Does $lim_{(x,y)to(0,0)}frac{6xy^2}{x^2+y^2}$ exist? [closed]
I am trying to solve the following
$$lim_{(x,y)to(0,0)}frac{6xy^2}{x^2+y^2}$$
This is basically $0/0$ form, but as I saw other post that L'Hôpital's rule does not work on multiple variables.
Then, how could we proceed to do this? Does the limit exist? If yes, what is it?
calculus limits multivariable-calculus limits-without-lhopital
edited Dec 8 at 22:37
amWhy
191k28224439
asked Dec 8 at 17:55
Adshead
51
closed as off-topic by Jyrki Lahtonen, amWhy, José Carlos Santos, RRL, KReiser Dec 9 at 1:56
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Jyrki Lahtonen, amWhy, José Carlos Santos, RRL, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
I am trying to solve the following
$$lim_{(x,y)to(0,0)}frac{6xy^2}{x^2+y^2}$$
This is basically $0/0$ form, but as I saw other post that L'Hôpital's rule does not work on multiple variables.
Then, how could we proceed to do this? Does the limit exist? If yes, what is it?
calculus limits multivariable-calculus limits-without-lhopital
edited Dec 8 at 22:37
amWhy
191k28224439
asked Dec 8 at 17:55
Adshead
51
closed as off-topic by Jyrki Lahtonen, amWhy, José Carlos Santos, RRL, KReiser Dec 9 at 1:56
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Jyrki Lahtonen, amWhy, José Carlos Santos, RRL, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
Here's a MathJax tutorial :)
– Shaun
Dec 8 at 18:02
3
One does not "solve" limits; rather, one finds or evaluates them.
– Shaun
Dec 8 at 18:03
How did you misspell L'hôpital when, clearly, you used thelimits-without-lhopitaltag?
– Shaun
Dec 8 at 18:07
1
@Shaun The OP probably typed in a keyword in the tag andlimits-without-lhopitalcame up as one of the options.
– Cookie
Dec 8 at 18:22
@Cookie, yeah, that might explain it.
– Shaun
Dec 8 at 18:24
add a comment |
I am trying to solve the following
$$lim_{(x,y)to(0,0)}frac{6xy^2}{x^2+y^2}$$
This is basically $0/0$ form, but as I saw other post that L'Hôpital's rule does not work on multiple variables.
Then, how could we proceed to do this? Does the limit exist? If yes, what is it?
calculus limits multivariable-calculus limits-without-lhopital
edited Dec 8 at 22:37
amWhy
191k28224439
asked Dec 8 at 17:55
Adshead
51
I am trying to solve the following
$$lim_{(x,y)to(0,0)}frac{6xy^2}{x^2+y^2}$$
This is basically $0/0$ form, but as I saw other post that L'Hôpital's rule does not work on multiple variables.
Then, how could we proceed to do this? Does the limit exist? If yes, what is it?
calculus limits multivariable-calculus limits-without-lhopital
calculus limits multivariable-calculus limits-without-lhopital
edited Dec 8 at 22:37
amWhy
191k28224439
asked Dec 8 at 17:55
Adshead
51
edited Dec 8 at 22:37
amWhy
191k28224439
asked Dec 8 at 17:55
Adshead
51
edited Dec 8 at 22:37
amWhy
191k28224439
edited Dec 8 at 22:37
amWhy
191k28224439
edited Dec 8 at 22:37
amWhy
191k28224439
191k28224439
asked Dec 8 at 17:55
Adshead
51
asked Dec 8 at 17:55
Adshead
51
asked Dec 8 at 17:55
Adshead
51
51
closed as off-topic by Jyrki Lahtonen, amWhy, José Carlos Santos, RRL, KReiser Dec 9 at 1:56
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Jyrki Lahtonen, amWhy, José Carlos Santos, RRL, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Jyrki Lahtonen, amWhy, José Carlos Santos, RRL, KReiser Dec 9 at 1:56
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Jyrki Lahtonen, amWhy, José Carlos Santos, RRL, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
Here's a MathJax tutorial :)
– Shaun
Dec 8 at 18:02
3
One does not "solve" limits; rather, one finds or evaluates them.
– Shaun
Dec 8 at 18:03
How did you misspell L'hôpital when, clearly, you used thelimits-without-lhopitaltag?
– Shaun
Dec 8 at 18:07
1
@Shaun The OP probably typed in a keyword in the tag andlimits-without-lhopitalcame up as one of the options.
– Cookie
Dec 8 at 18:22
@Cookie, yeah, that might explain it.
– Shaun
Dec 8 at 18:24
add a comment |
Here's a MathJax tutorial :)
– Shaun
Dec 8 at 18:02
3
One does not "solve" limits; rather, one finds or evaluates them.
– Shaun
Dec 8 at 18:03
How did you misspell L'hôpital when, clearly, you used thelimits-without-lhopitaltag?
– Shaun
Dec 8 at 18:07
1
@Shaun The OP probably typed in a keyword in the tag andlimits-without-lhopitalcame up as one of the options.
– Cookie
Dec 8 at 18:22
@Cookie, yeah, that might explain it.
– Shaun
Dec 8 at 18:24
Here's a MathJax tutorial :)
– Shaun
Dec 8 at 18:02
Here's a MathJax tutorial :)
– Shaun
Dec 8 at 18:02
3
3
One does not "solve" limits; rather, one finds or evaluates them.
– Shaun
Dec 8 at 18:03
One does not "solve" limits; rather, one finds or evaluates them.
– Shaun
Dec 8 at 18:03
How did you misspell L'hôpital when, clearly, you used the
limits-without-lhopital tag?– Shaun
Dec 8 at 18:07
How did you misspell L'hôpital when, clearly, you used the
limits-without-lhopital tag?– Shaun
Dec 8 at 18:07
1
1
@Shaun The OP probably typed in a keyword in the tag and
limits-without-lhopital came up as one of the options.– Cookie
Dec 8 at 18:22
@Shaun The OP probably typed in a keyword in the tag and
limits-without-lhopital came up as one of the options.– Cookie
Dec 8 at 18:22
@Cookie, yeah, that might explain it.
– Shaun
Dec 8 at 18:24
@Cookie, yeah, that might explain it.
– Shaun
Dec 8 at 18:24
add a comment |
4 Answers
4
active
oldest
votes
Using polar coordinates $$ begin{cases} x =rho cos(theta)&\ y=rho sin(theta)&end{cases}$$
the equation becomes:
$$lim_{rho to 0} frac{6rho^3 cos(theta)sin(theta)^2}{rho^2}=6rholeft(cos(theta)sin(theta)^2right).$$
Since $cos(theta)sin(theta)^2$ is bounded, the limit exists and is indeed 0.
answered Dec 8 at 18:13
Lord_Banquo
1017
add a comment |
Hint: $,,left | dfrac{6xy^2}{x^2+y^2}right | = |6x|cdot dfrac{y^2}{x^2+y^2}.$
answered Dec 8 at 18:23
zhw.
71.5k43075
add a comment |
Consider the following useful inequality (Cauchy-Schwartz, I believe):
$$x^2+y^2 geq 2 cdot |xy|$$
Using this, we see that:
$$0 leq frac{6|x|y^2}{x^2+y^2} = 6|y| cdot frac{|xy|}{x^2+y^2} leq 6|y| frac{1}{2}$$
The $frac{1}{2}$ is coming from the inequality I have written above. Taking the limit, you see that the right side goes to zero. Hence by squeeze theorem, the limit goes to 0.
answered Dec 8 at 18:26
jonan
244
Or more also AM-GM for the inequality, or $(x-y)^2ge 0$.
– gimusi
Dec 8 at 18:31
add a comment |
As an alternative we have
$$0le left|frac{6xy^2}{x^2+y^2}right|le frac{6|x|y^2}{x^2+y^2}+frac{6|x^3|}{x^2+y^2}=6|x|to 0$$
answered Dec 8 at 18:25
gimusi
1
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
Using polar coordinates $$ begin{cases} x =rho cos(theta)&\ y=rho sin(theta)&end{cases}$$
the equation becomes:
$$lim_{rho to 0} frac{6rho^3 cos(theta)sin(theta)^2}{rho^2}=6rholeft(cos(theta)sin(theta)^2right).$$
Since $cos(theta)sin(theta)^2$ is bounded, the limit exists and is indeed 0.
answered Dec 8 at 18:13
Lord_Banquo
1017
add a comment |
Using polar coordinates $$ begin{cases} x =rho cos(theta)&\ y=rho sin(theta)&end{cases}$$
the equation becomes:
$$lim_{rho to 0} frac{6rho^3 cos(theta)sin(theta)^2}{rho^2}=6rholeft(cos(theta)sin(theta)^2right).$$
Since $cos(theta)sin(theta)^2$ is bounded, the limit exists and is indeed 0.
answered Dec 8 at 18:13
Lord_Banquo
1017
add a comment |
Using polar coordinates $$ begin{cases} x =rho cos(theta)&\ y=rho sin(theta)&end{cases}$$
the equation becomes:
$$lim_{rho to 0} frac{6rho^3 cos(theta)sin(theta)^2}{rho^2}=6rholeft(cos(theta)sin(theta)^2right).$$
Since $cos(theta)sin(theta)^2$ is bounded, the limit exists and is indeed 0.
answered Dec 8 at 18:13
Lord_Banquo
1017
Using polar coordinates $$ begin{cases} x =rho cos(theta)&\ y=rho sin(theta)&end{cases}$$
the equation becomes:
$$lim_{rho to 0} frac{6rho^3 cos(theta)sin(theta)^2}{rho^2}=6rholeft(cos(theta)sin(theta)^2right).$$
Since $cos(theta)sin(theta)^2$ is bounded, the limit exists and is indeed 0.
answered Dec 8 at 18:13
Lord_Banquo
1017
edited Dec 9 at 14:00
answered Dec 8 at 18:13
Lord_Banquo
1017
answered Dec 8 at 18:13
Lord_Banquo
1017
answered Dec 8 at 18:13
Lord_Banquo
1017
1017
add a comment |
add a comment |
Hint: $,,left | dfrac{6xy^2}{x^2+y^2}right | = |6x|cdot dfrac{y^2}{x^2+y^2}.$
answered Dec 8 at 18:23
zhw.
71.5k43075
add a comment |
Hint: $,,left | dfrac{6xy^2}{x^2+y^2}right | = |6x|cdot dfrac{y^2}{x^2+y^2}.$
answered Dec 8 at 18:23
zhw.
71.5k43075
add a comment |
Hint: $,,left | dfrac{6xy^2}{x^2+y^2}right | = |6x|cdot dfrac{y^2}{x^2+y^2}.$
answered Dec 8 at 18:23
zhw.
71.5k43075
Hint: $,,left | dfrac{6xy^2}{x^2+y^2}right | = |6x|cdot dfrac{y^2}{x^2+y^2}.$
answered Dec 8 at 18:23
zhw.
71.5k43075
answered Dec 8 at 18:23
zhw.
71.5k43075
answered Dec 8 at 18:23
zhw.
71.5k43075
answered Dec 8 at 18:23
zhw.
71.5k43075
71.5k43075
add a comment |
add a comment |
Consider the following useful inequality (Cauchy-Schwartz, I believe):
$$x^2+y^2 geq 2 cdot |xy|$$
Using this, we see that:
$$0 leq frac{6|x|y^2}{x^2+y^2} = 6|y| cdot frac{|xy|}{x^2+y^2} leq 6|y| frac{1}{2}$$
The $frac{1}{2}$ is coming from the inequality I have written above. Taking the limit, you see that the right side goes to zero. Hence by squeeze theorem, the limit goes to 0.
answered Dec 8 at 18:26
jonan
244
Or more also AM-GM for the inequality, or $(x-y)^2ge 0$.
– gimusi
Dec 8 at 18:31
add a comment |
Consider the following useful inequality (Cauchy-Schwartz, I believe):
$$x^2+y^2 geq 2 cdot |xy|$$
Using this, we see that:
$$0 leq frac{6|x|y^2}{x^2+y^2} = 6|y| cdot frac{|xy|}{x^2+y^2} leq 6|y| frac{1}{2}$$
The $frac{1}{2}$ is coming from the inequality I have written above. Taking the limit, you see that the right side goes to zero. Hence by squeeze theorem, the limit goes to 0.
answered Dec 8 at 18:26
jonan
244
Or more also AM-GM for the inequality, or $(x-y)^2ge 0$.
– gimusi
Dec 8 at 18:31
add a comment |
Consider the following useful inequality (Cauchy-Schwartz, I believe):
$$x^2+y^2 geq 2 cdot |xy|$$
Using this, we see that:
$$0 leq frac{6|x|y^2}{x^2+y^2} = 6|y| cdot frac{|xy|}{x^2+y^2} leq 6|y| frac{1}{2}$$
The $frac{1}{2}$ is coming from the inequality I have written above. Taking the limit, you see that the right side goes to zero. Hence by squeeze theorem, the limit goes to 0.
answered Dec 8 at 18:26
jonan
244
Consider the following useful inequality (Cauchy-Schwartz, I believe):
$$x^2+y^2 geq 2 cdot |xy|$$
Using this, we see that:
$$0 leq frac{6|x|y^2}{x^2+y^2} = 6|y| cdot frac{|xy|}{x^2+y^2} leq 6|y| frac{1}{2}$$
The $frac{1}{2}$ is coming from the inequality I have written above. Taking the limit, you see that the right side goes to zero. Hence by squeeze theorem, the limit goes to 0.
answered Dec 8 at 18:26
jonan
244
answered Dec 8 at 18:26
jonan
244
answered Dec 8 at 18:26
jonan
244
answered Dec 8 at 18:26
jonan
244
244
Or more also AM-GM for the inequality, or $(x-y)^2ge 0$.
– gimusi
Dec 8 at 18:31
add a comment |
Or more also AM-GM for the inequality, or $(x-y)^2ge 0$.
– gimusi
Dec 8 at 18:31
Or more also AM-GM for the inequality, or $(x-y)^2ge 0$.
– gimusi
Dec 8 at 18:31
Or more also AM-GM for the inequality, or $(x-y)^2ge 0$.
– gimusi
Dec 8 at 18:31
add a comment |
As an alternative we have
$$0le left|frac{6xy^2}{x^2+y^2}right|le frac{6|x|y^2}{x^2+y^2}+frac{6|x^3|}{x^2+y^2}=6|x|to 0$$
answered Dec 8 at 18:25
gimusi
1
add a comment |
As an alternative we have
$$0le left|frac{6xy^2}{x^2+y^2}right|le frac{6|x|y^2}{x^2+y^2}+frac{6|x^3|}{x^2+y^2}=6|x|to 0$$
answered Dec 8 at 18:25
gimusi
1
add a comment |
As an alternative we have
$$0le left|frac{6xy^2}{x^2+y^2}right|le frac{6|x|y^2}{x^2+y^2}+frac{6|x^3|}{x^2+y^2}=6|x|to 0$$
answered Dec 8 at 18:25
gimusi
1
As an alternative we have
$$0le left|frac{6xy^2}{x^2+y^2}right|le frac{6|x|y^2}{x^2+y^2}+frac{6|x^3|}{x^2+y^2}=6|x|to 0$$
answered Dec 8 at 18:25
gimusi
1
answered Dec 8 at 18:25
gimusi
1
answered Dec 8 at 18:25
gimusi
1
answered Dec 8 at 18:25
gimusi
1
1
add a comment |
add a comment |
Here's a MathJax tutorial :)
– Shaun
Dec 8 at 18:02
3
One does not "solve" limits; rather, one finds or evaluates them.
– Shaun
Dec 8 at 18:03
How did you misspell L'hôpital when, clearly, you used the
limits-without-lhopitaltag?– Shaun
Dec 8 at 18:07
1
@Shaun The OP probably typed in a keyword in the tag and
limits-without-lhopitalcame up as one of the options.– Cookie
Dec 8 at 18:22
@Cookie, yeah, that might explain it.
– Shaun
Dec 8 at 18:24