Limit of a real valued function of two variables.
$lim_{(x,y)to(0,0)} frac {x²y}{x⁴+y²} = $?
If we take $y = mx^2$,
it comes out to be $frac {m^2}{1+m^2}$ which depends on $m$, so limit doesn't exist.
How to prove the non existence of limit by using polar form
What I tried –
$lim_{rto 0}frac{ r³cos²(t)sin(t)}{r³cos²(t) + rsin(t)}$
I don't know how to proceed.
I know that limit exists if the result I get is independent of $t$.
calculus
edited Nov 29 '18 at 4:24
Thomas Shelby
1,667216
asked Nov 29 '18 at 3:20
Mathsaddict
2608
add a comment |
$lim_{(x,y)to(0,0)} frac {x²y}{x⁴+y²} = $?
If we take $y = mx^2$,
it comes out to be $frac {m^2}{1+m^2}$ which depends on $m$, so limit doesn't exist.
How to prove the non existence of limit by using polar form
What I tried –
$lim_{rto 0}frac{ r³cos²(t)sin(t)}{r³cos²(t) + rsin(t)}$
I don't know how to proceed.
I know that limit exists if the result I get is independent of $t$.
calculus
edited Nov 29 '18 at 4:24
Thomas Shelby
1,667216
asked Nov 29 '18 at 3:20
Mathsaddict
2608
add a comment |
$lim_{(x,y)to(0,0)} frac {x²y}{x⁴+y²} = $?
If we take $y = mx^2$,
it comes out to be $frac {m^2}{1+m^2}$ which depends on $m$, so limit doesn't exist.
How to prove the non existence of limit by using polar form
What I tried –
$lim_{rto 0}frac{ r³cos²(t)sin(t)}{r³cos²(t) + rsin(t)}$
I don't know how to proceed.
I know that limit exists if the result I get is independent of $t$.
calculus
edited Nov 29 '18 at 4:24
Thomas Shelby
1,667216
asked Nov 29 '18 at 3:20
Mathsaddict
2608
$lim_{(x,y)to(0,0)} frac {x²y}{x⁴+y²} = $?
If we take $y = mx^2$,
it comes out to be $frac {m^2}{1+m^2}$ which depends on $m$, so limit doesn't exist.
How to prove the non existence of limit by using polar form
What I tried –
$lim_{rto 0}frac{ r³cos²(t)sin(t)}{r³cos²(t) + rsin(t)}$
I don't know how to proceed.
I know that limit exists if the result I get is independent of $t$.
calculus
calculus
edited Nov 29 '18 at 4:24
Thomas Shelby
1,667216
asked Nov 29 '18 at 3:20
Mathsaddict
2608
edited Nov 29 '18 at 4:24
Thomas Shelby
1,667216
asked Nov 29 '18 at 3:20
Mathsaddict
2608
edited Nov 29 '18 at 4:24
Thomas Shelby
1,667216
edited Nov 29 '18 at 4:24
Thomas Shelby
1,667216
edited Nov 29 '18 at 4:24
Thomas Shelby
1,667216
1,667216
asked Nov 29 '18 at 3:20
Mathsaddict
2608
asked Nov 29 '18 at 3:20
Mathsaddict
2608
asked Nov 29 '18 at 3:20
Mathsaddict
2608
2608
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Polar form? Don't go polar in terms of $x$ and $y$; substitute $u=x^2$ first. Then it's $lim_{(u,y)to (0,0),uge 0} frac{uy}{u^2+y^2}$. Convert that to polar coordinates (the $uge 0$ restriction becomes an angle restriction $-frac{pi}{2}le thetale frac{pi}{2}$) and work from there.
After all, the point of polar coordinates in this sort of question is to make the denominator nice.
answered Nov 29 '18 at 6:52
jmerry
2,076210
I did as you explained, and I got sin(t)cos(t), which depends on t. But, without converting x^2 to u, why do i get 0 in polar form? Limit should be same.
– Mathsaddict
Nov 29 '18 at 8:07
The limit along any straight-line path in x and y is zero. Converting to polar right away and looking at constant angles looks only at these straight-line paths - but you already noted that following a parabolic path gets you something else. It's just not enough on its own.
– jmerry
Nov 29 '18 at 8:28
add a comment |
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1 Answer
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1 Answer
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oldest
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votes
Polar form? Don't go polar in terms of $x$ and $y$; substitute $u=x^2$ first. Then it's $lim_{(u,y)to (0,0),uge 0} frac{uy}{u^2+y^2}$. Convert that to polar coordinates (the $uge 0$ restriction becomes an angle restriction $-frac{pi}{2}le thetale frac{pi}{2}$) and work from there.
After all, the point of polar coordinates in this sort of question is to make the denominator nice.
answered Nov 29 '18 at 6:52
jmerry
2,076210
I did as you explained, and I got sin(t)cos(t), which depends on t. But, without converting x^2 to u, why do i get 0 in polar form? Limit should be same.
– Mathsaddict
Nov 29 '18 at 8:07
The limit along any straight-line path in x and y is zero. Converting to polar right away and looking at constant angles looks only at these straight-line paths - but you already noted that following a parabolic path gets you something else. It's just not enough on its own.
– jmerry
Nov 29 '18 at 8:28
add a comment |
Polar form? Don't go polar in terms of $x$ and $y$; substitute $u=x^2$ first. Then it's $lim_{(u,y)to (0,0),uge 0} frac{uy}{u^2+y^2}$. Convert that to polar coordinates (the $uge 0$ restriction becomes an angle restriction $-frac{pi}{2}le thetale frac{pi}{2}$) and work from there.
After all, the point of polar coordinates in this sort of question is to make the denominator nice.
answered Nov 29 '18 at 6:52
jmerry
2,076210
I did as you explained, and I got sin(t)cos(t), which depends on t. But, without converting x^2 to u, why do i get 0 in polar form? Limit should be same.
– Mathsaddict
Nov 29 '18 at 8:07
The limit along any straight-line path in x and y is zero. Converting to polar right away and looking at constant angles looks only at these straight-line paths - but you already noted that following a parabolic path gets you something else. It's just not enough on its own.
– jmerry
Nov 29 '18 at 8:28
add a comment |
Polar form? Don't go polar in terms of $x$ and $y$; substitute $u=x^2$ first. Then it's $lim_{(u,y)to (0,0),uge 0} frac{uy}{u^2+y^2}$. Convert that to polar coordinates (the $uge 0$ restriction becomes an angle restriction $-frac{pi}{2}le thetale frac{pi}{2}$) and work from there.
After all, the point of polar coordinates in this sort of question is to make the denominator nice.
answered Nov 29 '18 at 6:52
jmerry
2,076210
Polar form? Don't go polar in terms of $x$ and $y$; substitute $u=x^2$ first. Then it's $lim_{(u,y)to (0,0),uge 0} frac{uy}{u^2+y^2}$. Convert that to polar coordinates (the $uge 0$ restriction becomes an angle restriction $-frac{pi}{2}le thetale frac{pi}{2}$) and work from there.
After all, the point of polar coordinates in this sort of question is to make the denominator nice.
answered Nov 29 '18 at 6:52
jmerry
2,076210
answered Nov 29 '18 at 6:52
jmerry
2,076210
answered Nov 29 '18 at 6:52
jmerry
2,076210
answered Nov 29 '18 at 6:52
jmerry
2,076210
2,076210
I did as you explained, and I got sin(t)cos(t), which depends on t. But, without converting x^2 to u, why do i get 0 in polar form? Limit should be same.
– Mathsaddict
Nov 29 '18 at 8:07
The limit along any straight-line path in x and y is zero. Converting to polar right away and looking at constant angles looks only at these straight-line paths - but you already noted that following a parabolic path gets you something else. It's just not enough on its own.
– jmerry
Nov 29 '18 at 8:28
add a comment |
I did as you explained, and I got sin(t)cos(t), which depends on t. But, without converting x^2 to u, why do i get 0 in polar form? Limit should be same.
– Mathsaddict
Nov 29 '18 at 8:07
The limit along any straight-line path in x and y is zero. Converting to polar right away and looking at constant angles looks only at these straight-line paths - but you already noted that following a parabolic path gets you something else. It's just not enough on its own.
– jmerry
Nov 29 '18 at 8:28
I did as you explained, and I got sin(t)cos(t), which depends on t. But, without converting x^2 to u, why do i get 0 in polar form? Limit should be same.
– Mathsaddict
Nov 29 '18 at 8:07
I did as you explained, and I got sin(t)cos(t), which depends on t. But, without converting x^2 to u, why do i get 0 in polar form? Limit should be same.
– Mathsaddict
Nov 29 '18 at 8:07
The limit along any straight-line path in x and y is zero. Converting to polar right away and looking at constant angles looks only at these straight-line paths - but you already noted that following a parabolic path gets you something else. It's just not enough on its own.
– jmerry
Nov 29 '18 at 8:28
The limit along any straight-line path in x and y is zero. Converting to polar right away and looking at constant angles looks only at these straight-line paths - but you already noted that following a parabolic path gets you something else. It's just not enough on its own.
– jmerry
Nov 29 '18 at 8:28
add a comment |
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