Existence of the Limit of a Two-Variable Function
$begingroup$
This problem is an example in my calculus textbook. Let:
$$
f(x,y)=frac{y^2sin^2(x)}{x^4+y^4}
$$
My textbook says that
$$
lim_{(x,y)to(0,0)}f(x,y) text{ does not exist.}
$$
Questions:
How do we know the limit does not exist?
In general, suppose the limit of a function exists but we do not know the value of such limit, how do we find it?
limits multivariable-calculus
edited Jan 18 at 0:51
Git Gud
28.9k1050101
asked Jan 18 at 0:37
A Slow LearnerA Slow Learner
468313
$endgroup$
add a comment |
$begingroup$
This problem is an example in my calculus textbook. Let:
$$
f(x,y)=frac{y^2sin^2(x)}{x^4+y^4}
$$
My textbook says that
$$
lim_{(x,y)to(0,0)}f(x,y) text{ does not exist.}
$$
Questions:
How do we know the limit does not exist?
In general, suppose the limit of a function exists but we do not know the value of such limit, how do we find it?
limits multivariable-calculus
edited Jan 18 at 0:51
Git Gud
28.9k1050101
asked Jan 18 at 0:37
A Slow LearnerA Slow Learner
468313
$endgroup$
$begingroup$
Typically, to prove that the limit does not exist, people find distinct sublimits. If you know the limit exists and just want to find it, then you just need to consider an appropriate (read easy) sublimit.
$endgroup$
– Git Gud
Jan 18 at 0:49
$begingroup$
@GitGud Does this mean that proving a limit does not exist, as in this question, might be very difficult if we cannot identify the proper sublimits, which more or less boils down to trial & error?
$endgroup$
– A Slow Learner
Jan 18 at 0:59
$begingroup$
There are strategies you can use. For rational functions in two variables you can try looking at sublimits of the form $tmapsto (t, kt)$, $tmapsto (t, kt^2)$, $tmapsto (t, kt^3)$ and so on, with varying values of $k$. If you're unable to find different sublimits, then perhaps the sublimit exists and you can try to focus on proving it does exist. When you don't have a rational function, you can probably use Taylor's theorem to get a similarly behaved rational function.
$endgroup$
– Git Gud
Jan 18 at 8:38
add a comment |
$begingroup$
This problem is an example in my calculus textbook. Let:
$$
f(x,y)=frac{y^2sin^2(x)}{x^4+y^4}
$$
My textbook says that
$$
lim_{(x,y)to(0,0)}f(x,y) text{ does not exist.}
$$
Questions:
How do we know the limit does not exist?
In general, suppose the limit of a function exists but we do not know the value of such limit, how do we find it?
limits multivariable-calculus
edited Jan 18 at 0:51
Git Gud
28.9k1050101
asked Jan 18 at 0:37
A Slow LearnerA Slow Learner
468313
$endgroup$
This problem is an example in my calculus textbook. Let:
$$
f(x,y)=frac{y^2sin^2(x)}{x^4+y^4}
$$
My textbook says that
$$
lim_{(x,y)to(0,0)}f(x,y) text{ does not exist.}
$$
Questions:
How do we know the limit does not exist?
In general, suppose the limit of a function exists but we do not know the value of such limit, how do we find it?
limits multivariable-calculus
limits multivariable-calculus
edited Jan 18 at 0:51
Git Gud
28.9k1050101
asked Jan 18 at 0:37
A Slow LearnerA Slow Learner
468313
edited Jan 18 at 0:51
Git Gud
28.9k1050101
asked Jan 18 at 0:37
A Slow LearnerA Slow Learner
468313
edited Jan 18 at 0:51
Git Gud
28.9k1050101
edited Jan 18 at 0:51
Git Gud
28.9k1050101
edited Jan 18 at 0:51
Git Gud
28.9k1050101
28.9k1050101
asked Jan 18 at 0:37
A Slow LearnerA Slow Learner
468313
asked Jan 18 at 0:37
A Slow LearnerA Slow Learner
468313
asked Jan 18 at 0:37
A Slow LearnerA Slow Learner
468313
468313
$begingroup$
Typically, to prove that the limit does not exist, people find distinct sublimits. If you know the limit exists and just want to find it, then you just need to consider an appropriate (read easy) sublimit.
$endgroup$
– Git Gud
Jan 18 at 0:49
$begingroup$
@GitGud Does this mean that proving a limit does not exist, as in this question, might be very difficult if we cannot identify the proper sublimits, which more or less boils down to trial & error?
$endgroup$
– A Slow Learner
Jan 18 at 0:59
$begingroup$
There are strategies you can use. For rational functions in two variables you can try looking at sublimits of the form $tmapsto (t, kt)$, $tmapsto (t, kt^2)$, $tmapsto (t, kt^3)$ and so on, with varying values of $k$. If you're unable to find different sublimits, then perhaps the sublimit exists and you can try to focus on proving it does exist. When you don't have a rational function, you can probably use Taylor's theorem to get a similarly behaved rational function.
$endgroup$
– Git Gud
Jan 18 at 8:38
add a comment |
$begingroup$
Typically, to prove that the limit does not exist, people find distinct sublimits. If you know the limit exists and just want to find it, then you just need to consider an appropriate (read easy) sublimit.
$endgroup$
– Git Gud
Jan 18 at 0:49
$begingroup$
@GitGud Does this mean that proving a limit does not exist, as in this question, might be very difficult if we cannot identify the proper sublimits, which more or less boils down to trial & error?
$endgroup$
– A Slow Learner
Jan 18 at 0:59
$begingroup$
There are strategies you can use. For rational functions in two variables you can try looking at sublimits of the form $tmapsto (t, kt)$, $tmapsto (t, kt^2)$, $tmapsto (t, kt^3)$ and so on, with varying values of $k$. If you're unable to find different sublimits, then perhaps the sublimit exists and you can try to focus on proving it does exist. When you don't have a rational function, you can probably use Taylor's theorem to get a similarly behaved rational function.
$endgroup$
– Git Gud
Jan 18 at 8:38
$begingroup$
Typically, to prove that the limit does not exist, people find distinct sublimits. If you know the limit exists and just want to find it, then you just need to consider an appropriate (read easy) sublimit.
$endgroup$
– Git Gud
Jan 18 at 0:49
$begingroup$
Typically, to prove that the limit does not exist, people find distinct sublimits. If you know the limit exists and just want to find it, then you just need to consider an appropriate (read easy) sublimit.
$endgroup$
– Git Gud
Jan 18 at 0:49
$begingroup$
@GitGud Does this mean that proving a limit does not exist, as in this question, might be very difficult if we cannot identify the proper sublimits, which more or less boils down to trial & error?
$endgroup$
– A Slow Learner
Jan 18 at 0:59
$begingroup$
@GitGud Does this mean that proving a limit does not exist, as in this question, might be very difficult if we cannot identify the proper sublimits, which more or less boils down to trial & error?
$endgroup$
– A Slow Learner
Jan 18 at 0:59
$begingroup$
There are strategies you can use. For rational functions in two variables you can try looking at sublimits of the form $tmapsto (t, kt)$, $tmapsto (t, kt^2)$, $tmapsto (t, kt^3)$ and so on, with varying values of $k$. If you're unable to find different sublimits, then perhaps the sublimit exists and you can try to focus on proving it does exist. When you don't have a rational function, you can probably use Taylor's theorem to get a similarly behaved rational function.
$endgroup$
– Git Gud
Jan 18 at 8:38
$begingroup$
There are strategies you can use. For rational functions in two variables you can try looking at sublimits of the form $tmapsto (t, kt)$, $tmapsto (t, kt^2)$, $tmapsto (t, kt^3)$ and so on, with varying values of $k$. If you're unable to find different sublimits, then perhaps the sublimit exists and you can try to focus on proving it does exist. When you don't have a rational function, you can probably use Taylor's theorem to get a similarly behaved rational function.
$endgroup$
– Git Gud
Jan 18 at 8:38
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Consider the curves $gamma_{1} = (t,t)$ and $gamma_{2}(t) = (t,t^{2})$. Thus we have
begin{align*}f(gamma_{1}(t)) = frac{t^{2}sin^{2}(t)}{t^{4} + t^{4}} = frac{sin^{2}(t)}{2t^{2}} Longrightarrow lim_{trightarrow 0}f(gamma_{1}(t)) = lim_{trightarrow 0}left[frac{1}{2}left(frac{sin(t)}{t}right)^{2}right] = frac{1}{2}
end{align*}
On the other hand, we have
begin{align*}
f(gamma_{2}(t)) = frac{t^{8}sin^{2}(t)}{t^{4} + t^{8}} = frac{t^{4}sin^{2}(t)}{1 + t^{4}} Longrightarrow lim_{trightarrow 0}f(gamma_{2}(t)) = lim_{trightarrow 0}frac{t^{4}sin^{2}(t)}{1+t^{4}} = frac{0times 0}{1+0} = 0
end{align*}
If the given limit existed, we should have $displaystylelim_{trightarrow0}f(gamma_{1}(t)) = lim_{trightarrow 0}f(gamma_{2}(t))$. Hence $displaystylelim_{textbf{x}rightarrowtextbf{0}} f(textbf{x})$ does not exist. Hope this helps.
answered Jan 18 at 0:48
user1337user1337
48210
$endgroup$
$begingroup$
Thank you! What is your general strategy for finding these two curves as in this example?
$endgroup$
– A Slow Learner
Jan 18 at 1:00
$begingroup$
As previously mentioned, there is no general strategy. In this particular case, such choices were made because they "break the symmetry".
$endgroup$
– user1337
Jan 18 at 1:07
$begingroup$
For this example, could you explain to me more how the two curves "break the symmetry"? I never heard of that before.
$endgroup$
– A Slow Learner
Jan 18 at 1:08
$begingroup$
Sorry, I am not an English native speaker. What I actually meant is: choose two curves whose limits of $f(gamma_{k}(t))$ are different as $trightarrow 0$.
$endgroup$
– user1337
Jan 18 at 1:12
$begingroup$
What if finding such two curves is difficult? For example, what do you do if the limit doesn't exist but all the curves you have found produce the same limit? Is keeping trying our only option?
$endgroup$
– A Slow Learner
Jan 18 at 1:15
|
show 3 more comments
$begingroup$
1: Use the approximation (for small $x), sin(x)approx x$. Then $f(x,y)approx frac{x^2y^2}{x^4+y^4}$. Now consider two cases:
(a) $x=y$ then the function $=frac{1}{2}$.
(b) $x=0$ or $y=0$ where the function $=0$, when the other variable $ne 0$.
Therefore the limit does not exist.
2: There is no general solution. Each problem must be evaluated on its own.
answered Jan 18 at 0:56
herb steinbergherb steinberg
3,1982311
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider the curves $gamma_{1} = (t,t)$ and $gamma_{2}(t) = (t,t^{2})$. Thus we have
begin{align*}f(gamma_{1}(t)) = frac{t^{2}sin^{2}(t)}{t^{4} + t^{4}} = frac{sin^{2}(t)}{2t^{2}} Longrightarrow lim_{trightarrow 0}f(gamma_{1}(t)) = lim_{trightarrow 0}left[frac{1}{2}left(frac{sin(t)}{t}right)^{2}right] = frac{1}{2}
end{align*}
On the other hand, we have
begin{align*}
f(gamma_{2}(t)) = frac{t^{8}sin^{2}(t)}{t^{4} + t^{8}} = frac{t^{4}sin^{2}(t)}{1 + t^{4}} Longrightarrow lim_{trightarrow 0}f(gamma_{2}(t)) = lim_{trightarrow 0}frac{t^{4}sin^{2}(t)}{1+t^{4}} = frac{0times 0}{1+0} = 0
end{align*}
If the given limit existed, we should have $displaystylelim_{trightarrow0}f(gamma_{1}(t)) = lim_{trightarrow 0}f(gamma_{2}(t))$. Hence $displaystylelim_{textbf{x}rightarrowtextbf{0}} f(textbf{x})$ does not exist. Hope this helps.
answered Jan 18 at 0:48
user1337user1337
48210
$endgroup$
$begingroup$
Thank you! What is your general strategy for finding these two curves as in this example?
$endgroup$
– A Slow Learner
Jan 18 at 1:00
$begingroup$
As previously mentioned, there is no general strategy. In this particular case, such choices were made because they "break the symmetry".
$endgroup$
– user1337
Jan 18 at 1:07
$begingroup$
For this example, could you explain to me more how the two curves "break the symmetry"? I never heard of that before.
$endgroup$
– A Slow Learner
Jan 18 at 1:08
$begingroup$
Sorry, I am not an English native speaker. What I actually meant is: choose two curves whose limits of $f(gamma_{k}(t))$ are different as $trightarrow 0$.
$endgroup$
– user1337
Jan 18 at 1:12
$begingroup$
What if finding such two curves is difficult? For example, what do you do if the limit doesn't exist but all the curves you have found produce the same limit? Is keeping trying our only option?
$endgroup$
– A Slow Learner
Jan 18 at 1:15
|
show 3 more comments
$begingroup$
Consider the curves $gamma_{1} = (t,t)$ and $gamma_{2}(t) = (t,t^{2})$. Thus we have
begin{align*}f(gamma_{1}(t)) = frac{t^{2}sin^{2}(t)}{t^{4} + t^{4}} = frac{sin^{2}(t)}{2t^{2}} Longrightarrow lim_{trightarrow 0}f(gamma_{1}(t)) = lim_{trightarrow 0}left[frac{1}{2}left(frac{sin(t)}{t}right)^{2}right] = frac{1}{2}
end{align*}
On the other hand, we have
begin{align*}
f(gamma_{2}(t)) = frac{t^{8}sin^{2}(t)}{t^{4} + t^{8}} = frac{t^{4}sin^{2}(t)}{1 + t^{4}} Longrightarrow lim_{trightarrow 0}f(gamma_{2}(t)) = lim_{trightarrow 0}frac{t^{4}sin^{2}(t)}{1+t^{4}} = frac{0times 0}{1+0} = 0
end{align*}
If the given limit existed, we should have $displaystylelim_{trightarrow0}f(gamma_{1}(t)) = lim_{trightarrow 0}f(gamma_{2}(t))$. Hence $displaystylelim_{textbf{x}rightarrowtextbf{0}} f(textbf{x})$ does not exist. Hope this helps.
answered Jan 18 at 0:48
user1337user1337
48210
$endgroup$
$begingroup$
Thank you! What is your general strategy for finding these two curves as in this example?
$endgroup$
– A Slow Learner
Jan 18 at 1:00
$begingroup$
As previously mentioned, there is no general strategy. In this particular case, such choices were made because they "break the symmetry".
$endgroup$
– user1337
Jan 18 at 1:07
$begingroup$
For this example, could you explain to me more how the two curves "break the symmetry"? I never heard of that before.
$endgroup$
– A Slow Learner
Jan 18 at 1:08
$begingroup$
Sorry, I am not an English native speaker. What I actually meant is: choose two curves whose limits of $f(gamma_{k}(t))$ are different as $trightarrow 0$.
$endgroup$
– user1337
Jan 18 at 1:12
$begingroup$
What if finding such two curves is difficult? For example, what do you do if the limit doesn't exist but all the curves you have found produce the same limit? Is keeping trying our only option?
$endgroup$
– A Slow Learner
Jan 18 at 1:15
|
show 3 more comments
$begingroup$
Consider the curves $gamma_{1} = (t,t)$ and $gamma_{2}(t) = (t,t^{2})$. Thus we have
begin{align*}f(gamma_{1}(t)) = frac{t^{2}sin^{2}(t)}{t^{4} + t^{4}} = frac{sin^{2}(t)}{2t^{2}} Longrightarrow lim_{trightarrow 0}f(gamma_{1}(t)) = lim_{trightarrow 0}left[frac{1}{2}left(frac{sin(t)}{t}right)^{2}right] = frac{1}{2}
end{align*}
On the other hand, we have
begin{align*}
f(gamma_{2}(t)) = frac{t^{8}sin^{2}(t)}{t^{4} + t^{8}} = frac{t^{4}sin^{2}(t)}{1 + t^{4}} Longrightarrow lim_{trightarrow 0}f(gamma_{2}(t)) = lim_{trightarrow 0}frac{t^{4}sin^{2}(t)}{1+t^{4}} = frac{0times 0}{1+0} = 0
end{align*}
If the given limit existed, we should have $displaystylelim_{trightarrow0}f(gamma_{1}(t)) = lim_{trightarrow 0}f(gamma_{2}(t))$. Hence $displaystylelim_{textbf{x}rightarrowtextbf{0}} f(textbf{x})$ does not exist. Hope this helps.
answered Jan 18 at 0:48
user1337user1337
48210
$endgroup$
Consider the curves $gamma_{1} = (t,t)$ and $gamma_{2}(t) = (t,t^{2})$. Thus we have
begin{align*}f(gamma_{1}(t)) = frac{t^{2}sin^{2}(t)}{t^{4} + t^{4}} = frac{sin^{2}(t)}{2t^{2}} Longrightarrow lim_{trightarrow 0}f(gamma_{1}(t)) = lim_{trightarrow 0}left[frac{1}{2}left(frac{sin(t)}{t}right)^{2}right] = frac{1}{2}
end{align*}
On the other hand, we have
begin{align*}
f(gamma_{2}(t)) = frac{t^{8}sin^{2}(t)}{t^{4} + t^{8}} = frac{t^{4}sin^{2}(t)}{1 + t^{4}} Longrightarrow lim_{trightarrow 0}f(gamma_{2}(t)) = lim_{trightarrow 0}frac{t^{4}sin^{2}(t)}{1+t^{4}} = frac{0times 0}{1+0} = 0
end{align*}
If the given limit existed, we should have $displaystylelim_{trightarrow0}f(gamma_{1}(t)) = lim_{trightarrow 0}f(gamma_{2}(t))$. Hence $displaystylelim_{textbf{x}rightarrowtextbf{0}} f(textbf{x})$ does not exist. Hope this helps.
answered Jan 18 at 0:48
user1337user1337
48210
answered Jan 18 at 0:48
user1337user1337
48210
answered Jan 18 at 0:48
user1337user1337
48210
answered Jan 18 at 0:48
user1337user1337
48210
48210
$begingroup$
Thank you! What is your general strategy for finding these two curves as in this example?
$endgroup$
– A Slow Learner
Jan 18 at 1:00
$begingroup$
As previously mentioned, there is no general strategy. In this particular case, such choices were made because they "break the symmetry".
$endgroup$
– user1337
Jan 18 at 1:07
$begingroup$
For this example, could you explain to me more how the two curves "break the symmetry"? I never heard of that before.
$endgroup$
– A Slow Learner
Jan 18 at 1:08
$begingroup$
Sorry, I am not an English native speaker. What I actually meant is: choose two curves whose limits of $f(gamma_{k}(t))$ are different as $trightarrow 0$.
$endgroup$
– user1337
Jan 18 at 1:12
$begingroup$
What if finding such two curves is difficult? For example, what do you do if the limit doesn't exist but all the curves you have found produce the same limit? Is keeping trying our only option?
$endgroup$
– A Slow Learner
Jan 18 at 1:15
|
show 3 more comments
$begingroup$
Thank you! What is your general strategy for finding these two curves as in this example?
$endgroup$
– A Slow Learner
Jan 18 at 1:00
$begingroup$
As previously mentioned, there is no general strategy. In this particular case, such choices were made because they "break the symmetry".
$endgroup$
– user1337
Jan 18 at 1:07
$begingroup$
For this example, could you explain to me more how the two curves "break the symmetry"? I never heard of that before.
$endgroup$
– A Slow Learner
Jan 18 at 1:08
$begingroup$
Sorry, I am not an English native speaker. What I actually meant is: choose two curves whose limits of $f(gamma_{k}(t))$ are different as $trightarrow 0$.
$endgroup$
– user1337
Jan 18 at 1:12
$begingroup$
What if finding such two curves is difficult? For example, what do you do if the limit doesn't exist but all the curves you have found produce the same limit? Is keeping trying our only option?
$endgroup$
– A Slow Learner
Jan 18 at 1:15
$begingroup$
Thank you! What is your general strategy for finding these two curves as in this example?
$endgroup$
– A Slow Learner
Jan 18 at 1:00
$begingroup$
Thank you! What is your general strategy for finding these two curves as in this example?
$endgroup$
– A Slow Learner
Jan 18 at 1:00
$begingroup$
As previously mentioned, there is no general strategy. In this particular case, such choices were made because they "break the symmetry".
$endgroup$
– user1337
Jan 18 at 1:07
$begingroup$
As previously mentioned, there is no general strategy. In this particular case, such choices were made because they "break the symmetry".
$endgroup$
– user1337
Jan 18 at 1:07
$begingroup$
For this example, could you explain to me more how the two curves "break the symmetry"? I never heard of that before.
$endgroup$
– A Slow Learner
Jan 18 at 1:08
$begingroup$
For this example, could you explain to me more how the two curves "break the symmetry"? I never heard of that before.
$endgroup$
– A Slow Learner
Jan 18 at 1:08
$begingroup$
Sorry, I am not an English native speaker. What I actually meant is: choose two curves whose limits of $f(gamma_{k}(t))$ are different as $trightarrow 0$.
$endgroup$
– user1337
Jan 18 at 1:12
$begingroup$
Sorry, I am not an English native speaker. What I actually meant is: choose two curves whose limits of $f(gamma_{k}(t))$ are different as $trightarrow 0$.
$endgroup$
– user1337
Jan 18 at 1:12
$begingroup$
What if finding such two curves is difficult? For example, what do you do if the limit doesn't exist but all the curves you have found produce the same limit? Is keeping trying our only option?
$endgroup$
– A Slow Learner
Jan 18 at 1:15
$begingroup$
What if finding such two curves is difficult? For example, what do you do if the limit doesn't exist but all the curves you have found produce the same limit? Is keeping trying our only option?
$endgroup$
– A Slow Learner
Jan 18 at 1:15
|
show 3 more comments
$begingroup$
1: Use the approximation (for small $x), sin(x)approx x$. Then $f(x,y)approx frac{x^2y^2}{x^4+y^4}$. Now consider two cases:
(a) $x=y$ then the function $=frac{1}{2}$.
(b) $x=0$ or $y=0$ where the function $=0$, when the other variable $ne 0$.
Therefore the limit does not exist.
2: There is no general solution. Each problem must be evaluated on its own.
answered Jan 18 at 0:56
herb steinbergherb steinberg
3,1982311
$endgroup$
add a comment |
$begingroup$
1: Use the approximation (for small $x), sin(x)approx x$. Then $f(x,y)approx frac{x^2y^2}{x^4+y^4}$. Now consider two cases:
(a) $x=y$ then the function $=frac{1}{2}$.
(b) $x=0$ or $y=0$ where the function $=0$, when the other variable $ne 0$.
Therefore the limit does not exist.
2: There is no general solution. Each problem must be evaluated on its own.
answered Jan 18 at 0:56
herb steinbergherb steinberg
3,1982311
$endgroup$
add a comment |
$begingroup$
1: Use the approximation (for small $x), sin(x)approx x$. Then $f(x,y)approx frac{x^2y^2}{x^4+y^4}$. Now consider two cases:
(a) $x=y$ then the function $=frac{1}{2}$.
(b) $x=0$ or $y=0$ where the function $=0$, when the other variable $ne 0$.
Therefore the limit does not exist.
2: There is no general solution. Each problem must be evaluated on its own.
answered Jan 18 at 0:56
herb steinbergherb steinberg
3,1982311
$endgroup$
1: Use the approximation (for small $x), sin(x)approx x$. Then $f(x,y)approx frac{x^2y^2}{x^4+y^4}$. Now consider two cases:
(a) $x=y$ then the function $=frac{1}{2}$.
(b) $x=0$ or $y=0$ where the function $=0$, when the other variable $ne 0$.
Therefore the limit does not exist.
2: There is no general solution. Each problem must be evaluated on its own.
answered Jan 18 at 0:56
herb steinbergherb steinberg
3,1982311
answered Jan 18 at 0:56
herb steinbergherb steinberg
3,1982311
answered Jan 18 at 0:56
herb steinbergherb steinberg
3,1982311
answered Jan 18 at 0:56
herb steinbergherb steinberg
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Typically, to prove that the limit does not exist, people find distinct sublimits. If you know the limit exists and just want to find it, then you just need to consider an appropriate (read easy) sublimit.
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– Git Gud
Jan 18 at 0:49
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@GitGud Does this mean that proving a limit does not exist, as in this question, might be very difficult if we cannot identify the proper sublimits, which more or less boils down to trial & error?
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– A Slow Learner
Jan 18 at 0:59
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There are strategies you can use. For rational functions in two variables you can try looking at sublimits of the form $tmapsto (t, kt)$, $tmapsto (t, kt^2)$, $tmapsto (t, kt^3)$ and so on, with varying values of $k$. If you're unable to find different sublimits, then perhaps the sublimit exists and you can try to focus on proving it does exist. When you don't have a rational function, you can probably use Taylor's theorem to get a similarly behaved rational function.
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– Git Gud
Jan 18 at 8:38