Not Continuity of a multivariable function [duplicate]
This question is an exact duplicate of:
Urgent with the continuity: Existence of directional derivatives at all directions of a $f:mathbb{R}^{2} to mathbb{R}$.
1 answer
Let $f(x,y)=
begin{cases}
frac{2x^2y}{x^2+y^2} & text{if } (x,y) neq (0,0) \
0 & text{if } (x,y)=(0,0) .
end{cases}$
(i) Prove the directional derivatives of $f$ exist in any direction at the point $(0,0)$.
(ii)¿Is $f$ continuous on $(0,0)$?
For (i) I took $u=(u_{1},u_{2}) in mathbb{R^{2}}$ such $||u||=1$ and $0=(0,0)$. So $$lim_{t to 0} frac{f(0+ t u)-f(0)}{t}=lim_{t to 0}frac{2(tu_{1})^2 (tu_{2})}{(tu_{1})^4+(tu_{2})^2}=lim_{t to 0}frac{(t^2 u_{1}^2)(t^2 u_{1}^{2}) tu_{2}}{t^{4}u_{1}^{4}+t^{2} u_{2}^{2}}$$. But I cannot find the limit when $t$ aproaches to $0$. Which by finding them I would prove the directional derivative exist at every direction right?
For (ii) I got the intuition $f$ is not continuous at $(0,0)$ so I took $lbrace (frac{1}{k},frac{1}{k}) rbrace_{k in mathbb{N}}$ and $lbrace (frac{1}{k},0) rbrace_{k in mathbb{N}}$ which are two different sequences in $mathbb{R}^{2}$ converging to $(0,0)$. However, $lbrace f(frac{1}{k},frac{1}{k}) rbrace_{k in mathbb{N}}=lbrace frac{1}{k^{2}} rbrace_{n in mathbb{N}} to 0$ and $lbrace f(frac{1}{k},0) rbrace_{k in mathbb{N}} to 0$. So maybe my intuition was not right?
Also I've have approach $(0,0)$ through $(x,mx)$ and when this and gives me 0 as this value approaches to zero bit I cannot find another trajectory approaching trought $(0,0)$ which gives me another value distinct to $0$. :(
Can anyone help
me end the proof of continuity or not continuity , please?
multivariable-calculus continuity
marked as duplicate by spaceisdarkgreen, KReiser, Jean-Claude Arbaut, Did, Jyrki Lahtonen Nov 26 at 13:03
This question was marked as an exact duplicate of an existing question.
add a comment |
This question is an exact duplicate of:
Urgent with the continuity: Existence of directional derivatives at all directions of a $f:mathbb{R}^{2} to mathbb{R}$.
1 answer
Let $f(x,y)=
begin{cases}
frac{2x^2y}{x^2+y^2} & text{if } (x,y) neq (0,0) \
0 & text{if } (x,y)=(0,0) .
end{cases}$
(i) Prove the directional derivatives of $f$ exist in any direction at the point $(0,0)$.
(ii)¿Is $f$ continuous on $(0,0)$?
For (i) I took $u=(u_{1},u_{2}) in mathbb{R^{2}}$ such $||u||=1$ and $0=(0,0)$. So $$lim_{t to 0} frac{f(0+ t u)-f(0)}{t}=lim_{t to 0}frac{2(tu_{1})^2 (tu_{2})}{(tu_{1})^4+(tu_{2})^2}=lim_{t to 0}frac{(t^2 u_{1}^2)(t^2 u_{1}^{2}) tu_{2}}{t^{4}u_{1}^{4}+t^{2} u_{2}^{2}}$$. But I cannot find the limit when $t$ aproaches to $0$. Which by finding them I would prove the directional derivative exist at every direction right?
For (ii) I got the intuition $f$ is not continuous at $(0,0)$ so I took $lbrace (frac{1}{k},frac{1}{k}) rbrace_{k in mathbb{N}}$ and $lbrace (frac{1}{k},0) rbrace_{k in mathbb{N}}$ which are two different sequences in $mathbb{R}^{2}$ converging to $(0,0)$. However, $lbrace f(frac{1}{k},frac{1}{k}) rbrace_{k in mathbb{N}}=lbrace frac{1}{k^{2}} rbrace_{n in mathbb{N}} to 0$ and $lbrace f(frac{1}{k},0) rbrace_{k in mathbb{N}} to 0$. So maybe my intuition was not right?
Also I've have approach $(0,0)$ through $(x,mx)$ and when this and gives me 0 as this value approaches to zero bit I cannot find another trajectory approaching trought $(0,0)$ which gives me another value distinct to $0$. :(
Can anyone help
me end the proof of continuity or not continuity , please?
multivariable-calculus continuity
marked as duplicate by spaceisdarkgreen, KReiser, Jean-Claude Arbaut, Did, Jyrki Lahtonen Nov 26 at 13:03
This question was marked as an exact duplicate of an existing question.
add a comment |
This question is an exact duplicate of:
Urgent with the continuity: Existence of directional derivatives at all directions of a $f:mathbb{R}^{2} to mathbb{R}$.
1 answer
Let $f(x,y)=
begin{cases}
frac{2x^2y}{x^2+y^2} & text{if } (x,y) neq (0,0) \
0 & text{if } (x,y)=(0,0) .
end{cases}$
(i) Prove the directional derivatives of $f$ exist in any direction at the point $(0,0)$.
(ii)¿Is $f$ continuous on $(0,0)$?
For (i) I took $u=(u_{1},u_{2}) in mathbb{R^{2}}$ such $||u||=1$ and $0=(0,0)$. So $$lim_{t to 0} frac{f(0+ t u)-f(0)}{t}=lim_{t to 0}frac{2(tu_{1})^2 (tu_{2})}{(tu_{1})^4+(tu_{2})^2}=lim_{t to 0}frac{(t^2 u_{1}^2)(t^2 u_{1}^{2}) tu_{2}}{t^{4}u_{1}^{4}+t^{2} u_{2}^{2}}$$. But I cannot find the limit when $t$ aproaches to $0$. Which by finding them I would prove the directional derivative exist at every direction right?
For (ii) I got the intuition $f$ is not continuous at $(0,0)$ so I took $lbrace (frac{1}{k},frac{1}{k}) rbrace_{k in mathbb{N}}$ and $lbrace (frac{1}{k},0) rbrace_{k in mathbb{N}}$ which are two different sequences in $mathbb{R}^{2}$ converging to $(0,0)$. However, $lbrace f(frac{1}{k},frac{1}{k}) rbrace_{k in mathbb{N}}=lbrace frac{1}{k^{2}} rbrace_{n in mathbb{N}} to 0$ and $lbrace f(frac{1}{k},0) rbrace_{k in mathbb{N}} to 0$. So maybe my intuition was not right?
Also I've have approach $(0,0)$ through $(x,mx)$ and when this and gives me 0 as this value approaches to zero bit I cannot find another trajectory approaching trought $(0,0)$ which gives me another value distinct to $0$. :(
Can anyone help
me end the proof of continuity or not continuity , please?
multivariable-calculus continuity
This question is an exact duplicate of:
Urgent with the continuity: Existence of directional derivatives at all directions of a $f:mathbb{R}^{2} to mathbb{R}$.
1 answer
Let $f(x,y)=
begin{cases}
frac{2x^2y}{x^2+y^2} & text{if } (x,y) neq (0,0) \
0 & text{if } (x,y)=(0,0) .
end{cases}$
(i) Prove the directional derivatives of $f$ exist in any direction at the point $(0,0)$.
(ii)¿Is $f$ continuous on $(0,0)$?
For (i) I took $u=(u_{1},u_{2}) in mathbb{R^{2}}$ such $||u||=1$ and $0=(0,0)$. So $$lim_{t to 0} frac{f(0+ t u)-f(0)}{t}=lim_{t to 0}frac{2(tu_{1})^2 (tu_{2})}{(tu_{1})^4+(tu_{2})^2}=lim_{t to 0}frac{(t^2 u_{1}^2)(t^2 u_{1}^{2}) tu_{2}}{t^{4}u_{1}^{4}+t^{2} u_{2}^{2}}$$. But I cannot find the limit when $t$ aproaches to $0$. Which by finding them I would prove the directional derivative exist at every direction right?
For (ii) I got the intuition $f$ is not continuous at $(0,0)$ so I took $lbrace (frac{1}{k},frac{1}{k}) rbrace_{k in mathbb{N}}$ and $lbrace (frac{1}{k},0) rbrace_{k in mathbb{N}}$ which are two different sequences in $mathbb{R}^{2}$ converging to $(0,0)$. However, $lbrace f(frac{1}{k},frac{1}{k}) rbrace_{k in mathbb{N}}=lbrace frac{1}{k^{2}} rbrace_{n in mathbb{N}} to 0$ and $lbrace f(frac{1}{k},0) rbrace_{k in mathbb{N}} to 0$. So maybe my intuition was not right?
Also I've have approach $(0,0)$ through $(x,mx)$ and when this and gives me 0 as this value approaches to zero bit I cannot find another trajectory approaching trought $(0,0)$ which gives me another value distinct to $0$. :(
Can anyone help
me end the proof of continuity or not continuity , please?
This question is an exact duplicate of:
Urgent with the continuity: Existence of directional derivatives at all directions of a $f:mathbb{R}^{2} to mathbb{R}$.
1 answer
multivariable-calculus continuity
multivariable-calculus continuity
edited Nov 26 at 9:17
Tianlalu
3,05521038
3,05521038
asked Nov 26 at 3:44
Cos
1126
1126
marked as duplicate by spaceisdarkgreen, KReiser, Jean-Claude Arbaut, Did, Jyrki Lahtonen Nov 26 at 13:03
This question was marked as an exact duplicate of an existing question.
marked as duplicate by spaceisdarkgreen, KReiser, Jean-Claude Arbaut, Did, Jyrki Lahtonen Nov 26 at 13:03
This question was marked as an exact duplicate of an existing question.
add a comment |
add a comment |
2 Answers
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active
oldest
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Using polar coordinates,
$midfrac{2x^2y}{x^2+y^2}mid=midfrac{2r^2cos^2theta rsintheta}{r^2cos^2theta +r^2sin^2theta}mid=midfrac{2r^3cos^2thetasintheta}{r^2}midle 2rto0$ as $rto0$.
Hence $f$ is continuous at the origin.
add a comment |
Using AM-GM inequality, x^2+y^2 >= 2xy. 1/(x^2+y^2) =< 1/2xy.
So |2x^2y/(x^2+y^2)| =< |(2x^2y/2xy = x )| which approaches to zero as x approaches to zero.
Hence it is continuous at (0,0).
Hi and welcome to math.SE. Please use MathJax formatting to improve readability of your answer.
– francescop21
Nov 26 at 12:02
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Using polar coordinates,
$midfrac{2x^2y}{x^2+y^2}mid=midfrac{2r^2cos^2theta rsintheta}{r^2cos^2theta +r^2sin^2theta}mid=midfrac{2r^3cos^2thetasintheta}{r^2}midle 2rto0$ as $rto0$.
Hence $f$ is continuous at the origin.
add a comment |
Using polar coordinates,
$midfrac{2x^2y}{x^2+y^2}mid=midfrac{2r^2cos^2theta rsintheta}{r^2cos^2theta +r^2sin^2theta}mid=midfrac{2r^3cos^2thetasintheta}{r^2}midle 2rto0$ as $rto0$.
Hence $f$ is continuous at the origin.
add a comment |
Using polar coordinates,
$midfrac{2x^2y}{x^2+y^2}mid=midfrac{2r^2cos^2theta rsintheta}{r^2cos^2theta +r^2sin^2theta}mid=midfrac{2r^3cos^2thetasintheta}{r^2}midle 2rto0$ as $rto0$.
Hence $f$ is continuous at the origin.
Using polar coordinates,
$midfrac{2x^2y}{x^2+y^2}mid=midfrac{2r^2cos^2theta rsintheta}{r^2cos^2theta +r^2sin^2theta}mid=midfrac{2r^3cos^2thetasintheta}{r^2}midle 2rto0$ as $rto0$.
Hence $f$ is continuous at the origin.
edited Nov 26 at 5:36
answered Nov 26 at 5:13
Chris Custer
10.7k3724
10.7k3724
add a comment |
add a comment |
Using AM-GM inequality, x^2+y^2 >= 2xy. 1/(x^2+y^2) =< 1/2xy.
So |2x^2y/(x^2+y^2)| =< |(2x^2y/2xy = x )| which approaches to zero as x approaches to zero.
Hence it is continuous at (0,0).
Hi and welcome to math.SE. Please use MathJax formatting to improve readability of your answer.
– francescop21
Nov 26 at 12:02
add a comment |
Using AM-GM inequality, x^2+y^2 >= 2xy. 1/(x^2+y^2) =< 1/2xy.
So |2x^2y/(x^2+y^2)| =< |(2x^2y/2xy = x )| which approaches to zero as x approaches to zero.
Hence it is continuous at (0,0).
Hi and welcome to math.SE. Please use MathJax formatting to improve readability of your answer.
– francescop21
Nov 26 at 12:02
add a comment |
Using AM-GM inequality, x^2+y^2 >= 2xy. 1/(x^2+y^2) =< 1/2xy.
So |2x^2y/(x^2+y^2)| =< |(2x^2y/2xy = x )| which approaches to zero as x approaches to zero.
Hence it is continuous at (0,0).
Using AM-GM inequality, x^2+y^2 >= 2xy. 1/(x^2+y^2) =< 1/2xy.
So |2x^2y/(x^2+y^2)| =< |(2x^2y/2xy = x )| which approaches to zero as x approaches to zero.
Hence it is continuous at (0,0).
answered Nov 26 at 11:57
Souvik
1
1
Hi and welcome to math.SE. Please use MathJax formatting to improve readability of your answer.
– francescop21
Nov 26 at 12:02
add a comment |
Hi and welcome to math.SE. Please use MathJax formatting to improve readability of your answer.
– francescop21
Nov 26 at 12:02
Hi and welcome to math.SE. Please use MathJax formatting to improve readability of your answer.
– francescop21
Nov 26 at 12:02
Hi and welcome to math.SE. Please use MathJax formatting to improve readability of your answer.
– francescop21
Nov 26 at 12:02
add a comment |