Show that $limlimits_(x,y)to(0,0)fracx^3y-xy^3x^4+2y^4$ does not exist.
Clash Royale CLAN TAG#URR8PPP
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Show that $$lim_(x,y)to(0,0)fracx^3y-xy^3x^4+2y^4$$ does not exist.
I'm not even sure how to approach this. I tried factoring out $xy$ in the numerator to get $xy(x^2 - y^2)$, but I don't think that gets me anywhere with the denominator.
calculus limits multivariable-calculus
edited Feb 13 at 9:49
rtybase
11.3k21533
asked Feb 13 at 4:31
krauser126krauser126
475
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add a comment |
$begingroup$
Show that $$lim_(x,y)to(0,0)fracx^3y-xy^3x^4+2y^4$$ does not exist.
I'm not even sure how to approach this. I tried factoring out $xy$ in the numerator to get $xy(x^2 - y^2)$, but I don't think that gets me anywhere with the denominator.
calculus limits multivariable-calculus
edited Feb 13 at 9:49
rtybase
11.3k21533
asked Feb 13 at 4:31
krauser126krauser126
475
$endgroup$
add a comment |
$begingroup$
Show that $$lim_(x,y)to(0,0)fracx^3y-xy^3x^4+2y^4$$ does not exist.
I'm not even sure how to approach this. I tried factoring out $xy$ in the numerator to get $xy(x^2 - y^2)$, but I don't think that gets me anywhere with the denominator.
calculus limits multivariable-calculus
edited Feb 13 at 9:49
rtybase
11.3k21533
asked Feb 13 at 4:31
krauser126krauser126
475
$endgroup$
Show that $$lim_(x,y)to(0,0)fracx^3y-xy^3x^4+2y^4$$ does not exist.
I'm not even sure how to approach this. I tried factoring out $xy$ in the numerator to get $xy(x^2 - y^2)$, but I don't think that gets me anywhere with the denominator.
calculus limits multivariable-calculus
calculus limits multivariable-calculus
edited Feb 13 at 9:49
rtybase
11.3k21533
asked Feb 13 at 4:31
krauser126krauser126
475
edited Feb 13 at 9:49
rtybase
11.3k21533
asked Feb 13 at 4:31
krauser126krauser126
475
edited Feb 13 at 9:49
rtybase
11.3k21533
edited Feb 13 at 9:49
rtybase
11.3k21533
edited Feb 13 at 9:49
rtybase
11.3k21533
11.3k21533
asked Feb 13 at 4:31
krauser126krauser126
475
asked Feb 13 at 4:31
krauser126krauser126
475
asked Feb 13 at 4:31
krauser126krauser126
475
475
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
HINT:
What happens if the limit is taken along $y=2x$? What happens when the limit is taken along $y=0$? Are these equal? If not, what can one conclude?
answered Feb 13 at 4:35
Mark ViolaMark Viola
133k1277176
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add a comment |
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Let's approach the limit along the line $y=mx.$
$beginalign
&lim_(x,y)to (0,0)dfracx^3y-xy^3x^4+2y^4\
&=lim_xto 0dfracx^3mx-x(mx)^3x^4+2(mx)^4\
&=lim_xto 0dfracx^4m-m^3x^4x^4+2m^4x^4\
&=lim_xto 0dfracm-m^31+2m^4\
&=dfracm-m^31+2m^4\
endalign$
So what can you conclude about the limit ?
edited Feb 13 at 5:11
Javi maxwell
878
answered Feb 13 at 4:40
Thomas ShelbyThomas Shelby
3,7342625
$endgroup$
add a comment |
$begingroup$
Putting $x=y$ your expression vanishes, and for$x=2y$, the limit will be $1/3$. Therefore the limit does not exist
answered Feb 13 at 4:35
HAMIDINE SOUMAREHAMIDINE SOUMARE
93329
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add a comment |
$begingroup$
Doing the change to polar coordinates:
$$
fracx^3y - xy^3x^4 + 2y^4 =
frac
r^3cos^3theta,rsintheta - rcostheta,r^3sin^3theta
r^4cos^4theta + 2r^4sin^4theta
= frac
cos^3thetasintheta - costhetasin^3theta
cos^4theta + 2sin^4theta,
$$
dependent of $theta$ (and independent of $r$), so the limit does not exists.
answered Feb 13 at 9:58
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.5k42971
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
HINT:
What happens if the limit is taken along $y=2x$? What happens when the limit is taken along $y=0$? Are these equal? If not, what can one conclude?
answered Feb 13 at 4:35
Mark ViolaMark Viola
133k1277176
$endgroup$
add a comment |
$begingroup$
HINT:
What happens if the limit is taken along $y=2x$? What happens when the limit is taken along $y=0$? Are these equal? If not, what can one conclude?
answered Feb 13 at 4:35
Mark ViolaMark Viola
133k1277176
$endgroup$
add a comment |
$begingroup$
HINT:
What happens if the limit is taken along $y=2x$? What happens when the limit is taken along $y=0$? Are these equal? If not, what can one conclude?
answered Feb 13 at 4:35
Mark ViolaMark Viola
133k1277176
$endgroup$
HINT:
What happens if the limit is taken along $y=2x$? What happens when the limit is taken along $y=0$? Are these equal? If not, what can one conclude?
answered Feb 13 at 4:35
Mark ViolaMark Viola
133k1277176
answered Feb 13 at 4:35
Mark ViolaMark Viola
133k1277176
answered Feb 13 at 4:35
Mark ViolaMark Viola
133k1277176
answered Feb 13 at 4:35
Mark ViolaMark Viola
133k1277176
133k1277176
add a comment |
add a comment |
$begingroup$
Let's approach the limit along the line $y=mx.$
$beginalign
&lim_(x,y)to (0,0)dfracx^3y-xy^3x^4+2y^4\
&=lim_xto 0dfracx^3mx-x(mx)^3x^4+2(mx)^4\
&=lim_xto 0dfracx^4m-m^3x^4x^4+2m^4x^4\
&=lim_xto 0dfracm-m^31+2m^4\
&=dfracm-m^31+2m^4\
endalign$
So what can you conclude about the limit ?
edited Feb 13 at 5:11
Javi maxwell
878
answered Feb 13 at 4:40
Thomas ShelbyThomas Shelby
3,7342625
$endgroup$
add a comment |
$begingroup$
Let's approach the limit along the line $y=mx.$
$beginalign
&lim_(x,y)to (0,0)dfracx^3y-xy^3x^4+2y^4\
&=lim_xto 0dfracx^3mx-x(mx)^3x^4+2(mx)^4\
&=lim_xto 0dfracx^4m-m^3x^4x^4+2m^4x^4\
&=lim_xto 0dfracm-m^31+2m^4\
&=dfracm-m^31+2m^4\
endalign$
So what can you conclude about the limit ?
edited Feb 13 at 5:11
Javi maxwell
878
answered Feb 13 at 4:40
Thomas ShelbyThomas Shelby
3,7342625
$endgroup$
add a comment |
$begingroup$
Let's approach the limit along the line $y=mx.$
$beginalign
&lim_(x,y)to (0,0)dfracx^3y-xy^3x^4+2y^4\
&=lim_xto 0dfracx^3mx-x(mx)^3x^4+2(mx)^4\
&=lim_xto 0dfracx^4m-m^3x^4x^4+2m^4x^4\
&=lim_xto 0dfracm-m^31+2m^4\
&=dfracm-m^31+2m^4\
endalign$
So what can you conclude about the limit ?
edited Feb 13 at 5:11
Javi maxwell
878
answered Feb 13 at 4:40
Thomas ShelbyThomas Shelby
3,7342625
$endgroup$
Let's approach the limit along the line $y=mx.$
$beginalign
&lim_(x,y)to (0,0)dfracx^3y-xy^3x^4+2y^4\
&=lim_xto 0dfracx^3mx-x(mx)^3x^4+2(mx)^4\
&=lim_xto 0dfracx^4m-m^3x^4x^4+2m^4x^4\
&=lim_xto 0dfracm-m^31+2m^4\
&=dfracm-m^31+2m^4\
endalign$
So what can you conclude about the limit ?
edited Feb 13 at 5:11
Javi maxwell
878
answered Feb 13 at 4:40
Thomas ShelbyThomas Shelby
3,7342625
edited Feb 13 at 5:11
Javi maxwell
878
edited Feb 13 at 5:11
Javi maxwell
878
edited Feb 13 at 5:11
Javi maxwell
878
878
answered Feb 13 at 4:40
Thomas ShelbyThomas Shelby
3,7342625
answered Feb 13 at 4:40
Thomas ShelbyThomas Shelby
3,7342625
answered Feb 13 at 4:40
Thomas ShelbyThomas Shelby
3,7342625
3,7342625
add a comment |
add a comment |
$begingroup$
Putting $x=y$ your expression vanishes, and for$x=2y$, the limit will be $1/3$. Therefore the limit does not exist
answered Feb 13 at 4:35
HAMIDINE SOUMAREHAMIDINE SOUMARE
93329
$endgroup$
add a comment |
$begingroup$
Putting $x=y$ your expression vanishes, and for$x=2y$, the limit will be $1/3$. Therefore the limit does not exist
answered Feb 13 at 4:35
HAMIDINE SOUMAREHAMIDINE SOUMARE
93329
$endgroup$
add a comment |
$begingroup$
Putting $x=y$ your expression vanishes, and for$x=2y$, the limit will be $1/3$. Therefore the limit does not exist
answered Feb 13 at 4:35
HAMIDINE SOUMAREHAMIDINE SOUMARE
93329
$endgroup$
Putting $x=y$ your expression vanishes, and for$x=2y$, the limit will be $1/3$. Therefore the limit does not exist
answered Feb 13 at 4:35
HAMIDINE SOUMAREHAMIDINE SOUMARE
93329
answered Feb 13 at 4:35
HAMIDINE SOUMAREHAMIDINE SOUMARE
93329
answered Feb 13 at 4:35
HAMIDINE SOUMAREHAMIDINE SOUMARE
93329
answered Feb 13 at 4:35
HAMIDINE SOUMAREHAMIDINE SOUMARE
93329
93329
add a comment |
add a comment |
$begingroup$
Doing the change to polar coordinates:
$$
fracx^3y - xy^3x^4 + 2y^4 =
frac
r^3cos^3theta,rsintheta - rcostheta,r^3sin^3theta
r^4cos^4theta + 2r^4sin^4theta
= frac
cos^3thetasintheta - costhetasin^3theta
cos^4theta + 2sin^4theta,
$$
dependent of $theta$ (and independent of $r$), so the limit does not exists.
answered Feb 13 at 9:58
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.5k42971
$endgroup$
add a comment |
$begingroup$
Doing the change to polar coordinates:
$$
fracx^3y - xy^3x^4 + 2y^4 =
frac
r^3cos^3theta,rsintheta - rcostheta,r^3sin^3theta
r^4cos^4theta + 2r^4sin^4theta
= frac
cos^3thetasintheta - costhetasin^3theta
cos^4theta + 2sin^4theta,
$$
dependent of $theta$ (and independent of $r$), so the limit does not exists.
answered Feb 13 at 9:58
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.5k42971
$endgroup$
add a comment |
$begingroup$
Doing the change to polar coordinates:
$$
fracx^3y - xy^3x^4 + 2y^4 =
frac
r^3cos^3theta,rsintheta - rcostheta,r^3sin^3theta
r^4cos^4theta + 2r^4sin^4theta
= frac
cos^3thetasintheta - costhetasin^3theta
cos^4theta + 2sin^4theta,
$$
dependent of $theta$ (and independent of $r$), so the limit does not exists.
answered Feb 13 at 9:58
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.5k42971
$endgroup$
Doing the change to polar coordinates:
$$
fracx^3y - xy^3x^4 + 2y^4 =
frac
r^3cos^3theta,rsintheta - rcostheta,r^3sin^3theta
r^4cos^4theta + 2r^4sin^4theta
= frac
cos^3thetasintheta - costhetasin^3theta
cos^4theta + 2sin^4theta,
$$
dependent of $theta$ (and independent of $r$), so the limit does not exists.
answered Feb 13 at 9:58
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.5k42971
answered Feb 13 at 9:58
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.5k42971
answered Feb 13 at 9:58
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.5k42971
answered Feb 13 at 9:58
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.5k42971
34.5k42971
add a comment |
add a comment |
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