now choose
$$g(x) = begin{cases}1,& x inmathbb{Q};\
0,& xin mathbb{R}setminus mathbb{Q}.end{cases}$$
answered Dec 2 '18 at 15:02
Michael HoppeMichael Hoppe
10.8k31834
$endgroup$
add a comment |
$begingroup$
It's very easy to show that if $g$ is a bounded function defined on an interval containing zero then $f(x):=x^2g(x)$ is differentiable at $x=0$ and $f'(0)=0$. Now choose
$$g(x) = begin{cases}1,& x inmathbb{Q};\
0,& xin mathbb{R}setminus mathbb{Q}.end{cases}$$
answered Dec 2 '18 at 15:02
Michael HoppeMichael Hoppe
10.8k31834
$endgroup$
add a comment |
$begingroup$
It's very easy to show that if $g$ is a bounded function defined on an interval containing zero then $f(x):=x^2g(x)$ is differentiable at $x=0$ and $f'(0)=0$. Now choose
$$g(x) = begin{cases}1,& x inmathbb{Q};\
0,& xin mathbb{R}setminus mathbb{Q}.end{cases}$$
answered Dec 2 '18 at 15:02
Michael HoppeMichael Hoppe
10.8k31834
$endgroup$
It's very easy to show that if $g$ is a bounded function defined on an interval containing zero then $f(x):=x^2g(x)$ is differentiable at $x=0$ and $f'(0)=0$. Now choose
$$g(x) = begin{cases}1,& x inmathbb{Q};\
0,& xin mathbb{R}setminus mathbb{Q}.end{cases}$$
answered Dec 2 '18 at 15:02
Michael HoppeMichael Hoppe
10.8k31834
answered Dec 2 '18 at 15:02
Michael HoppeMichael Hoppe
10.8k31834
answered Dec 2 '18 at 15:02
Michael HoppeMichael Hoppe
10.8k31834
answered Dec 2 '18 at 15:02
Michael HoppeMichael Hoppe
10.8k31834
10.8k31834
add a comment |
add a comment |
