For which values of $a, b$ and $c in mathbb{R}$ does the limit $lim_{(x, y) to (0, 0)} frac{x y}{ax^2 + bxy +...
When going from the $y$ direction, that is $x=0$ and $y to 0$, I get $0$ for the limit and when going from $x$ direction ($y=0$ and $x to 0$) I also get $0$ for the limit. I don't know what to do with these values and how to proceed in order to get values for $a, b$ and $c$.
limits
asked 2 days ago
mrMoonpenguinmrMoonpenguin
132
add a comment |
When going from the $y$ direction, that is $x=0$ and $y to 0$, I get $0$ for the limit and when going from $x$ direction ($y=0$ and $x to 0$) I also get $0$ for the limit. I don't know what to do with these values and how to proceed in order to get values for $a, b$ and $c$.
limits
asked 2 days ago
mrMoonpenguinmrMoonpenguin
132
What happens if you proceed along the line $y=x$? Along the curve $y=x^2$? Along $x=y^2$? Remember that we have to be able to approach $(0,0)$ via any path and get the same limit, not just along the axes.
– Cameron Buie
2 days ago
Does your result hold when $c=0?$
– saulspatz
2 days ago
add a comment |
When going from the $y$ direction, that is $x=0$ and $y to 0$, I get $0$ for the limit and when going from $x$ direction ($y=0$ and $x to 0$) I also get $0$ for the limit. I don't know what to do with these values and how to proceed in order to get values for $a, b$ and $c$.
limits
asked 2 days ago
mrMoonpenguinmrMoonpenguin
132
When going from the $y$ direction, that is $x=0$ and $y to 0$, I get $0$ for the limit and when going from $x$ direction ($y=0$ and $x to 0$) I also get $0$ for the limit. I don't know what to do with these values and how to proceed in order to get values for $a, b$ and $c$.
limits
limits
asked 2 days ago
mrMoonpenguinmrMoonpenguin
132
asked 2 days ago
mrMoonpenguinmrMoonpenguin
132
asked 2 days ago
mrMoonpenguinmrMoonpenguin
132
asked 2 days ago
mrMoonpenguinmrMoonpenguin
132
asked 2 days ago
mrMoonpenguinmrMoonpenguin
132
132
What happens if you proceed along the line $y=x$? Along the curve $y=x^2$? Along $x=y^2$? Remember that we have to be able to approach $(0,0)$ via any path and get the same limit, not just along the axes.
– Cameron Buie
2 days ago
Does your result hold when $c=0?$
– saulspatz
2 days ago
add a comment |
What happens if you proceed along the line $y=x$? Along the curve $y=x^2$? Along $x=y^2$? Remember that we have to be able to approach $(0,0)$ via any path and get the same limit, not just along the axes.
– Cameron Buie
2 days ago
Does your result hold when $c=0?$
– saulspatz
2 days ago
What happens if you proceed along the line $y=x$? Along the curve $y=x^2$? Along $x=y^2$? Remember that we have to be able to approach $(0,0)$ via any path and get the same limit, not just along the axes.
– Cameron Buie
2 days ago
What happens if you proceed along the line $y=x$? Along the curve $y=x^2$? Along $x=y^2$? Remember that we have to be able to approach $(0,0)$ via any path and get the same limit, not just along the axes.
– Cameron Buie
2 days ago
Does your result hold when $c=0?$
– saulspatz
2 days ago
Does your result hold when $c=0?$
– saulspatz
2 days ago
add a comment |
1 Answer
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Let’s note $f(x,y)= frac{x y}{ax^2 + bxy + cy^2}$. Note that we can’t have $(a,b,c)=(0,0,0)$.
If $a=c=0$, then the limit exists and is equal to $1/b$.
For $(a,c) neq (0,0)$, by symmetry we can suppose $aneq 0$.
Take $y = lambda x$
Then
$$ f(x,lambda x)= frac{x lambda}{ax^2 + bxlambda x + c lambda^2 x^2}= frac{lambda}{a + blambda + c lambda^2}$$
For $lambda =0$, this is always equal to $0$. Then pick-up a value $lambda$ large enough in order to have $a + blambda + c lambda^2neq 0$ and $lambda neq 0$. You get another limit.
Finally, the limit exists if and only if $a=c=0$.
answered 2 days ago
mathcounterexamples.netmathcounterexamples.net
25.2k21953
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
Let’s note $f(x,y)= frac{x y}{ax^2 + bxy + cy^2}$. Note that we can’t have $(a,b,c)=(0,0,0)$.
If $a=c=0$, then the limit exists and is equal to $1/b$.
For $(a,c) neq (0,0)$, by symmetry we can suppose $aneq 0$.
Take $y = lambda x$
Then
$$ f(x,lambda x)= frac{x lambda}{ax^2 + bxlambda x + c lambda^2 x^2}= frac{lambda}{a + blambda + c lambda^2}$$
For $lambda =0$, this is always equal to $0$. Then pick-up a value $lambda$ large enough in order to have $a + blambda + c lambda^2neq 0$ and $lambda neq 0$. You get another limit.
Finally, the limit exists if and only if $a=c=0$.
answered 2 days ago
mathcounterexamples.netmathcounterexamples.net
25.2k21953
add a comment |
Let’s note $f(x,y)= frac{x y}{ax^2 + bxy + cy^2}$. Note that we can’t have $(a,b,c)=(0,0,0)$.
If $a=c=0$, then the limit exists and is equal to $1/b$.
For $(a,c) neq (0,0)$, by symmetry we can suppose $aneq 0$.
Take $y = lambda x$
Then
$$ f(x,lambda x)= frac{x lambda}{ax^2 + bxlambda x + c lambda^2 x^2}= frac{lambda}{a + blambda + c lambda^2}$$
For $lambda =0$, this is always equal to $0$. Then pick-up a value $lambda$ large enough in order to have $a + blambda + c lambda^2neq 0$ and $lambda neq 0$. You get another limit.
Finally, the limit exists if and only if $a=c=0$.
answered 2 days ago
mathcounterexamples.netmathcounterexamples.net
25.2k21953
add a comment |
Let’s note $f(x,y)= frac{x y}{ax^2 + bxy + cy^2}$. Note that we can’t have $(a,b,c)=(0,0,0)$.
If $a=c=0$, then the limit exists and is equal to $1/b$.
For $(a,c) neq (0,0)$, by symmetry we can suppose $aneq 0$.
Take $y = lambda x$
Then
$$ f(x,lambda x)= frac{x lambda}{ax^2 + bxlambda x + c lambda^2 x^2}= frac{lambda}{a + blambda + c lambda^2}$$
For $lambda =0$, this is always equal to $0$. Then pick-up a value $lambda$ large enough in order to have $a + blambda + c lambda^2neq 0$ and $lambda neq 0$. You get another limit.
Finally, the limit exists if and only if $a=c=0$.
answered 2 days ago
mathcounterexamples.netmathcounterexamples.net
25.2k21953
Let’s note $f(x,y)= frac{x y}{ax^2 + bxy + cy^2}$. Note that we can’t have $(a,b,c)=(0,0,0)$.
If $a=c=0$, then the limit exists and is equal to $1/b$.
For $(a,c) neq (0,0)$, by symmetry we can suppose $aneq 0$.
Take $y = lambda x$
Then
$$ f(x,lambda x)= frac{x lambda}{ax^2 + bxlambda x + c lambda^2 x^2}= frac{lambda}{a + blambda + c lambda^2}$$
For $lambda =0$, this is always equal to $0$. Then pick-up a value $lambda$ large enough in order to have $a + blambda + c lambda^2neq 0$ and $lambda neq 0$. You get another limit.
Finally, the limit exists if and only if $a=c=0$.
answered 2 days ago
mathcounterexamples.netmathcounterexamples.net
25.2k21953
edited 2 days ago
answered 2 days ago
mathcounterexamples.netmathcounterexamples.net
25.2k21953
answered 2 days ago
mathcounterexamples.netmathcounterexamples.net
25.2k21953
answered 2 days ago
mathcounterexamples.netmathcounterexamples.net
25.2k21953
25.2k21953
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What happens if you proceed along the line $y=x$? Along the curve $y=x^2$? Along $x=y^2$? Remember that we have to be able to approach $(0,0)$ via any path and get the same limit, not just along the axes.
– Cameron Buie
2 days ago
Does your result hold when $c=0?$
– saulspatz
2 days ago