How to find the all possible values of an undefined limit?Question about finding the limit at an undefined point.Limit definition applicationsThe Limit of an Integral Containing ExponentialsHow can I use the Limit Laws to solve this limit?Does the limit $lim_(x,y)to (0,0) frac sin^2 (x-y)$ exist?Proving the limit exists or not for multivariable functions.WolframAlpha says limit exists when it doesn't?Compute the limit without L'Hospital's ruleWhat is the limit of zero times x, as x approaches infinity?To evaluate the limit as h approaches 0, how do I rewrite the limit in terms of theta?
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How to find the all possible values of an undefined limit?
Question about finding the limit at an undefined point.Limit definition applicationsThe Limit of an Integral Containing ExponentialsHow can I use the Limit Laws to solve this limit?Does the limit $lim_(x,y)to (0,0) frac sin^2 (x-y)y$ exist?Proving the limit exists or not for multivariable functions.WolframAlpha says limit exists when it doesn't?Compute the limit without L'Hospital's ruleWhat is the limit of zero times x, as x approaches infinity?To evaluate the limit as h approaches 0, how do I rewrite the limit in terms of theta?
$begingroup$
The following limit is not defined:
$$ lim_(x,y)rightarrow (0,0) fracxyx^2(1+y).$$
However, I am interested in a way to find all possible values as $(x,y)$ approaches $(0,0)$. My intuition is that the answer are $pminfty$ and $0$. And the method of approaching this is probably by representing $x,y$ as complex number. But I am unsure of how to proceed. Any hint or reference would be much appreciated.
limits analysis complex-numbers
edited Mar 11 at 20:34
Michael Rozenberg
108k1895200
asked Mar 11 at 20:23
PaichuPaichu
773616
$endgroup$
add a comment |
$begingroup$
The following limit is not defined:
$$ lim_(x,y)rightarrow (0,0) fracxyx^2(1+y).$$
However, I am interested in a way to find all possible values as $(x,y)$ approaches $(0,0)$. My intuition is that the answer are $pminfty$ and $0$. And the method of approaching this is probably by representing $x,y$ as complex number. But I am unsure of how to proceed. Any hint or reference would be much appreciated.
limits analysis complex-numbers
edited Mar 11 at 20:34
Michael Rozenberg
108k1895200
asked Mar 11 at 20:23
PaichuPaichu
773616
$endgroup$
$begingroup$
You surely mean the limit approach the number 0, not $(0,0)$ as $(x,y) to (0,0)$, right? Also, since $x$ and $y$ are complex numbers, they cannot be $pm infty$, since $pm infty notin mathbbC$.
$endgroup$
– Viktor Glombik
Mar 11 at 20:25
$begingroup$
I mean the limit as both $x$ and $y$ approach 0.
$endgroup$
– Paichu
Mar 11 at 20:27
1
$begingroup$
If $x=y$ it goes to one.
$endgroup$
– hamam_Abdallah
Mar 11 at 20:29
add a comment |
$begingroup$
The following limit is not defined:
$$ lim_(x,y)rightarrow (0,0) fracxyx^2(1+y).$$
However, I am interested in a way to find all possible values as $(x,y)$ approaches $(0,0)$. My intuition is that the answer are $pminfty$ and $0$. And the method of approaching this is probably by representing $x,y$ as complex number. But I am unsure of how to proceed. Any hint or reference would be much appreciated.
limits analysis complex-numbers
edited Mar 11 at 20:34
Michael Rozenberg
108k1895200
asked Mar 11 at 20:23
PaichuPaichu
773616
$endgroup$
The following limit is not defined:
$$ lim_(x,y)rightarrow (0,0) fracxyx^2(1+y).$$
However, I am interested in a way to find all possible values as $(x,y)$ approaches $(0,0)$. My intuition is that the answer are $pminfty$ and $0$. And the method of approaching this is probably by representing $x,y$ as complex number. But I am unsure of how to proceed. Any hint or reference would be much appreciated.
limits analysis complex-numbers
limits analysis complex-numbers
edited Mar 11 at 20:34
Michael Rozenberg
108k1895200
asked Mar 11 at 20:23
PaichuPaichu
773616
edited Mar 11 at 20:34
Michael Rozenberg
108k1895200
asked Mar 11 at 20:23
PaichuPaichu
773616
edited Mar 11 at 20:34
Michael Rozenberg
108k1895200
edited Mar 11 at 20:34
Michael Rozenberg
108k1895200
edited Mar 11 at 20:34
Michael Rozenberg
108k1895200
108k1895200
asked Mar 11 at 20:23
PaichuPaichu
773616
asked Mar 11 at 20:23
PaichuPaichu
773616
asked Mar 11 at 20:23
PaichuPaichu
773616
773616
$begingroup$
You surely mean the limit approach the number 0, not $(0,0)$ as $(x,y) to (0,0)$, right? Also, since $x$ and $y$ are complex numbers, they cannot be $pm infty$, since $pm infty notin mathbbC$.
$endgroup$
– Viktor Glombik
Mar 11 at 20:25
$begingroup$
I mean the limit as both $x$ and $y$ approach 0.
$endgroup$
– Paichu
Mar 11 at 20:27
1
$begingroup$
If $x=y$ it goes to one.
$endgroup$
– hamam_Abdallah
Mar 11 at 20:29
add a comment |
$begingroup$
You surely mean the limit approach the number 0, not $(0,0)$ as $(x,y) to (0,0)$, right? Also, since $x$ and $y$ are complex numbers, they cannot be $pm infty$, since $pm infty notin mathbbC$.
$endgroup$
– Viktor Glombik
Mar 11 at 20:25
$begingroup$
I mean the limit as both $x$ and $y$ approach 0.
$endgroup$
– Paichu
Mar 11 at 20:27
1
$begingroup$
If $x=y$ it goes to one.
$endgroup$
– hamam_Abdallah
Mar 11 at 20:29
$begingroup$
You surely mean the limit approach the number 0, not $(0,0)$ as $(x,y) to (0,0)$, right? Also, since $x$ and $y$ are complex numbers, they cannot be $pm infty$, since $pm infty notin mathbbC$.
$endgroup$
– Viktor Glombik
Mar 11 at 20:25
$begingroup$
You surely mean the limit approach the number 0, not $(0,0)$ as $(x,y) to (0,0)$, right? Also, since $x$ and $y$ are complex numbers, they cannot be $pm infty$, since $pm infty notin mathbbC$.
$endgroup$
– Viktor Glombik
Mar 11 at 20:25
$begingroup$
I mean the limit as both $x$ and $y$ approach 0.
$endgroup$
– Paichu
Mar 11 at 20:27
$begingroup$
I mean the limit as both $x$ and $y$ approach 0.
$endgroup$
– Paichu
Mar 11 at 20:27
1
1
$begingroup$
If $x=y$ it goes to one.
$endgroup$
– hamam_Abdallah
Mar 11 at 20:29
$begingroup$
If $x=y$ it goes to one.
$endgroup$
– hamam_Abdallah
Mar 11 at 20:29
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Take $(x,y)=(t,kt),$ where $kinmathbb R$ and $trightarrow0$.
Also, take $(t^2,t)$.
answered Mar 11 at 20:28
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
$begingroup$
Ignore the factor $1+y$, which plays no role, and simplify the $x$. Now
$$frac yx$$ can take any value.
answered Mar 11 at 20:38
Yves DaoustYves Daoust
130k676229
$endgroup$
add a comment |
$begingroup$
Let $y=x$, then $$fracx^2x^2+x^3=frac11+x$$ as $x$ tends to $0$, the limit $f(x,y)$ converges to $1$.
For the second limit, let $y=x^2$, then $$f(x,y)=fracy^3/2y+y^2$$ which simplifies to:
$$fracy^1/21+y$$ and as you let $y$ tend to $0$ the function converges to $0$
Therefore the limit does not exist.
answered Mar 11 at 20:36
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
483215
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Take $(x,y)=(t,kt),$ where $kinmathbb R$ and $trightarrow0$.
Also, take $(t^2,t)$.
answered Mar 11 at 20:28
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
$begingroup$
Take $(x,y)=(t,kt),$ where $kinmathbb R$ and $trightarrow0$.
Also, take $(t^2,t)$.
answered Mar 11 at 20:28
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
$begingroup$
Take $(x,y)=(t,kt),$ where $kinmathbb R$ and $trightarrow0$.
Also, take $(t^2,t)$.
answered Mar 11 at 20:28
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
Take $(x,y)=(t,kt),$ where $kinmathbb R$ and $trightarrow0$.
Also, take $(t^2,t)$.
answered Mar 11 at 20:28
Michael RozenbergMichael Rozenberg
108k1895200
edited Mar 11 at 20:33
answered Mar 11 at 20:28
Michael RozenbergMichael Rozenberg
108k1895200
answered Mar 11 at 20:28
Michael RozenbergMichael Rozenberg
108k1895200
answered Mar 11 at 20:28
Michael RozenbergMichael Rozenberg
108k1895200
108k1895200
add a comment |
add a comment |
$begingroup$
Ignore the factor $1+y$, which plays no role, and simplify the $x$. Now
$$frac yx$$ can take any value.
answered Mar 11 at 20:38
Yves DaoustYves Daoust
130k676229
$endgroup$
add a comment |
$begingroup$
Ignore the factor $1+y$, which plays no role, and simplify the $x$. Now
$$frac yx$$ can take any value.
answered Mar 11 at 20:38
Yves DaoustYves Daoust
130k676229
$endgroup$
add a comment |
$begingroup$
Ignore the factor $1+y$, which plays no role, and simplify the $x$. Now
$$frac yx$$ can take any value.
answered Mar 11 at 20:38
Yves DaoustYves Daoust
130k676229
$endgroup$
Ignore the factor $1+y$, which plays no role, and simplify the $x$. Now
$$frac yx$$ can take any value.
answered Mar 11 at 20:38
Yves DaoustYves Daoust
130k676229
answered Mar 11 at 20:38
Yves DaoustYves Daoust
130k676229
answered Mar 11 at 20:38
Yves DaoustYves Daoust
130k676229
answered Mar 11 at 20:38
Yves DaoustYves Daoust
130k676229
130k676229
add a comment |
add a comment |
$begingroup$
Let $y=x$, then $$fracx^2x^2+x^3=frac11+x$$ as $x$ tends to $0$, the limit $f(x,y)$ converges to $1$.
For the second limit, let $y=x^2$, then $$f(x,y)=fracy^3/2y+y^2$$ which simplifies to:
$$fracy^1/21+y$$ and as you let $y$ tend to $0$ the function converges to $0$
Therefore the limit does not exist.
answered Mar 11 at 20:36
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
483215
$endgroup$
add a comment |
$begingroup$
Let $y=x$, then $$fracx^2x^2+x^3=frac11+x$$ as $x$ tends to $0$, the limit $f(x,y)$ converges to $1$.
For the second limit, let $y=x^2$, then $$f(x,y)=fracy^3/2y+y^2$$ which simplifies to:
$$fracy^1/21+y$$ and as you let $y$ tend to $0$ the function converges to $0$
Therefore the limit does not exist.
answered Mar 11 at 20:36
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
483215
$endgroup$
add a comment |
$begingroup$
Let $y=x$, then $$fracx^2x^2+x^3=frac11+x$$ as $x$ tends to $0$, the limit $f(x,y)$ converges to $1$.
For the second limit, let $y=x^2$, then $$f(x,y)=fracy^3/2y+y^2$$ which simplifies to:
$$fracy^1/21+y$$ and as you let $y$ tend to $0$ the function converges to $0$
Therefore the limit does not exist.
answered Mar 11 at 20:36
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
483215
$endgroup$
Let $y=x$, then $$fracx^2x^2+x^3=frac11+x$$ as $x$ tends to $0$, the limit $f(x,y)$ converges to $1$.
For the second limit, let $y=x^2$, then $$f(x,y)=fracy^3/2y+y^2$$ which simplifies to:
$$fracy^1/21+y$$ and as you let $y$ tend to $0$ the function converges to $0$
Therefore the limit does not exist.
answered Mar 11 at 20:36
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
483215
answered Mar 11 at 20:36
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
483215
answered Mar 11 at 20:36
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
483215
answered Mar 11 at 20:36
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
483215
483215
add a comment |
add a comment |
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$begingroup$
You surely mean the limit approach the number 0, not $(0,0)$ as $(x,y) to (0,0)$, right? Also, since $x$ and $y$ are complex numbers, they cannot be $pm infty$, since $pm infty notin mathbbC$.
$endgroup$
– Viktor Glombik
Mar 11 at 20:25
$begingroup$
I mean the limit as both $x$ and $y$ approach 0.
$endgroup$
– Paichu
Mar 11 at 20:27
1
$begingroup$
If $x=y$ it goes to one.
$endgroup$
– hamam_Abdallah
Mar 11 at 20:29