Prove $lim_(x,y) to (0,0) fracexp(xy)cdot xycdot(x^2-y^2)x^2+y^2 =0$.
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
I want to show that
$$lim_(x,y) to (0,0) fracexp(xy)cdot xycdot(x^2-y^2)x^2+y^2 =0.$$
Is it valid to do it like this:
$$lim_(x,y) to (0,0) left|fracexp(xy)cdot xycdot(x^2-y^2)x^2+y^2right| leq lim_(x,y) to (0,0) |exp(xy)cdot xycdot (x^2-y^2)|=0$$
calculus limits limits-without-lhopital
edited Aug 8 at 20:02
Nosrati
20.2k41644
asked Aug 8 at 18:21
Ferdi K
1
add a comment |Â
up vote
0
down vote
favorite
I want to show that
$$lim_(x,y) to (0,0) fracexp(xy)cdot xycdot(x^2-y^2)x^2+y^2 =0.$$
Is it valid to do it like this:
$$lim_(x,y) to (0,0) left|fracexp(xy)cdot xycdot(x^2-y^2)x^2+y^2right| leq lim_(x,y) to (0,0) |exp(xy)cdot xycdot (x^2-y^2)|=0$$
calculus limits limits-without-lhopital
edited Aug 8 at 20:02
Nosrati
20.2k41644
asked Aug 8 at 18:21
Ferdi K
1
3
How do you conclude that $left|frac1x^2+y^2right|leq1$? Since $(x,y)to(0,0)$, it blows up to $infty$ which causes your inequality to fail.
– Clayton
Aug 8 at 18:24
oh, yes! thanks a lot!
– Ferdi K
Aug 8 at 18:28
Regarding the title, the exponential factor is actually pretty much irrelevant here since it just goes to one at the origin. As a hint to getting on the right track with this problem, it's often helpful to express things in polar coordinates
– spaceisdarkgreen
Aug 8 at 18:32
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I want to show that
$$lim_(x,y) to (0,0) fracexp(xy)cdot xycdot(x^2-y^2)x^2+y^2 =0.$$
Is it valid to do it like this:
$$lim_(x,y) to (0,0) left|fracexp(xy)cdot xycdot(x^2-y^2)x^2+y^2right| leq lim_(x,y) to (0,0) |exp(xy)cdot xycdot (x^2-y^2)|=0$$
calculus limits limits-without-lhopital
edited Aug 8 at 20:02
Nosrati
20.2k41644
asked Aug 8 at 18:21
Ferdi K
1
I want to show that
$$lim_(x,y) to (0,0) fracexp(xy)cdot xycdot(x^2-y^2)x^2+y^2 =0.$$
Is it valid to do it like this:
$$lim_(x,y) to (0,0) left|fracexp(xy)cdot xycdot(x^2-y^2)x^2+y^2right| leq lim_(x,y) to (0,0) |exp(xy)cdot xycdot (x^2-y^2)|=0$$
calculus limits limits-without-lhopital
edited Aug 8 at 20:02
Nosrati
20.2k41644
asked Aug 8 at 18:21
Ferdi K
1
edited Aug 8 at 20:02
Nosrati
20.2k41644
edited Aug 8 at 20:02
Nosrati
20.2k41644
edited Aug 8 at 20:02
Nosrati
20.2k41644
20.2k41644
asked Aug 8 at 18:21
Ferdi K
1
asked Aug 8 at 18:21
Ferdi K
1
asked Aug 8 at 18:21
Ferdi K
1
1
3
How do you conclude that $left|frac1x^2+y^2right|leq1$? Since $(x,y)to(0,0)$, it blows up to $infty$ which causes your inequality to fail.
– Clayton
Aug 8 at 18:24
oh, yes! thanks a lot!
– Ferdi K
Aug 8 at 18:28
Regarding the title, the exponential factor is actually pretty much irrelevant here since it just goes to one at the origin. As a hint to getting on the right track with this problem, it's often helpful to express things in polar coordinates
– spaceisdarkgreen
Aug 8 at 18:32
add a comment |Â
3
How do you conclude that $left|frac1x^2+y^2right|leq1$? Since $(x,y)to(0,0)$, it blows up to $infty$ which causes your inequality to fail.
– Clayton
Aug 8 at 18:24
oh, yes! thanks a lot!
– Ferdi K
Aug 8 at 18:28
Regarding the title, the exponential factor is actually pretty much irrelevant here since it just goes to one at the origin. As a hint to getting on the right track with this problem, it's often helpful to express things in polar coordinates
– spaceisdarkgreen
Aug 8 at 18:32
3
3
How do you conclude that $left|frac1x^2+y^2right|leq1$? Since $(x,y)to(0,0)$, it blows up to $infty$ which causes your inequality to fail.
– Clayton
Aug 8 at 18:24
How do you conclude that $left|frac1x^2+y^2right|leq1$? Since $(x,y)to(0,0)$, it blows up to $infty$ which causes your inequality to fail.
– Clayton
Aug 8 at 18:24
oh, yes! thanks a lot!
– Ferdi K
Aug 8 at 18:28
oh, yes! thanks a lot!
– Ferdi K
Aug 8 at 18:28
Regarding the title, the exponential factor is actually pretty much irrelevant here since it just goes to one at the origin. As a hint to getting on the right track with this problem, it's often helpful to express things in polar coordinates
– spaceisdarkgreen
Aug 8 at 18:32
Regarding the title, the exponential factor is actually pretty much irrelevant here since it just goes to one at the origin. As a hint to getting on the right track with this problem, it's often helpful to express things in polar coordinates
– spaceisdarkgreen
Aug 8 at 18:32
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
We have that $e^xyto 1$ and by polar coordinates
$$left|fracxycdot(x^2-y^2)x^2+y^2right|=r^2|(cos thetasin theta)(cos^2theta-sin^2theta)|le r^2to 0$$
answered Aug 8 at 18:34
gimusi
65.8k73684
Why go to polar coordinates? Seems unnecessary.
– zhw.
Aug 8 at 19:58
@zhw. I don't claim it is necessary but I think that in this way it is more clear to see. But of course it is only a personal preference.
– gimusi
Aug 8 at 20:00
add a comment |Â
up vote
0
down vote
We may restrict the neibourhood to $x^2+y^2<1$, then $e^xy<2$ and
$$left|dfrace^xyxy(x^2-y^2)x^2+y^2right|leqslantdfracyx^2+y^2leqslant x^2+y^2$$
answered Aug 8 at 19:51
Nosrati
20.2k41644
I would have thought the last term would be $2|xy|$ as it's simpler to think about.
– zhw.
Aug 8 at 19:58
@zhw. $|x^2+y^2|<delta^2$, here $delta=min1,sqrtvarepsilon$.
– Nosrati
Aug 8 at 20:21
What I'm saying is why not just cancel the $x^2+y^2$ that is both upstairs and downstairs?
– zhw.
Aug 8 at 20:24
add a comment |Â
up vote
0
down vote
No, that's not valid. We have a/b 0 and b>1, but in this case we have the denominator going to less than 1, so that doesn't work. The $x^2+y^2$ suggests polar coordinates; that gives $$frace^r^2sin(theta)cos(theta)r^2sin(theta)cos(theta)r^2(cos^2(theta)-sin^2(theta))r^2=$$$$e^r^2sin(theta)cos(theta)r^2sin(theta)cos(theta)(cos^2(theta)-sin^2(theta))=$$$$frace^r^2sin(theta)cos(theta)r^2sin(2theta)cos(2theta)2=$$$$frace^r^2sin(2theta)/2r^2sin(4theta)4 $$
edited Aug 8 at 20:06
zhw.
66.1k42870
answered Aug 8 at 18:38
Acccumulation
4,5382314
I've corrected the Mathjax in the first line. Try it on the other lines.
– zhw.
Aug 8 at 20:02
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
We have that $e^xyto 1$ and by polar coordinates
$$left|fracxycdot(x^2-y^2)x^2+y^2right|=r^2|(cos thetasin theta)(cos^2theta-sin^2theta)|le r^2to 0$$
answered Aug 8 at 18:34
gimusi
65.8k73684
Why go to polar coordinates? Seems unnecessary.
– zhw.
Aug 8 at 19:58
@zhw. I don't claim it is necessary but I think that in this way it is more clear to see. But of course it is only a personal preference.
– gimusi
Aug 8 at 20:00
add a comment |Â
up vote
2
down vote
We have that $e^xyto 1$ and by polar coordinates
$$left|fracxycdot(x^2-y^2)x^2+y^2right|=r^2|(cos thetasin theta)(cos^2theta-sin^2theta)|le r^2to 0$$
answered Aug 8 at 18:34
gimusi
65.8k73684
Why go to polar coordinates? Seems unnecessary.
– zhw.
Aug 8 at 19:58
@zhw. I don't claim it is necessary but I think that in this way it is more clear to see. But of course it is only a personal preference.
– gimusi
Aug 8 at 20:00
add a comment |Â
up vote
2
down vote
up vote
2
down vote
We have that $e^xyto 1$ and by polar coordinates
$$left|fracxycdot(x^2-y^2)x^2+y^2right|=r^2|(cos thetasin theta)(cos^2theta-sin^2theta)|le r^2to 0$$
answered Aug 8 at 18:34
gimusi
65.8k73684
We have that $e^xyto 1$ and by polar coordinates
$$left|fracxycdot(x^2-y^2)x^2+y^2right|=r^2|(cos thetasin theta)(cos^2theta-sin^2theta)|le r^2to 0$$
answered Aug 8 at 18:34
gimusi
65.8k73684
answered Aug 8 at 18:34
gimusi
65.8k73684
answered Aug 8 at 18:34
gimusi
65.8k73684
answered Aug 8 at 18:34
gimusi
65.8k73684
65.8k73684
Why go to polar coordinates? Seems unnecessary.
– zhw.
Aug 8 at 19:58
@zhw. I don't claim it is necessary but I think that in this way it is more clear to see. But of course it is only a personal preference.
– gimusi
Aug 8 at 20:00
add a comment |Â
Why go to polar coordinates? Seems unnecessary.
– zhw.
Aug 8 at 19:58
@zhw. I don't claim it is necessary but I think that in this way it is more clear to see. But of course it is only a personal preference.
– gimusi
Aug 8 at 20:00
Why go to polar coordinates? Seems unnecessary.
– zhw.
Aug 8 at 19:58
Why go to polar coordinates? Seems unnecessary.
– zhw.
Aug 8 at 19:58
@zhw. I don't claim it is necessary but I think that in this way it is more clear to see. But of course it is only a personal preference.
– gimusi
Aug 8 at 20:00
@zhw. I don't claim it is necessary but I think that in this way it is more clear to see. But of course it is only a personal preference.
– gimusi
Aug 8 at 20:00
add a comment |Â
up vote
0
down vote
We may restrict the neibourhood to $x^2+y^2<1$, then $e^xy<2$ and
$$left|dfrace^xyxy(x^2-y^2)x^2+y^2right|leqslantdfracyx^2+y^2leqslant x^2+y^2$$
answered Aug 8 at 19:51
Nosrati
20.2k41644
I would have thought the last term would be $2|xy|$ as it's simpler to think about.
– zhw.
Aug 8 at 19:58
@zhw. $|x^2+y^2|<delta^2$, here $delta=min1,sqrtvarepsilon$.
– Nosrati
Aug 8 at 20:21
What I'm saying is why not just cancel the $x^2+y^2$ that is both upstairs and downstairs?
– zhw.
Aug 8 at 20:24
add a comment |Â
up vote
0
down vote
We may restrict the neibourhood to $x^2+y^2<1$, then $e^xy<2$ and
$$left|dfrace^xyxy(x^2-y^2)x^2+y^2right|leqslantdfracyx^2+y^2leqslant x^2+y^2$$
answered Aug 8 at 19:51
Nosrati
20.2k41644
I would have thought the last term would be $2|xy|$ as it's simpler to think about.
– zhw.
Aug 8 at 19:58
@zhw. $|x^2+y^2|<delta^2$, here $delta=min1,sqrtvarepsilon$.
– Nosrati
Aug 8 at 20:21
What I'm saying is why not just cancel the $x^2+y^2$ that is both upstairs and downstairs?
– zhw.
Aug 8 at 20:24
add a comment |Â
up vote
0
down vote
up vote
0
down vote
We may restrict the neibourhood to $x^2+y^2<1$, then $e^xy<2$ and
$$left|dfrace^xyxy(x^2-y^2)x^2+y^2right|leqslantdfracyx^2+y^2leqslant x^2+y^2$$
answered Aug 8 at 19:51
Nosrati
20.2k41644
We may restrict the neibourhood to $x^2+y^2<1$, then $e^xy<2$ and
$$left|dfrace^xyxy(x^2-y^2)x^2+y^2right|leqslantdfracyx^2+y^2leqslant x^2+y^2$$
answered Aug 8 at 19:51
Nosrati
20.2k41644
answered Aug 8 at 19:51
Nosrati
20.2k41644
answered Aug 8 at 19:51
Nosrati
20.2k41644
answered Aug 8 at 19:51
Nosrati
20.2k41644
20.2k41644
I would have thought the last term would be $2|xy|$ as it's simpler to think about.
– zhw.
Aug 8 at 19:58
@zhw. $|x^2+y^2|<delta^2$, here $delta=min1,sqrtvarepsilon$.
– Nosrati
Aug 8 at 20:21
What I'm saying is why not just cancel the $x^2+y^2$ that is both upstairs and downstairs?
– zhw.
Aug 8 at 20:24
add a comment |Â
I would have thought the last term would be $2|xy|$ as it's simpler to think about.
– zhw.
Aug 8 at 19:58
@zhw. $|x^2+y^2|<delta^2$, here $delta=min1,sqrtvarepsilon$.
– Nosrati
Aug 8 at 20:21
What I'm saying is why not just cancel the $x^2+y^2$ that is both upstairs and downstairs?
– zhw.
Aug 8 at 20:24
I would have thought the last term would be $2|xy|$ as it's simpler to think about.
– zhw.
Aug 8 at 19:58
I would have thought the last term would be $2|xy|$ as it's simpler to think about.
– zhw.
Aug 8 at 19:58
@zhw. $|x^2+y^2|<delta^2$, here $delta=min1,sqrtvarepsilon$.
– Nosrati
Aug 8 at 20:21
@zhw. $|x^2+y^2|<delta^2$, here $delta=min1,sqrtvarepsilon$.
– Nosrati
Aug 8 at 20:21
What I'm saying is why not just cancel the $x^2+y^2$ that is both upstairs and downstairs?
– zhw.
Aug 8 at 20:24
What I'm saying is why not just cancel the $x^2+y^2$ that is both upstairs and downstairs?
– zhw.
Aug 8 at 20:24
add a comment |Â
up vote
0
down vote
No, that's not valid. We have a/b 0 and b>1, but in this case we have the denominator going to less than 1, so that doesn't work. The $x^2+y^2$ suggests polar coordinates; that gives $$frace^r^2sin(theta)cos(theta)r^2sin(theta)cos(theta)r^2(cos^2(theta)-sin^2(theta))r^2=$$$$e^r^2sin(theta)cos(theta)r^2sin(theta)cos(theta)(cos^2(theta)-sin^2(theta))=$$$$frace^r^2sin(theta)cos(theta)r^2sin(2theta)cos(2theta)2=$$$$frace^r^2sin(2theta)/2r^2sin(4theta)4 $$
edited Aug 8 at 20:06
zhw.
66.1k42870
answered Aug 8 at 18:38
Acccumulation
4,5382314
I've corrected the Mathjax in the first line. Try it on the other lines.
– zhw.
Aug 8 at 20:02
add a comment |Â
up vote
0
down vote
No, that's not valid. We have a/b 0 and b>1, but in this case we have the denominator going to less than 1, so that doesn't work. The $x^2+y^2$ suggests polar coordinates; that gives $$frace^r^2sin(theta)cos(theta)r^2sin(theta)cos(theta)r^2(cos^2(theta)-sin^2(theta))r^2=$$$$e^r^2sin(theta)cos(theta)r^2sin(theta)cos(theta)(cos^2(theta)-sin^2(theta))=$$$$frace^r^2sin(theta)cos(theta)r^2sin(2theta)cos(2theta)2=$$$$frace^r^2sin(2theta)/2r^2sin(4theta)4 $$
edited Aug 8 at 20:06
zhw.
66.1k42870
answered Aug 8 at 18:38
Acccumulation
4,5382314
I've corrected the Mathjax in the first line. Try it on the other lines.
– zhw.
Aug 8 at 20:02
add a comment |Â
up vote
0
down vote
up vote
0
down vote
No, that's not valid. We have a/b 0 and b>1, but in this case we have the denominator going to less than 1, so that doesn't work. The $x^2+y^2$ suggests polar coordinates; that gives $$frace^r^2sin(theta)cos(theta)r^2sin(theta)cos(theta)r^2(cos^2(theta)-sin^2(theta))r^2=$$$$e^r^2sin(theta)cos(theta)r^2sin(theta)cos(theta)(cos^2(theta)-sin^2(theta))=$$$$frace^r^2sin(theta)cos(theta)r^2sin(2theta)cos(2theta)2=$$$$frace^r^2sin(2theta)/2r^2sin(4theta)4 $$
edited Aug 8 at 20:06
zhw.
66.1k42870
answered Aug 8 at 18:38
Acccumulation
4,5382314
No, that's not valid. We have a/b 0 and b>1, but in this case we have the denominator going to less than 1, so that doesn't work. The $x^2+y^2$ suggests polar coordinates; that gives $$frace^r^2sin(theta)cos(theta)r^2sin(theta)cos(theta)r^2(cos^2(theta)-sin^2(theta))r^2=$$$$e^r^2sin(theta)cos(theta)r^2sin(theta)cos(theta)(cos^2(theta)-sin^2(theta))=$$$$frace^r^2sin(theta)cos(theta)r^2sin(2theta)cos(2theta)2=$$$$frace^r^2sin(2theta)/2r^2sin(4theta)4 $$
edited Aug 8 at 20:06
zhw.
66.1k42870
answered Aug 8 at 18:38
Acccumulation
4,5382314
edited Aug 8 at 20:06
zhw.
66.1k42870
edited Aug 8 at 20:06
zhw.
66.1k42870
edited Aug 8 at 20:06
zhw.
66.1k42870
66.1k42870
answered Aug 8 at 18:38
Acccumulation
4,5382314
answered Aug 8 at 18:38
Acccumulation
4,5382314
answered Aug 8 at 18:38
Acccumulation
4,5382314
4,5382314
I've corrected the Mathjax in the first line. Try it on the other lines.
– zhw.
Aug 8 at 20:02
add a comment |Â
I've corrected the Mathjax in the first line. Try it on the other lines.
– zhw.
Aug 8 at 20:02
I've corrected the Mathjax in the first line. Try it on the other lines.
– zhw.
Aug 8 at 20:02
I've corrected the Mathjax in the first line. Try it on the other lines.
– zhw.
Aug 8 at 20:02
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2876409%2fprove-lim-x-y-to-0-0-frac-expxy-cdot-xy-cdotx2-y2x2y2-0%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
3
How do you conclude that $left|frac1x^2+y^2right|leq1$? Since $(x,y)to(0,0)$, it blows up to $infty$ which causes your inequality to fail.
– Clayton
Aug 8 at 18:24
oh, yes! thanks a lot!
– Ferdi K
Aug 8 at 18:28
Regarding the title, the exponential factor is actually pretty much irrelevant here since it just goes to one at the origin. As a hint to getting on the right track with this problem, it's often helpful to express things in polar coordinates
– spaceisdarkgreen
Aug 8 at 18:32