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Question regarding limits of multivariable functions - is it really zero?
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I am given the following problem in a textbook, and it's a solved problem.
Find $displaystyle lim_(x,y) to (0,0) fracx^2 y^2x^4 + 3y^4$ or justify its nonexistence.
The author justifies the nonexistence of the limit in the following way: he uses the paths $x = 0$ then $y = 0$ and then, to justify its nonexistence, $y = x$. These are the author's notes:
Since
beginalign
lim_(x,0) to (0,0) fracx^2 y^2x^4 + 3y^4 = lim_(x,0) to (0,0) fracx^2 cdot 0^2x^4 + 3 0^4 = lim_(x,0) to (0,0) frac0x^4 = 0
endalign
and
beginalign
lim_(0,y) to (0,0) fracx^2 y^2x^4 + 3y^4 = lim_(0,y) to (0,0) frac0^2 cdot y^20^4 + 3 y^4 = lim_(0,y) to (0,0) frac03y^4 = 0
endalign
but
beginalign
lim_(x,x) to (0,0) fracx^2 y^2x^4 + 3y^4 = lim_(x,x) to (0,0) fracx^2 cdot x^2x^4 + 3 x^4 = lim_(x,x) to (0,0) fracx^44x^4 = frac14
endalign
then the limit does not exist.
So my question is: are these limits
beginalign
lim_(0,y) to (0,0) frac03y^4 qquad lim_(x,0) to (0,0) frac0x^4
endalign
really equal to zero? Shouldn't they be indeterminate, given that the variable on the denominator approaches zero?
limits multivariable-calculus limits-without-lhopital
asked Aug 2 at 22:04
bru1987
930818
add a comment |Â
up vote
0
down vote
favorite
I am given the following problem in a textbook, and it's a solved problem.
Find $displaystyle lim_(x,y) to (0,0) fracx^2 y^2x^4 + 3y^4$ or justify its nonexistence.
The author justifies the nonexistence of the limit in the following way: he uses the paths $x = 0$ then $y = 0$ and then, to justify its nonexistence, $y = x$. These are the author's notes:
Since
beginalign
lim_(x,0) to (0,0) fracx^2 y^2x^4 + 3y^4 = lim_(x,0) to (0,0) fracx^2 cdot 0^2x^4 + 3 0^4 = lim_(x,0) to (0,0) frac0x^4 = 0
endalign
and
beginalign
lim_(0,y) to (0,0) fracx^2 y^2x^4 + 3y^4 = lim_(0,y) to (0,0) frac0^2 cdot y^20^4 + 3 y^4 = lim_(0,y) to (0,0) frac03y^4 = 0
endalign
but
beginalign
lim_(x,x) to (0,0) fracx^2 y^2x^4 + 3y^4 = lim_(x,x) to (0,0) fracx^2 cdot x^2x^4 + 3 x^4 = lim_(x,x) to (0,0) fracx^44x^4 = frac14
endalign
then the limit does not exist.
So my question is: are these limits
beginalign
lim_(0,y) to (0,0) frac03y^4 qquad lim_(x,0) to (0,0) frac0x^4
endalign
really equal to zero? Shouldn't they be indeterminate, given that the variable on the denominator approaches zero?
limits multivariable-calculus limits-without-lhopital
asked Aug 2 at 22:04
bru1987
930818
Well they approach to $0$, but aren't $0$, what you should have written in the next step is that e.g. $lim 0/3y^4 = lim 0 = 0$
– Nameless
Aug 3 at 9:37
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am given the following problem in a textbook, and it's a solved problem.
Find $displaystyle lim_(x,y) to (0,0) fracx^2 y^2x^4 + 3y^4$ or justify its nonexistence.
The author justifies the nonexistence of the limit in the following way: he uses the paths $x = 0$ then $y = 0$ and then, to justify its nonexistence, $y = x$. These are the author's notes:
Since
beginalign
lim_(x,0) to (0,0) fracx^2 y^2x^4 + 3y^4 = lim_(x,0) to (0,0) fracx^2 cdot 0^2x^4 + 3 0^4 = lim_(x,0) to (0,0) frac0x^4 = 0
endalign
and
beginalign
lim_(0,y) to (0,0) fracx^2 y^2x^4 + 3y^4 = lim_(0,y) to (0,0) frac0^2 cdot y^20^4 + 3 y^4 = lim_(0,y) to (0,0) frac03y^4 = 0
endalign
but
beginalign
lim_(x,x) to (0,0) fracx^2 y^2x^4 + 3y^4 = lim_(x,x) to (0,0) fracx^2 cdot x^2x^4 + 3 x^4 = lim_(x,x) to (0,0) fracx^44x^4 = frac14
endalign
then the limit does not exist.
So my question is: are these limits
beginalign
lim_(0,y) to (0,0) frac03y^4 qquad lim_(x,0) to (0,0) frac0x^4
endalign
really equal to zero? Shouldn't they be indeterminate, given that the variable on the denominator approaches zero?
limits multivariable-calculus limits-without-lhopital
asked Aug 2 at 22:04
bru1987
930818
I am given the following problem in a textbook, and it's a solved problem.
Find $displaystyle lim_(x,y) to (0,0) fracx^2 y^2x^4 + 3y^4$ or justify its nonexistence.
The author justifies the nonexistence of the limit in the following way: he uses the paths $x = 0$ then $y = 0$ and then, to justify its nonexistence, $y = x$. These are the author's notes:
Since
beginalign
lim_(x,0) to (0,0) fracx^2 y^2x^4 + 3y^4 = lim_(x,0) to (0,0) fracx^2 cdot 0^2x^4 + 3 0^4 = lim_(x,0) to (0,0) frac0x^4 = 0
endalign
and
beginalign
lim_(0,y) to (0,0) fracx^2 y^2x^4 + 3y^4 = lim_(0,y) to (0,0) frac0^2 cdot y^20^4 + 3 y^4 = lim_(0,y) to (0,0) frac03y^4 = 0
endalign
but
beginalign
lim_(x,x) to (0,0) fracx^2 y^2x^4 + 3y^4 = lim_(x,x) to (0,0) fracx^2 cdot x^2x^4 + 3 x^4 = lim_(x,x) to (0,0) fracx^44x^4 = frac14
endalign
then the limit does not exist.
So my question is: are these limits
beginalign
lim_(0,y) to (0,0) frac03y^4 qquad lim_(x,0) to (0,0) frac0x^4
endalign
really equal to zero? Shouldn't they be indeterminate, given that the variable on the denominator approaches zero?
limits multivariable-calculus limits-without-lhopital
asked Aug 2 at 22:04
bru1987
930818
asked Aug 2 at 22:04
bru1987
930818
asked Aug 2 at 22:04
bru1987
930818
asked Aug 2 at 22:04
bru1987
930818
930818
Well they approach to $0$, but aren't $0$, what you should have written in the next step is that e.g. $lim 0/3y^4 = lim 0 = 0$
– Nameless
Aug 3 at 9:37
add a comment |Â
Well they approach to $0$, but aren't $0$, what you should have written in the next step is that e.g. $lim 0/3y^4 = lim 0 = 0$
– Nameless
Aug 3 at 9:37
Well they approach to $0$, but aren't $0$, what you should have written in the next step is that e.g. $lim 0/3y^4 = lim 0 = 0$
– Nameless
Aug 3 at 9:37
Well they approach to $0$, but aren't $0$, what you should have written in the next step is that e.g. $lim 0/3y^4 = lim 0 = 0$
– Nameless
Aug 3 at 9:37
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
No, both those limits are indeed defined and equal to zero.
The multivariable context is adding unnecessary "noise" here. Consider the simpler expression $$lim_xrightarrow 00over x,$$ which already captures what's going on.
Remember the precise definition of a limit. Colloquially, we often say something like "$lim_xrightarrow af(x)=L$ iff I can get $f$ to be as close as I want to $L$ just by requiring that $x$ be sufficiently close to $a$." However, this misses a very important point which is explicit in the precise formal definition: $$lim_xrightarrow af(x)=Liff forall epsilon>0exists delta>0forall x((colorred0<vert x-avert<delta)implies (vert f(x)-Lvert<epsilon)).$$ Note the "$0<$" clause in the left hand side: what this says is that we ignore the specific behavior of $f$ at exactly $a$.
This is why $lim_xrightarrow 00over x=0$, despite the bad behavior when $x$ is exactly $0$: for any $epsilon>0$, let $delta=17$ (say); then it's easy to check that for any $x$ within $17$ of $0$ but not actually equal to $0$, we have $vert0over x-0vert<epsilon$.
A reasonable question at this point is why we define limits this way. Ultimately that's a question that deserves a serious answer, but very briefly the point is that we're interested in what $f$ "ought to be" at $a$, not what it literally is. Both the derivative and the integral give great examples of this: in each case, we really want to divide by zero but that gives nonsense, so instead we look at what happens when the change in $x$ or the width of our rectangles gets really close to zero without actually being zero. This also gives a nice spin on continuity: intuitively, a function is continuous if it always is what it ought to be.
answered Aug 2 at 22:15
Noah Schweber
110k9138259
add a comment |Â
up vote
1
down vote
Indeed, $lim_limits(0,y)to (0,0) frac 03y^4 = 0$ since for all $y$ in the neighborhood of $0, f(0,y) < epsilon$
However that is not sufficient for $lim_limits(x,y)to (0,0) frac x^2y^2x^4+3y^4$ . Because
$lim_limits(y,y)to (0,0) = frac y^44y^4 = frac 14.$
If the limit exists, it must be have the same value (be within $epsilon$ of it at least) for all $x,y$ in the neighborhood.
answered Aug 2 at 22:14
Doug M
39k31748
add a comment |Â
up vote
-1
down vote
Both limit
$$beginalign
lim_(0,y) to (0,0) frac03y^4=0 qquad lim_(x,0) to (0,0) frac0x^4=0
endalign$$
are correct since the function $f(x,y)$ is identically zero along the trajectories with either $x=0$ or $y=0$.
In other words, what we cannot consider is $x=y=0$ at the same time but all the trajectories with either $x=0$ or $y=0$ are admissible and consistent with the definition of limit.
To clarify better that point see also the related Why are we allowed to cancel fractions in limits?
answered Aug 2 at 22:14
gimusi
63.8k73480
1
This is not answering the actual question, introduced in bold at the end of the OP: 'So my question is: are these limits [...] really equal to zero? Shouldn't they be indeterminate, given that the variable on the denominator approaches zero?"
– Clement C.
Aug 2 at 22:16
@ClementC. The answer was already contained in the point $$x=0 implies fracx^2 y^2x^4 + 3y^4=0$$ Moreover I've added a supplementary comment at the end to better aswer the main question.
– gimusi
Aug 2 at 22:19
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
No, both those limits are indeed defined and equal to zero.
The multivariable context is adding unnecessary "noise" here. Consider the simpler expression $$lim_xrightarrow 00over x,$$ which already captures what's going on.
Remember the precise definition of a limit. Colloquially, we often say something like "$lim_xrightarrow af(x)=L$ iff I can get $f$ to be as close as I want to $L$ just by requiring that $x$ be sufficiently close to $a$." However, this misses a very important point which is explicit in the precise formal definition: $$lim_xrightarrow af(x)=Liff forall epsilon>0exists delta>0forall x((colorred0<vert x-avert<delta)implies (vert f(x)-Lvert<epsilon)).$$ Note the "$0<$" clause in the left hand side: what this says is that we ignore the specific behavior of $f$ at exactly $a$.
This is why $lim_xrightarrow 00over x=0$, despite the bad behavior when $x$ is exactly $0$: for any $epsilon>0$, let $delta=17$ (say); then it's easy to check that for any $x$ within $17$ of $0$ but not actually equal to $0$, we have $vert0over x-0vert<epsilon$.
A reasonable question at this point is why we define limits this way. Ultimately that's a question that deserves a serious answer, but very briefly the point is that we're interested in what $f$ "ought to be" at $a$, not what it literally is. Both the derivative and the integral give great examples of this: in each case, we really want to divide by zero but that gives nonsense, so instead we look at what happens when the change in $x$ or the width of our rectangles gets really close to zero without actually being zero. This also gives a nice spin on continuity: intuitively, a function is continuous if it always is what it ought to be.
answered Aug 2 at 22:15
Noah Schweber
110k9138259
add a comment |Â
up vote
2
down vote
accepted
No, both those limits are indeed defined and equal to zero.
The multivariable context is adding unnecessary "noise" here. Consider the simpler expression $$lim_xrightarrow 00over x,$$ which already captures what's going on.
Remember the precise definition of a limit. Colloquially, we often say something like "$lim_xrightarrow af(x)=L$ iff I can get $f$ to be as close as I want to $L$ just by requiring that $x$ be sufficiently close to $a$." However, this misses a very important point which is explicit in the precise formal definition: $$lim_xrightarrow af(x)=Liff forall epsilon>0exists delta>0forall x((colorred0<vert x-avert<delta)implies (vert f(x)-Lvert<epsilon)).$$ Note the "$0<$" clause in the left hand side: what this says is that we ignore the specific behavior of $f$ at exactly $a$.
This is why $lim_xrightarrow 00over x=0$, despite the bad behavior when $x$ is exactly $0$: for any $epsilon>0$, let $delta=17$ (say); then it's easy to check that for any $x$ within $17$ of $0$ but not actually equal to $0$, we have $vert0over x-0vert<epsilon$.
A reasonable question at this point is why we define limits this way. Ultimately that's a question that deserves a serious answer, but very briefly the point is that we're interested in what $f$ "ought to be" at $a$, not what it literally is. Both the derivative and the integral give great examples of this: in each case, we really want to divide by zero but that gives nonsense, so instead we look at what happens when the change in $x$ or the width of our rectangles gets really close to zero without actually being zero. This also gives a nice spin on continuity: intuitively, a function is continuous if it always is what it ought to be.
answered Aug 2 at 22:15
Noah Schweber
110k9138259
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
No, both those limits are indeed defined and equal to zero.
The multivariable context is adding unnecessary "noise" here. Consider the simpler expression $$lim_xrightarrow 00over x,$$ which already captures what's going on.
Remember the precise definition of a limit. Colloquially, we often say something like "$lim_xrightarrow af(x)=L$ iff I can get $f$ to be as close as I want to $L$ just by requiring that $x$ be sufficiently close to $a$." However, this misses a very important point which is explicit in the precise formal definition: $$lim_xrightarrow af(x)=Liff forall epsilon>0exists delta>0forall x((colorred0<vert x-avert<delta)implies (vert f(x)-Lvert<epsilon)).$$ Note the "$0<$" clause in the left hand side: what this says is that we ignore the specific behavior of $f$ at exactly $a$.
This is why $lim_xrightarrow 00over x=0$, despite the bad behavior when $x$ is exactly $0$: for any $epsilon>0$, let $delta=17$ (say); then it's easy to check that for any $x$ within $17$ of $0$ but not actually equal to $0$, we have $vert0over x-0vert<epsilon$.
A reasonable question at this point is why we define limits this way. Ultimately that's a question that deserves a serious answer, but very briefly the point is that we're interested in what $f$ "ought to be" at $a$, not what it literally is. Both the derivative and the integral give great examples of this: in each case, we really want to divide by zero but that gives nonsense, so instead we look at what happens when the change in $x$ or the width of our rectangles gets really close to zero without actually being zero. This also gives a nice spin on continuity: intuitively, a function is continuous if it always is what it ought to be.
answered Aug 2 at 22:15
Noah Schweber
110k9138259
No, both those limits are indeed defined and equal to zero.
The multivariable context is adding unnecessary "noise" here. Consider the simpler expression $$lim_xrightarrow 00over x,$$ which already captures what's going on.
Remember the precise definition of a limit. Colloquially, we often say something like "$lim_xrightarrow af(x)=L$ iff I can get $f$ to be as close as I want to $L$ just by requiring that $x$ be sufficiently close to $a$." However, this misses a very important point which is explicit in the precise formal definition: $$lim_xrightarrow af(x)=Liff forall epsilon>0exists delta>0forall x((colorred0<vert x-avert<delta)implies (vert f(x)-Lvert<epsilon)).$$ Note the "$0<$" clause in the left hand side: what this says is that we ignore the specific behavior of $f$ at exactly $a$.
This is why $lim_xrightarrow 00over x=0$, despite the bad behavior when $x$ is exactly $0$: for any $epsilon>0$, let $delta=17$ (say); then it's easy to check that for any $x$ within $17$ of $0$ but not actually equal to $0$, we have $vert0over x-0vert<epsilon$.
A reasonable question at this point is why we define limits this way. Ultimately that's a question that deserves a serious answer, but very briefly the point is that we're interested in what $f$ "ought to be" at $a$, not what it literally is. Both the derivative and the integral give great examples of this: in each case, we really want to divide by zero but that gives nonsense, so instead we look at what happens when the change in $x$ or the width of our rectangles gets really close to zero without actually being zero. This also gives a nice spin on continuity: intuitively, a function is continuous if it always is what it ought to be.
answered Aug 2 at 22:15
Noah Schweber
110k9138259
edited Aug 2 at 22:21
answered Aug 2 at 22:15
Noah Schweber
110k9138259
answered Aug 2 at 22:15
Noah Schweber
110k9138259
answered Aug 2 at 22:15
Noah Schweber
110k9138259
110k9138259
add a comment |Â
add a comment |Â
up vote
1
down vote
Indeed, $lim_limits(0,y)to (0,0) frac 03y^4 = 0$ since for all $y$ in the neighborhood of $0, f(0,y) < epsilon$
However that is not sufficient for $lim_limits(x,y)to (0,0) frac x^2y^2x^4+3y^4$ . Because
$lim_limits(y,y)to (0,0) = frac y^44y^4 = frac 14.$
If the limit exists, it must be have the same value (be within $epsilon$ of it at least) for all $x,y$ in the neighborhood.
answered Aug 2 at 22:14
Doug M
39k31748
add a comment |Â
up vote
1
down vote
Indeed, $lim_limits(0,y)to (0,0) frac 03y^4 = 0$ since for all $y$ in the neighborhood of $0, f(0,y) < epsilon$
However that is not sufficient for $lim_limits(x,y)to (0,0) frac x^2y^2x^4+3y^4$ . Because
$lim_limits(y,y)to (0,0) = frac y^44y^4 = frac 14.$
If the limit exists, it must be have the same value (be within $epsilon$ of it at least) for all $x,y$ in the neighborhood.
answered Aug 2 at 22:14
Doug M
39k31748
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Indeed, $lim_limits(0,y)to (0,0) frac 03y^4 = 0$ since for all $y$ in the neighborhood of $0, f(0,y) < epsilon$
However that is not sufficient for $lim_limits(x,y)to (0,0) frac x^2y^2x^4+3y^4$ . Because
$lim_limits(y,y)to (0,0) = frac y^44y^4 = frac 14.$
If the limit exists, it must be have the same value (be within $epsilon$ of it at least) for all $x,y$ in the neighborhood.
answered Aug 2 at 22:14
Doug M
39k31748
Indeed, $lim_limits(0,y)to (0,0) frac 03y^4 = 0$ since for all $y$ in the neighborhood of $0, f(0,y) < epsilon$
However that is not sufficient for $lim_limits(x,y)to (0,0) frac x^2y^2x^4+3y^4$ . Because
$lim_limits(y,y)to (0,0) = frac y^44y^4 = frac 14.$
If the limit exists, it must be have the same value (be within $epsilon$ of it at least) for all $x,y$ in the neighborhood.
answered Aug 2 at 22:14
Doug M
39k31748
answered Aug 2 at 22:14
Doug M
39k31748
answered Aug 2 at 22:14
Doug M
39k31748
answered Aug 2 at 22:14
Doug M
39k31748
39k31748
add a comment |Â
add a comment |Â
up vote
-1
down vote
Both limit
$$beginalign
lim_(0,y) to (0,0) frac03y^4=0 qquad lim_(x,0) to (0,0) frac0x^4=0
endalign$$
are correct since the function $f(x,y)$ is identically zero along the trajectories with either $x=0$ or $y=0$.
In other words, what we cannot consider is $x=y=0$ at the same time but all the trajectories with either $x=0$ or $y=0$ are admissible and consistent with the definition of limit.
To clarify better that point see also the related Why are we allowed to cancel fractions in limits?
answered Aug 2 at 22:14
gimusi
63.8k73480
1
This is not answering the actual question, introduced in bold at the end of the OP: 'So my question is: are these limits [...] really equal to zero? Shouldn't they be indeterminate, given that the variable on the denominator approaches zero?"
– Clement C.
Aug 2 at 22:16
@ClementC. The answer was already contained in the point $$x=0 implies fracx^2 y^2x^4 + 3y^4=0$$ Moreover I've added a supplementary comment at the end to better aswer the main question.
– gimusi
Aug 2 at 22:19
add a comment |Â
up vote
-1
down vote
Both limit
$$beginalign
lim_(0,y) to (0,0) frac03y^4=0 qquad lim_(x,0) to (0,0) frac0x^4=0
endalign$$
are correct since the function $f(x,y)$ is identically zero along the trajectories with either $x=0$ or $y=0$.
In other words, what we cannot consider is $x=y=0$ at the same time but all the trajectories with either $x=0$ or $y=0$ are admissible and consistent with the definition of limit.
To clarify better that point see also the related Why are we allowed to cancel fractions in limits?
answered Aug 2 at 22:14
gimusi
63.8k73480
1
This is not answering the actual question, introduced in bold at the end of the OP: 'So my question is: are these limits [...] really equal to zero? Shouldn't they be indeterminate, given that the variable on the denominator approaches zero?"
– Clement C.
Aug 2 at 22:16
@ClementC. The answer was already contained in the point $$x=0 implies fracx^2 y^2x^4 + 3y^4=0$$ Moreover I've added a supplementary comment at the end to better aswer the main question.
– gimusi
Aug 2 at 22:19
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
Both limit
$$beginalign
lim_(0,y) to (0,0) frac03y^4=0 qquad lim_(x,0) to (0,0) frac0x^4=0
endalign$$
are correct since the function $f(x,y)$ is identically zero along the trajectories with either $x=0$ or $y=0$.
In other words, what we cannot consider is $x=y=0$ at the same time but all the trajectories with either $x=0$ or $y=0$ are admissible and consistent with the definition of limit.
To clarify better that point see also the related Why are we allowed to cancel fractions in limits?
answered Aug 2 at 22:14
gimusi
63.8k73480
Both limit
$$beginalign
lim_(0,y) to (0,0) frac03y^4=0 qquad lim_(x,0) to (0,0) frac0x^4=0
endalign$$
are correct since the function $f(x,y)$ is identically zero along the trajectories with either $x=0$ or $y=0$.
In other words, what we cannot consider is $x=y=0$ at the same time but all the trajectories with either $x=0$ or $y=0$ are admissible and consistent with the definition of limit.
To clarify better that point see also the related Why are we allowed to cancel fractions in limits?
answered Aug 2 at 22:14
gimusi
63.8k73480
edited Aug 3 at 9:28
answered Aug 2 at 22:14
gimusi
63.8k73480
answered Aug 2 at 22:14
gimusi
63.8k73480
answered Aug 2 at 22:14
gimusi
63.8k73480
63.8k73480
1
This is not answering the actual question, introduced in bold at the end of the OP: 'So my question is: are these limits [...] really equal to zero? Shouldn't they be indeterminate, given that the variable on the denominator approaches zero?"
– Clement C.
Aug 2 at 22:16
@ClementC. The answer was already contained in the point $$x=0 implies fracx^2 y^2x^4 + 3y^4=0$$ Moreover I've added a supplementary comment at the end to better aswer the main question.
– gimusi
Aug 2 at 22:19
add a comment |Â
1
This is not answering the actual question, introduced in bold at the end of the OP: 'So my question is: are these limits [...] really equal to zero? Shouldn't they be indeterminate, given that the variable on the denominator approaches zero?"
– Clement C.
Aug 2 at 22:16
@ClementC. The answer was already contained in the point $$x=0 implies fracx^2 y^2x^4 + 3y^4=0$$ Moreover I've added a supplementary comment at the end to better aswer the main question.
– gimusi
Aug 2 at 22:19
1
1
This is not answering the actual question, introduced in bold at the end of the OP: 'So my question is: are these limits [...] really equal to zero? Shouldn't they be indeterminate, given that the variable on the denominator approaches zero?"
– Clement C.
Aug 2 at 22:16
This is not answering the actual question, introduced in bold at the end of the OP: 'So my question is: are these limits [...] really equal to zero? Shouldn't they be indeterminate, given that the variable on the denominator approaches zero?"
– Clement C.
Aug 2 at 22:16
@ClementC. The answer was already contained in the point $$x=0 implies fracx^2 y^2x^4 + 3y^4=0$$ Moreover I've added a supplementary comment at the end to better aswer the main question.
– gimusi
Aug 2 at 22:19
@ClementC. The answer was already contained in the point $$x=0 implies fracx^2 y^2x^4 + 3y^4=0$$ Moreover I've added a supplementary comment at the end to better aswer the main question.
– gimusi
Aug 2 at 22:19
add a comment |Â
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Well they approach to $0$, but aren't $0$, what you should have written in the next step is that e.g. $lim 0/3y^4 = lim 0 = 0$
– Nameless
Aug 3 at 9:37