Difficulty proving multivariate limit involving sin^2(x) does not exist
I am attempting to prove that
$$lim_{(x,y) to (0,0)} frac{x sin^{2}(y)}{x^{2} + y^{2}}$$
does not exist, but am having issues. I can't seem to find a path for which the value is not zero (to prove that the limit value depends on the given path) , however I do believe that it does not exist, as I can't seem to prove that it does either with the Squeeze theorem.
I have attempted approaching along $y=0$, $x=0$, $y=mx$, $y=mx^{2}$ to name a few, and the usual strategies of attempting to cancel out factors or splitting up my equation don't seem to help as they all just give me zero. Am I on the wrong track here with believing the limit does not exist in the first place?
calculus limits multivariable-calculus
asked Jun 4 '15 at 22:48
HavelTheGreat
15810
add a comment |
I am attempting to prove that
$$lim_{(x,y) to (0,0)} frac{x sin^{2}(y)}{x^{2} + y^{2}}$$
does not exist, but am having issues. I can't seem to find a path for which the value is not zero (to prove that the limit value depends on the given path) , however I do believe that it does not exist, as I can't seem to prove that it does either with the Squeeze theorem.
I have attempted approaching along $y=0$, $x=0$, $y=mx$, $y=mx^{2}$ to name a few, and the usual strategies of attempting to cancel out factors or splitting up my equation don't seem to help as they all just give me zero. Am I on the wrong track here with believing the limit does not exist in the first place?
calculus limits multivariable-calculus
asked Jun 4 '15 at 22:48
HavelTheGreat
15810
add a comment |
I am attempting to prove that
$$lim_{(x,y) to (0,0)} frac{x sin^{2}(y)}{x^{2} + y^{2}}$$
does not exist, but am having issues. I can't seem to find a path for which the value is not zero (to prove that the limit value depends on the given path) , however I do believe that it does not exist, as I can't seem to prove that it does either with the Squeeze theorem.
I have attempted approaching along $y=0$, $x=0$, $y=mx$, $y=mx^{2}$ to name a few, and the usual strategies of attempting to cancel out factors or splitting up my equation don't seem to help as they all just give me zero. Am I on the wrong track here with believing the limit does not exist in the first place?
calculus limits multivariable-calculus
asked Jun 4 '15 at 22:48
HavelTheGreat
15810
I am attempting to prove that
$$lim_{(x,y) to (0,0)} frac{x sin^{2}(y)}{x^{2} + y^{2}}$$
does not exist, but am having issues. I can't seem to find a path for which the value is not zero (to prove that the limit value depends on the given path) , however I do believe that it does not exist, as I can't seem to prove that it does either with the Squeeze theorem.
I have attempted approaching along $y=0$, $x=0$, $y=mx$, $y=mx^{2}$ to name a few, and the usual strategies of attempting to cancel out factors or splitting up my equation don't seem to help as they all just give me zero. Am I on the wrong track here with believing the limit does not exist in the first place?
calculus limits multivariable-calculus
calculus limits multivariable-calculus
asked Jun 4 '15 at 22:48
HavelTheGreat
15810
asked Jun 4 '15 at 22:48
HavelTheGreat
15810
asked Jun 4 '15 at 22:48
HavelTheGreat
15810
asked Jun 4 '15 at 22:48
HavelTheGreat
15810
asked Jun 4 '15 at 22:48
HavelTheGreat
15810
15810
add a comment |
add a comment |
3 Answers
3
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oldest
votes
Hint: What if it were $y^2$ instead of $sin^2 y$ in the numerator?
answered Jun 4 '15 at 22:55
zhw.
71.7k43075
Ahh, right the limit would approach zero. Thanks.
– HavelTheGreat
Jun 4 '15 at 22:58
1
Small suggestion: The limit doesn't approach anything, either it exists or it doesn't, and if it does, it's just a number and doesn't dance around or approach anything.
– zhw.
Jun 4 '15 at 23:05
Right, complete misuse of 'approach' on my part. Thanks again.
– HavelTheGreat
Jun 4 '15 at 23:11
add a comment |
$x^2+y^2 geq 2|x||y| to left|dfrac{xsin^2y}{x^2+y^2}right| leq dfrac{|x|sin^2y}{2|x||y|} leq dfrac{1}{2}|sin y|cdot left|dfrac{sin y}{y}right| to$ limit $= 0$ by squeeze theorem.
answered Jun 4 '15 at 22:55
DeepSea
70.9k54487
add a comment |
How about using polar coordinates?
Define $x^{2}+y^{2}=r$ ,
$x=rcos(beta)$,
$y=rsin(beta)$,
Substitute to get
$$lim_{rto 0}frac{rcos(beta)sin^{2}((rsin(beta))}{r}$$
The outer $r$ cancels out with the denominator and the other $r$ cause the function approaching to $0$.
edited Nov 28 '18 at 7:51
C.F.G
1,4261821
answered Jun 4 '15 at 23:08
Socre
469210
add a comment |
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3 Answers
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active
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3 Answers
3
active
oldest
votes
active
oldest
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active
oldest
votes
Hint: What if it were $y^2$ instead of $sin^2 y$ in the numerator?
answered Jun 4 '15 at 22:55
zhw.
71.7k43075
Ahh, right the limit would approach zero. Thanks.
– HavelTheGreat
Jun 4 '15 at 22:58
1
Small suggestion: The limit doesn't approach anything, either it exists or it doesn't, and if it does, it's just a number and doesn't dance around or approach anything.
– zhw.
Jun 4 '15 at 23:05
Right, complete misuse of 'approach' on my part. Thanks again.
– HavelTheGreat
Jun 4 '15 at 23:11
add a comment |
Hint: What if it were $y^2$ instead of $sin^2 y$ in the numerator?
answered Jun 4 '15 at 22:55
zhw.
71.7k43075
Ahh, right the limit would approach zero. Thanks.
– HavelTheGreat
Jun 4 '15 at 22:58
1
Small suggestion: The limit doesn't approach anything, either it exists or it doesn't, and if it does, it's just a number and doesn't dance around or approach anything.
– zhw.
Jun 4 '15 at 23:05
Right, complete misuse of 'approach' on my part. Thanks again.
– HavelTheGreat
Jun 4 '15 at 23:11
add a comment |
Hint: What if it were $y^2$ instead of $sin^2 y$ in the numerator?
answered Jun 4 '15 at 22:55
zhw.
71.7k43075
Hint: What if it were $y^2$ instead of $sin^2 y$ in the numerator?
answered Jun 4 '15 at 22:55
zhw.
71.7k43075
answered Jun 4 '15 at 22:55
zhw.
71.7k43075
answered Jun 4 '15 at 22:55
zhw.
71.7k43075
answered Jun 4 '15 at 22:55
zhw.
71.7k43075
71.7k43075
Ahh, right the limit would approach zero. Thanks.
– HavelTheGreat
Jun 4 '15 at 22:58
1
Small suggestion: The limit doesn't approach anything, either it exists or it doesn't, and if it does, it's just a number and doesn't dance around or approach anything.
– zhw.
Jun 4 '15 at 23:05
Right, complete misuse of 'approach' on my part. Thanks again.
– HavelTheGreat
Jun 4 '15 at 23:11
add a comment |
Ahh, right the limit would approach zero. Thanks.
– HavelTheGreat
Jun 4 '15 at 22:58
1
Small suggestion: The limit doesn't approach anything, either it exists or it doesn't, and if it does, it's just a number and doesn't dance around or approach anything.
– zhw.
Jun 4 '15 at 23:05
Right, complete misuse of 'approach' on my part. Thanks again.
– HavelTheGreat
Jun 4 '15 at 23:11
Ahh, right the limit would approach zero. Thanks.
– HavelTheGreat
Jun 4 '15 at 22:58
Ahh, right the limit would approach zero. Thanks.
– HavelTheGreat
Jun 4 '15 at 22:58
1
1
Small suggestion: The limit doesn't approach anything, either it exists or it doesn't, and if it does, it's just a number and doesn't dance around or approach anything.
– zhw.
Jun 4 '15 at 23:05
Small suggestion: The limit doesn't approach anything, either it exists or it doesn't, and if it does, it's just a number and doesn't dance around or approach anything.
– zhw.
Jun 4 '15 at 23:05
Right, complete misuse of 'approach' on my part. Thanks again.
– HavelTheGreat
Jun 4 '15 at 23:11
Right, complete misuse of 'approach' on my part. Thanks again.
– HavelTheGreat
Jun 4 '15 at 23:11
add a comment |
$x^2+y^2 geq 2|x||y| to left|dfrac{xsin^2y}{x^2+y^2}right| leq dfrac{|x|sin^2y}{2|x||y|} leq dfrac{1}{2}|sin y|cdot left|dfrac{sin y}{y}right| to$ limit $= 0$ by squeeze theorem.
answered Jun 4 '15 at 22:55
DeepSea
70.9k54487
add a comment |
$x^2+y^2 geq 2|x||y| to left|dfrac{xsin^2y}{x^2+y^2}right| leq dfrac{|x|sin^2y}{2|x||y|} leq dfrac{1}{2}|sin y|cdot left|dfrac{sin y}{y}right| to$ limit $= 0$ by squeeze theorem.
answered Jun 4 '15 at 22:55
DeepSea
70.9k54487
add a comment |
$x^2+y^2 geq 2|x||y| to left|dfrac{xsin^2y}{x^2+y^2}right| leq dfrac{|x|sin^2y}{2|x||y|} leq dfrac{1}{2}|sin y|cdot left|dfrac{sin y}{y}right| to$ limit $= 0$ by squeeze theorem.
answered Jun 4 '15 at 22:55
DeepSea
70.9k54487
$x^2+y^2 geq 2|x||y| to left|dfrac{xsin^2y}{x^2+y^2}right| leq dfrac{|x|sin^2y}{2|x||y|} leq dfrac{1}{2}|sin y|cdot left|dfrac{sin y}{y}right| to$ limit $= 0$ by squeeze theorem.
answered Jun 4 '15 at 22:55
DeepSea
70.9k54487
answered Jun 4 '15 at 22:55
DeepSea
70.9k54487
answered Jun 4 '15 at 22:55
DeepSea
70.9k54487
answered Jun 4 '15 at 22:55
DeepSea
70.9k54487
70.9k54487
add a comment |
add a comment |
How about using polar coordinates?
Define $x^{2}+y^{2}=r$ ,
$x=rcos(beta)$,
$y=rsin(beta)$,
Substitute to get
$$lim_{rto 0}frac{rcos(beta)sin^{2}((rsin(beta))}{r}$$
The outer $r$ cancels out with the denominator and the other $r$ cause the function approaching to $0$.
edited Nov 28 '18 at 7:51
C.F.G
1,4261821
answered Jun 4 '15 at 23:08
Socre
469210
add a comment |
How about using polar coordinates?
Define $x^{2}+y^{2}=r$ ,
$x=rcos(beta)$,
$y=rsin(beta)$,
Substitute to get
$$lim_{rto 0}frac{rcos(beta)sin^{2}((rsin(beta))}{r}$$
The outer $r$ cancels out with the denominator and the other $r$ cause the function approaching to $0$.
edited Nov 28 '18 at 7:51
C.F.G
1,4261821
answered Jun 4 '15 at 23:08
Socre
469210
add a comment |
How about using polar coordinates?
Define $x^{2}+y^{2}=r$ ,
$x=rcos(beta)$,
$y=rsin(beta)$,
Substitute to get
$$lim_{rto 0}frac{rcos(beta)sin^{2}((rsin(beta))}{r}$$
The outer $r$ cancels out with the denominator and the other $r$ cause the function approaching to $0$.
edited Nov 28 '18 at 7:51
C.F.G
1,4261821
answered Jun 4 '15 at 23:08
Socre
469210
How about using polar coordinates?
Define $x^{2}+y^{2}=r$ ,
$x=rcos(beta)$,
$y=rsin(beta)$,
Substitute to get
$$lim_{rto 0}frac{rcos(beta)sin^{2}((rsin(beta))}{r}$$
The outer $r$ cancels out with the denominator and the other $r$ cause the function approaching to $0$.
edited Nov 28 '18 at 7:51
C.F.G
1,4261821
answered Jun 4 '15 at 23:08
Socre
469210
edited Nov 28 '18 at 7:51
C.F.G
1,4261821
edited Nov 28 '18 at 7:51
C.F.G
1,4261821
edited Nov 28 '18 at 7:51
C.F.G
1,4261821
1,4261821
answered Jun 4 '15 at 23:08
Socre
469210
answered Jun 4 '15 at 23:08
Socre
469210
answered Jun 4 '15 at 23:08
Socre
469210
469210
add a comment |
add a comment |
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