About function of several variables
$begingroup$
Proof f(x,y)=$frac{x^2y}{x^4+y^2},$when $x^2+y^2ne 0$
f(x,y)=0,when $x^2+y^2=0$
is Continuity on (0,0) on half-line x=$t costheta,y=tsintheta ,0le tle+propto$
My attempt :
f(0,y)=0,f(x,0)=0, so it’s continuity on x-axis,y-axis,
let k=$tan theta$ ,then f=$frac{kx^3}{x^4+k^2x^2}$
= $frac{kx}{x^2+k^2}$
$lim_{k
real-analysis calculus sequences-and-series limits analysis
$endgroup$
add a comment |
$begingroup$
Proof f(x,y)=$frac{x^2y}{x^4+y^2},$when $x^2+y^2ne 0$
f(x,y)=0,when $x^2+y^2=0$
is Continuity on (0,0) on half-line x=$t costheta,y=tsintheta ,0le tle+propto$
My attempt :
f(0,y)=0,f(x,0)=0, so it’s continuity on x-axis,y-axis,
let k=$tan theta$ ,then f=$frac{kx^3}{x^4+k^2x^2}$
= $frac{kx}{x^2+k^2}$
$lim_{k
real-analysis calculus sequences-and-series limits analysis
$endgroup$
add a comment |
$begingroup$
Proof f(x,y)=$frac{x^2y}{x^4+y^2},$when $x^2+y^2ne 0$
f(x,y)=0,when $x^2+y^2=0$
is Continuity on (0,0) on half-line x=$t costheta,y=tsintheta ,0le tle+propto$
My attempt :
f(0,y)=0,f(x,0)=0, so it’s continuity on x-axis,y-axis,
let k=$tan theta$ ,then f=$frac{kx^3}{x^4+k^2x^2}$
= $frac{kx}{x^2+k^2}$
$lim_{k
real-analysis calculus sequences-and-series limits analysis
$endgroup$
Proof f(x,y)=$frac{x^2y}{x^4+y^2},$when $x^2+y^2ne 0$
f(x,y)=0,when $x^2+y^2=0$
is Continuity on (0,0) on half-line x=$t costheta,y=tsintheta ,0le tle+propto$
My attempt :
f(0,y)=0,f(x,0)=0, so it’s continuity on x-axis,y-axis,
let k=$tan theta$ ,then f=$frac{kx^3}{x^4+k^2x^2}$
= $frac{kx}{x^2+k^2}$
$lim_{k
real-analysis calculus sequences-and-series limits analysis
real-analysis calculus sequences-and-series limits analysis
edited Jan 8 at 11:10
jackson
asked Jan 1 at 15:30
jacksonjackson
1209
1209
add a comment |
add a comment |
1 Answer
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$begingroup$
If $k ne 0$, then $lim_{x to 0^+} frac{kx}{x^2+k^2}=0$ and $lim_{x to 0^-} frac{kx}{x^2+k^2}=0$
Now, let's consider the $x$-axis, that is when $y=0$ and along that line, $f(x,y)=0$ .
Hence, it is continuouson those half-lines.
$endgroup$
$begingroup$
So I almost right?
$endgroup$
– jackson
Jan 1 at 15:39
$begingroup$
is there any reason that stop you from taking the limit?
$endgroup$
– Siong Thye Goh
Jan 1 at 15:40
$begingroup$
Seems like no reason ...thanks
$endgroup$
– jackson
Jan 1 at 15:42
$begingroup$
But can is proper I use k to replace $theta$
$endgroup$
– jackson
Jan 1 at 15:46
$begingroup$
What about y-axis
$endgroup$
– jackson
Jan 1 at 15:47
|
show 4 more comments
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $k ne 0$, then $lim_{x to 0^+} frac{kx}{x^2+k^2}=0$ and $lim_{x to 0^-} frac{kx}{x^2+k^2}=0$
Now, let's consider the $x$-axis, that is when $y=0$ and along that line, $f(x,y)=0$ .
Hence, it is continuouson those half-lines.
$endgroup$
$begingroup$
So I almost right?
$endgroup$
– jackson
Jan 1 at 15:39
$begingroup$
is there any reason that stop you from taking the limit?
$endgroup$
– Siong Thye Goh
Jan 1 at 15:40
$begingroup$
Seems like no reason ...thanks
$endgroup$
– jackson
Jan 1 at 15:42
$begingroup$
But can is proper I use k to replace $theta$
$endgroup$
– jackson
Jan 1 at 15:46
$begingroup$
What about y-axis
$endgroup$
– jackson
Jan 1 at 15:47
|
show 4 more comments
$begingroup$
If $k ne 0$, then $lim_{x to 0^+} frac{kx}{x^2+k^2}=0$ and $lim_{x to 0^-} frac{kx}{x^2+k^2}=0$
Now, let's consider the $x$-axis, that is when $y=0$ and along that line, $f(x,y)=0$ .
Hence, it is continuouson those half-lines.
$endgroup$
$begingroup$
So I almost right?
$endgroup$
– jackson
Jan 1 at 15:39
$begingroup$
is there any reason that stop you from taking the limit?
$endgroup$
– Siong Thye Goh
Jan 1 at 15:40
$begingroup$
Seems like no reason ...thanks
$endgroup$
– jackson
Jan 1 at 15:42
$begingroup$
But can is proper I use k to replace $theta$
$endgroup$
– jackson
Jan 1 at 15:46
$begingroup$
What about y-axis
$endgroup$
– jackson
Jan 1 at 15:47
|
show 4 more comments
$begingroup$
If $k ne 0$, then $lim_{x to 0^+} frac{kx}{x^2+k^2}=0$ and $lim_{x to 0^-} frac{kx}{x^2+k^2}=0$
Now, let's consider the $x$-axis, that is when $y=0$ and along that line, $f(x,y)=0$ .
Hence, it is continuouson those half-lines.
$endgroup$
If $k ne 0$, then $lim_{x to 0^+} frac{kx}{x^2+k^2}=0$ and $lim_{x to 0^-} frac{kx}{x^2+k^2}=0$
Now, let's consider the $x$-axis, that is when $y=0$ and along that line, $f(x,y)=0$ .
Hence, it is continuouson those half-lines.
edited Jan 1 at 16:01
answered Jan 1 at 15:37
Siong Thye GohSiong Thye Goh
102k1466118
102k1466118
$begingroup$
So I almost right?
$endgroup$
– jackson
Jan 1 at 15:39
$begingroup$
is there any reason that stop you from taking the limit?
$endgroup$
– Siong Thye Goh
Jan 1 at 15:40
$begingroup$
Seems like no reason ...thanks
$endgroup$
– jackson
Jan 1 at 15:42
$begingroup$
But can is proper I use k to replace $theta$
$endgroup$
– jackson
Jan 1 at 15:46
$begingroup$
What about y-axis
$endgroup$
– jackson
Jan 1 at 15:47
|
show 4 more comments
$begingroup$
So I almost right?
$endgroup$
– jackson
Jan 1 at 15:39
$begingroup$
is there any reason that stop you from taking the limit?
$endgroup$
– Siong Thye Goh
Jan 1 at 15:40
$begingroup$
Seems like no reason ...thanks
$endgroup$
– jackson
Jan 1 at 15:42
$begingroup$
But can is proper I use k to replace $theta$
$endgroup$
– jackson
Jan 1 at 15:46
$begingroup$
What about y-axis
$endgroup$
– jackson
Jan 1 at 15:47
$begingroup$
So I almost right?
$endgroup$
– jackson
Jan 1 at 15:39
$begingroup$
So I almost right?
$endgroup$
– jackson
Jan 1 at 15:39
$begingroup$
is there any reason that stop you from taking the limit?
$endgroup$
– Siong Thye Goh
Jan 1 at 15:40
$begingroup$
is there any reason that stop you from taking the limit?
$endgroup$
– Siong Thye Goh
Jan 1 at 15:40
$begingroup$
Seems like no reason ...thanks
$endgroup$
– jackson
Jan 1 at 15:42
$begingroup$
Seems like no reason ...thanks
$endgroup$
– jackson
Jan 1 at 15:42
$begingroup$
But can is proper I use k to replace $theta$
$endgroup$
– jackson
Jan 1 at 15:46
$begingroup$
But can is proper I use k to replace $theta$
$endgroup$
– jackson
Jan 1 at 15:46
$begingroup$
What about y-axis
$endgroup$
– jackson
Jan 1 at 15:47
$begingroup$
What about y-axis
$endgroup$
– jackson
Jan 1 at 15:47
|
show 4 more comments
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