Find the limit of $ lim_{(x,y)rightarrow (0,0)} frac{x-sin x+y}{x^3 + 6y} $Multivariable limit $lim_{(x,y)...
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Find the limit of $ lim_{(x,y)rightarrow (0,0)} frac{x-sin x+y}{x^3 + 6y} $
Multivariable limit $lim_{(x,y) rightarrow (0,0)} frac { x sin(y/sqrt{x}) } {sqrt{ x^2 + y^2 }}$Limit of $lim_{(x,y)rightarrow(0,0)}frac{x^2y^2}{x^3+y^3}$Evaluate the limit $lim_{(x,y) to (0,0)}frac{cos(x) - 1 - frac{x^2}{2}}{x^4 + y^4}$Why doesn't the limit $lim_{(x,y) rightarrow (0,0)} frac{ e^{x+y} - x - y}{sqrt{x^2 + y^2}}$ exist?What is $lim_{(x,y)rightarrow(0,0)} frac{x^3-y^3}{x^2-y^2}$?Limit $lim_{(x,y)to (0,0)}frac{sin(xy)}{xy}$Prove $lim_{(x,y)rightarrow (0,0)}frac{x+sin y}{x+y}=1$ by definitionHow to demonstrate the limit $ lim_{(x,y) rightarrow (0,0)} frac {xsin(xy)}{x^2+y^2} $ with $epsilon$ and $delta$!what is limit of $lim_{(x,y)rightarrow(0,0)}frac{sin(2x+2y)-2x-2y}{rootof {{x^2+y^2}}}$$ lim_{(x,y)rightarrow (0,0)} frac{x^{5}y^{3}}{x^{6}+y^{4}}. $ Does it exist or not?
$begingroup$
I have this limit: $$ lim_{(x,y)rightarrow (0,0)} frac{x-sin x+y}{x^3 + 6y}. $$
Taking some different paths I always get the same anwser, $frac{1}{6} $. So I guess that's the limit and I try to prove it using the definition of limit with $ delta $ and $ varepsilon $ but I can't continue. Any tips/solutions?
algebra-precalculus limits multivariable-calculus
edited Mar 11 at 10:51
Robert Z
101k1069142
asked Mar 11 at 10:39
Dr.MathematicsDr.Mathematics
314
$endgroup$
add a comment |
$begingroup$
I have this limit: $$ lim_{(x,y)rightarrow (0,0)} frac{x-sin x+y}{x^3 + 6y}. $$
Taking some different paths I always get the same anwser, $frac{1}{6} $. So I guess that's the limit and I try to prove it using the definition of limit with $ delta $ and $ varepsilon $ but I can't continue. Any tips/solutions?
algebra-precalculus limits multivariable-calculus
edited Mar 11 at 10:51
Robert Z
101k1069142
asked Mar 11 at 10:39
Dr.MathematicsDr.Mathematics
314
$endgroup$
$begingroup$
Which paths did you take?
$endgroup$
– Michael Burr
Mar 11 at 10:43
$begingroup$
I tried $x=0$, $y=0$ and $y=x$
$endgroup$
– Dr.Mathematics
Mar 11 at 11:26
add a comment |
$begingroup$
I have this limit: $$ lim_{(x,y)rightarrow (0,0)} frac{x-sin x+y}{x^3 + 6y}. $$
Taking some different paths I always get the same anwser, $frac{1}{6} $. So I guess that's the limit and I try to prove it using the definition of limit with $ delta $ and $ varepsilon $ but I can't continue. Any tips/solutions?
algebra-precalculus limits multivariable-calculus
edited Mar 11 at 10:51
Robert Z
101k1069142
asked Mar 11 at 10:39
Dr.MathematicsDr.Mathematics
314
$endgroup$
I have this limit: $$ lim_{(x,y)rightarrow (0,0)} frac{x-sin x+y}{x^3 + 6y}. $$
Taking some different paths I always get the same anwser, $frac{1}{6} $. So I guess that's the limit and I try to prove it using the definition of limit with $ delta $ and $ varepsilon $ but I can't continue. Any tips/solutions?
algebra-precalculus limits multivariable-calculus
algebra-precalculus limits multivariable-calculus
edited Mar 11 at 10:51
Robert Z
101k1069142
asked Mar 11 at 10:39
Dr.MathematicsDr.Mathematics
314
edited Mar 11 at 10:51
Robert Z
101k1069142
asked Mar 11 at 10:39
Dr.MathematicsDr.Mathematics
314
edited Mar 11 at 10:51
Robert Z
101k1069142
edited Mar 11 at 10:51
Robert Z
101k1069142
edited Mar 11 at 10:51
Robert Z
101k1069142
101k1069142
asked Mar 11 at 10:39
Dr.MathematicsDr.Mathematics
314
asked Mar 11 at 10:39
Dr.MathematicsDr.Mathematics
314
asked Mar 11 at 10:39
Dr.MathematicsDr.Mathematics
314
314
$begingroup$
Which paths did you take?
$endgroup$
– Michael Burr
Mar 11 at 10:43
$begingroup$
I tried $x=0$, $y=0$ and $y=x$
$endgroup$
– Dr.Mathematics
Mar 11 at 11:26
add a comment |
$begingroup$
Which paths did you take?
$endgroup$
– Michael Burr
Mar 11 at 10:43
$begingroup$
I tried $x=0$, $y=0$ and $y=x$
$endgroup$
– Dr.Mathematics
Mar 11 at 11:26
$begingroup$
Which paths did you take?
$endgroup$
– Michael Burr
Mar 11 at 10:43
$begingroup$
Which paths did you take?
$endgroup$
– Michael Burr
Mar 11 at 10:43
$begingroup$
I tried $x=0$, $y=0$ and $y=x$
$endgroup$
– Dr.Mathematics
Mar 11 at 11:26
$begingroup$
I tried $x=0$, $y=0$ and $y=x$
$endgroup$
– Dr.Mathematics
Mar 11 at 11:26
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint. Recall that as $xto 0$, $sin(x)=x-frac{x^3}{3!}+frac{x^5}{5!}+o(x^5)$. Now try the path $y=-x^3/6+ax^5$ with $anot=0$. What do you get?
answered Mar 11 at 10:47
Robert ZRobert Z
101k1069142
$endgroup$
$begingroup$
The limit depends on $ alpha $ , so the limit does not exist. But when can I be sure and stop using paths? I tried 3 of them ( $x=0$, $y=0$, $y=x$) and I got $frac{1}{6}$ every time. So I tried to prove with definition. Can I show somehow with definition that the limit is not $frac{1}{6}$? Can I find a $varepsilon $ ?
$endgroup$
– Dr.Mathematics
Mar 11 at 11:27
$begingroup$
Unfortunately checking only a subfamily of paths does not guarantee that the limit exists. Once we have two paths with different limits, then the limit does not exist (this can be proved with $epsilon-delta$).
$endgroup$
– Robert Z
Mar 11 at 11:40
$begingroup$
Yeah I know that checking some paths does not guarantee that the limit exists, but we can assume that this is the limit's solution and then the definition will show us if we are right or not. Am i right? I guess it is not always easy to guess a path that gives us a different limit that the "easy" paths.
$endgroup$
– Dr.Mathematics
Mar 11 at 11:47
$begingroup$
No, it is not always easy to guess a "critical" path, but I don't think that a proof which uses the $epsilon-delta$ definition is an easier alternative.
$endgroup$
– Robert Z
Mar 11 at 12:10
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
Hint. Recall that as $xto 0$, $sin(x)=x-frac{x^3}{3!}+frac{x^5}{5!}+o(x^5)$. Now try the path $y=-x^3/6+ax^5$ with $anot=0$. What do you get?
answered Mar 11 at 10:47
Robert ZRobert Z
101k1069142
$endgroup$
$begingroup$
The limit depends on $ alpha $ , so the limit does not exist. But when can I be sure and stop using paths? I tried 3 of them ( $x=0$, $y=0$, $y=x$) and I got $frac{1}{6}$ every time. So I tried to prove with definition. Can I show somehow with definition that the limit is not $frac{1}{6}$? Can I find a $varepsilon $ ?
$endgroup$
– Dr.Mathematics
Mar 11 at 11:27
$begingroup$
Unfortunately checking only a subfamily of paths does not guarantee that the limit exists. Once we have two paths with different limits, then the limit does not exist (this can be proved with $epsilon-delta$).
$endgroup$
– Robert Z
Mar 11 at 11:40
$begingroup$
Yeah I know that checking some paths does not guarantee that the limit exists, but we can assume that this is the limit's solution and then the definition will show us if we are right or not. Am i right? I guess it is not always easy to guess a path that gives us a different limit that the "easy" paths.
$endgroup$
– Dr.Mathematics
Mar 11 at 11:47
$begingroup$
No, it is not always easy to guess a "critical" path, but I don't think that a proof which uses the $epsilon-delta$ definition is an easier alternative.
$endgroup$
– Robert Z
Mar 11 at 12:10
add a comment |
$begingroup$
Hint. Recall that as $xto 0$, $sin(x)=x-frac{x^3}{3!}+frac{x^5}{5!}+o(x^5)$. Now try the path $y=-x^3/6+ax^5$ with $anot=0$. What do you get?
answered Mar 11 at 10:47
Robert ZRobert Z
101k1069142
$endgroup$
$begingroup$
The limit depends on $ alpha $ , so the limit does not exist. But when can I be sure and stop using paths? I tried 3 of them ( $x=0$, $y=0$, $y=x$) and I got $frac{1}{6}$ every time. So I tried to prove with definition. Can I show somehow with definition that the limit is not $frac{1}{6}$? Can I find a $varepsilon $ ?
$endgroup$
– Dr.Mathematics
Mar 11 at 11:27
$begingroup$
Unfortunately checking only a subfamily of paths does not guarantee that the limit exists. Once we have two paths with different limits, then the limit does not exist (this can be proved with $epsilon-delta$).
$endgroup$
– Robert Z
Mar 11 at 11:40
$begingroup$
Yeah I know that checking some paths does not guarantee that the limit exists, but we can assume that this is the limit's solution and then the definition will show us if we are right or not. Am i right? I guess it is not always easy to guess a path that gives us a different limit that the "easy" paths.
$endgroup$
– Dr.Mathematics
Mar 11 at 11:47
$begingroup$
No, it is not always easy to guess a "critical" path, but I don't think that a proof which uses the $epsilon-delta$ definition is an easier alternative.
$endgroup$
– Robert Z
Mar 11 at 12:10
add a comment |
$begingroup$
Hint. Recall that as $xto 0$, $sin(x)=x-frac{x^3}{3!}+frac{x^5}{5!}+o(x^5)$. Now try the path $y=-x^3/6+ax^5$ with $anot=0$. What do you get?
answered Mar 11 at 10:47
Robert ZRobert Z
101k1069142
$endgroup$
Hint. Recall that as $xto 0$, $sin(x)=x-frac{x^3}{3!}+frac{x^5}{5!}+o(x^5)$. Now try the path $y=-x^3/6+ax^5$ with $anot=0$. What do you get?
answered Mar 11 at 10:47
Robert ZRobert Z
101k1069142
edited Mar 11 at 10:53
answered Mar 11 at 10:47
Robert ZRobert Z
101k1069142
answered Mar 11 at 10:47
Robert ZRobert Z
101k1069142
answered Mar 11 at 10:47
Robert ZRobert Z
101k1069142
101k1069142
$begingroup$
The limit depends on $ alpha $ , so the limit does not exist. But when can I be sure and stop using paths? I tried 3 of them ( $x=0$, $y=0$, $y=x$) and I got $frac{1}{6}$ every time. So I tried to prove with definition. Can I show somehow with definition that the limit is not $frac{1}{6}$? Can I find a $varepsilon $ ?
$endgroup$
– Dr.Mathematics
Mar 11 at 11:27
$begingroup$
Unfortunately checking only a subfamily of paths does not guarantee that the limit exists. Once we have two paths with different limits, then the limit does not exist (this can be proved with $epsilon-delta$).
$endgroup$
– Robert Z
Mar 11 at 11:40
$begingroup$
Yeah I know that checking some paths does not guarantee that the limit exists, but we can assume that this is the limit's solution and then the definition will show us if we are right or not. Am i right? I guess it is not always easy to guess a path that gives us a different limit that the "easy" paths.
$endgroup$
– Dr.Mathematics
Mar 11 at 11:47
$begingroup$
No, it is not always easy to guess a "critical" path, but I don't think that a proof which uses the $epsilon-delta$ definition is an easier alternative.
$endgroup$
– Robert Z
Mar 11 at 12:10
add a comment |
$begingroup$
The limit depends on $ alpha $ , so the limit does not exist. But when can I be sure and stop using paths? I tried 3 of them ( $x=0$, $y=0$, $y=x$) and I got $frac{1}{6}$ every time. So I tried to prove with definition. Can I show somehow with definition that the limit is not $frac{1}{6}$? Can I find a $varepsilon $ ?
$endgroup$
– Dr.Mathematics
Mar 11 at 11:27
$begingroup$
Unfortunately checking only a subfamily of paths does not guarantee that the limit exists. Once we have two paths with different limits, then the limit does not exist (this can be proved with $epsilon-delta$).
$endgroup$
– Robert Z
Mar 11 at 11:40
$begingroup$
Yeah I know that checking some paths does not guarantee that the limit exists, but we can assume that this is the limit's solution and then the definition will show us if we are right or not. Am i right? I guess it is not always easy to guess a path that gives us a different limit that the "easy" paths.
$endgroup$
– Dr.Mathematics
Mar 11 at 11:47
$begingroup$
No, it is not always easy to guess a "critical" path, but I don't think that a proof which uses the $epsilon-delta$ definition is an easier alternative.
$endgroup$
– Robert Z
Mar 11 at 12:10
$begingroup$
The limit depends on $ alpha $ , so the limit does not exist. But when can I be sure and stop using paths? I tried 3 of them ( $x=0$, $y=0$, $y=x$) and I got $frac{1}{6}$ every time. So I tried to prove with definition. Can I show somehow with definition that the limit is not $frac{1}{6}$? Can I find a $varepsilon $ ?
$endgroup$
– Dr.Mathematics
Mar 11 at 11:27
$begingroup$
The limit depends on $ alpha $ , so the limit does not exist. But when can I be sure and stop using paths? I tried 3 of them ( $x=0$, $y=0$, $y=x$) and I got $frac{1}{6}$ every time. So I tried to prove with definition. Can I show somehow with definition that the limit is not $frac{1}{6}$? Can I find a $varepsilon $ ?
$endgroup$
– Dr.Mathematics
Mar 11 at 11:27
$begingroup$
Unfortunately checking only a subfamily of paths does not guarantee that the limit exists. Once we have two paths with different limits, then the limit does not exist (this can be proved with $epsilon-delta$).
$endgroup$
– Robert Z
Mar 11 at 11:40
$begingroup$
Unfortunately checking only a subfamily of paths does not guarantee that the limit exists. Once we have two paths with different limits, then the limit does not exist (this can be proved with $epsilon-delta$).
$endgroup$
– Robert Z
Mar 11 at 11:40
$begingroup$
Yeah I know that checking some paths does not guarantee that the limit exists, but we can assume that this is the limit's solution and then the definition will show us if we are right or not. Am i right? I guess it is not always easy to guess a path that gives us a different limit that the "easy" paths.
$endgroup$
– Dr.Mathematics
Mar 11 at 11:47
$begingroup$
Yeah I know that checking some paths does not guarantee that the limit exists, but we can assume that this is the limit's solution and then the definition will show us if we are right or not. Am i right? I guess it is not always easy to guess a path that gives us a different limit that the "easy" paths.
$endgroup$
– Dr.Mathematics
Mar 11 at 11:47
$begingroup$
No, it is not always easy to guess a "critical" path, but I don't think that a proof which uses the $epsilon-delta$ definition is an easier alternative.
$endgroup$
– Robert Z
Mar 11 at 12:10
$begingroup$
No, it is not always easy to guess a "critical" path, but I don't think that a proof which uses the $epsilon-delta$ definition is an easier alternative.
$endgroup$
– Robert Z
Mar 11 at 12:10
add a comment |
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$begingroup$
Which paths did you take?
$endgroup$
– Michael Burr
Mar 11 at 10:43
$begingroup$
I tried $x=0$, $y=0$ and $y=x$
$endgroup$
– Dr.Mathematics
Mar 11 at 11:26