$lim_{(x,y)to(0,0)}frac{x^2y^2}{sin(x)cos(y)}$ is this done correctly?
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$lim_{(x,y)to(0,0)}frac{x^2y^2}{sin(x)cos(y)}$ is it allowed to split a multi-variable limit into its component variables as in the next step?
$= (lim_{xto0}frac{x^2}{sin(x)})(lim_{xto0}frac{y^2}{sin(y)})$ this is an indeterminate form and now I use L'Hopital
$=(lim_{xto0}frac{2x}{cos(x)})(lim_{xto0}frac{2y}{cos(y)})$
$=(frac{0}{1})(frac{0}{1})=0$
calculus limits multivariable-calculus
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favorite
$lim_{(x,y)to(0,0)}frac{x^2y^2}{sin(x)cos(y)}$ is it allowed to split a multi-variable limit into its component variables as in the next step?
$= (lim_{xto0}frac{x^2}{sin(x)})(lim_{xto0}frac{y^2}{sin(y)})$ this is an indeterminate form and now I use L'Hopital
$=(lim_{xto0}frac{2x}{cos(x)})(lim_{xto0}frac{2y}{cos(y)})$
$=(frac{0}{1})(frac{0}{1})=0$
calculus limits multivariable-calculus
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up vote
0
down vote
favorite
up vote
0
down vote
favorite
$lim_{(x,y)to(0,0)}frac{x^2y^2}{sin(x)cos(y)}$ is it allowed to split a multi-variable limit into its component variables as in the next step?
$= (lim_{xto0}frac{x^2}{sin(x)})(lim_{xto0}frac{y^2}{sin(y)})$ this is an indeterminate form and now I use L'Hopital
$=(lim_{xto0}frac{2x}{cos(x)})(lim_{xto0}frac{2y}{cos(y)})$
$=(frac{0}{1})(frac{0}{1})=0$
calculus limits multivariable-calculus
$lim_{(x,y)to(0,0)}frac{x^2y^2}{sin(x)cos(y)}$ is it allowed to split a multi-variable limit into its component variables as in the next step?
$= (lim_{xto0}frac{x^2}{sin(x)})(lim_{xto0}frac{y^2}{sin(y)})$ this is an indeterminate form and now I use L'Hopital
$=(lim_{xto0}frac{2x}{cos(x)})(lim_{xto0}frac{2y}{cos(y)})$
$=(frac{0}{1})(frac{0}{1})=0$
calculus limits multivariable-calculus
calculus limits multivariable-calculus
edited Nov 23 at 15:27
Martin Sleziak
44.5k7115268
44.5k7115268
asked Feb 7 '16 at 8:52
Gobabis
622517
622517
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3 Answers
3
active
oldest
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up vote
2
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accepted
Even switching the order of taking a limit can lead to problems.
As an example: $ limlimits_{m to infty} limlimits_{n to infty} {frac{1}{n}} ^{frac{1}{m}} = limlimits_{m to infty} 0 ^{frac{1}{m}} = 0 neq 1 = limlimits_{n to infty} 1 = limlimits_{n to infty} limlimits_{m to infty} frac{1}{n} ^frac{1}{m} $
While: $ limlimits_{(m,n) to (infty, infty)} {frac{1}{n}} ^{frac{1}{m}}$ does not exist.
At this point I would like to refer to this post well written post and hope this answers some further questions of yours: https://math.stackexchange.com/q/15257
add a comment |
up vote
3
down vote
No, it is generally not allowed as it may be false. But you can do as follows:
$$lim_{(x,y)to(0,0)};xcdotfrac x{sin x}cdotfrac{y^2}{cos y}=0cdot1cdotfrac01=0$$
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It is actually easier:
$$
left|frac{x^2 y^2}{sin x cos y}right| leq 2left|frac{x}{sin x} right| left| y^2 right| left| x right| leq
|xy^2| =o(1)
$$
as $(x,y) to (0,0)$.
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Even switching the order of taking a limit can lead to problems.
As an example: $ limlimits_{m to infty} limlimits_{n to infty} {frac{1}{n}} ^{frac{1}{m}} = limlimits_{m to infty} 0 ^{frac{1}{m}} = 0 neq 1 = limlimits_{n to infty} 1 = limlimits_{n to infty} limlimits_{m to infty} frac{1}{n} ^frac{1}{m} $
While: $ limlimits_{(m,n) to (infty, infty)} {frac{1}{n}} ^{frac{1}{m}}$ does not exist.
At this point I would like to refer to this post well written post and hope this answers some further questions of yours: https://math.stackexchange.com/q/15257
add a comment |
up vote
2
down vote
accepted
Even switching the order of taking a limit can lead to problems.
As an example: $ limlimits_{m to infty} limlimits_{n to infty} {frac{1}{n}} ^{frac{1}{m}} = limlimits_{m to infty} 0 ^{frac{1}{m}} = 0 neq 1 = limlimits_{n to infty} 1 = limlimits_{n to infty} limlimits_{m to infty} frac{1}{n} ^frac{1}{m} $
While: $ limlimits_{(m,n) to (infty, infty)} {frac{1}{n}} ^{frac{1}{m}}$ does not exist.
At this point I would like to refer to this post well written post and hope this answers some further questions of yours: https://math.stackexchange.com/q/15257
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Even switching the order of taking a limit can lead to problems.
As an example: $ limlimits_{m to infty} limlimits_{n to infty} {frac{1}{n}} ^{frac{1}{m}} = limlimits_{m to infty} 0 ^{frac{1}{m}} = 0 neq 1 = limlimits_{n to infty} 1 = limlimits_{n to infty} limlimits_{m to infty} frac{1}{n} ^frac{1}{m} $
While: $ limlimits_{(m,n) to (infty, infty)} {frac{1}{n}} ^{frac{1}{m}}$ does not exist.
At this point I would like to refer to this post well written post and hope this answers some further questions of yours: https://math.stackexchange.com/q/15257
Even switching the order of taking a limit can lead to problems.
As an example: $ limlimits_{m to infty} limlimits_{n to infty} {frac{1}{n}} ^{frac{1}{m}} = limlimits_{m to infty} 0 ^{frac{1}{m}} = 0 neq 1 = limlimits_{n to infty} 1 = limlimits_{n to infty} limlimits_{m to infty} frac{1}{n} ^frac{1}{m} $
While: $ limlimits_{(m,n) to (infty, infty)} {frac{1}{n}} ^{frac{1}{m}}$ does not exist.
At this point I would like to refer to this post well written post and hope this answers some further questions of yours: https://math.stackexchange.com/q/15257
edited Apr 13 '17 at 12:21
Community♦
1
1
answered Feb 7 '16 at 9:34
Imago
1,25411019
1,25411019
add a comment |
add a comment |
up vote
3
down vote
No, it is generally not allowed as it may be false. But you can do as follows:
$$lim_{(x,y)to(0,0)};xcdotfrac x{sin x}cdotfrac{y^2}{cos y}=0cdot1cdotfrac01=0$$
add a comment |
up vote
3
down vote
No, it is generally not allowed as it may be false. But you can do as follows:
$$lim_{(x,y)to(0,0)};xcdotfrac x{sin x}cdotfrac{y^2}{cos y}=0cdot1cdotfrac01=0$$
add a comment |
up vote
3
down vote
up vote
3
down vote
No, it is generally not allowed as it may be false. But you can do as follows:
$$lim_{(x,y)to(0,0)};xcdotfrac x{sin x}cdotfrac{y^2}{cos y}=0cdot1cdotfrac01=0$$
No, it is generally not allowed as it may be false. But you can do as follows:
$$lim_{(x,y)to(0,0)};xcdotfrac x{sin x}cdotfrac{y^2}{cos y}=0cdot1cdotfrac01=0$$
answered Feb 7 '16 at 9:01
DonAntonio
176k1491224
176k1491224
add a comment |
add a comment |
up vote
0
down vote
It is actually easier:
$$
left|frac{x^2 y^2}{sin x cos y}right| leq 2left|frac{x}{sin x} right| left| y^2 right| left| x right| leq
|xy^2| =o(1)
$$
as $(x,y) to (0,0)$.
add a comment |
up vote
0
down vote
It is actually easier:
$$
left|frac{x^2 y^2}{sin x cos y}right| leq 2left|frac{x}{sin x} right| left| y^2 right| left| x right| leq
|xy^2| =o(1)
$$
as $(x,y) to (0,0)$.
add a comment |
up vote
0
down vote
up vote
0
down vote
It is actually easier:
$$
left|frac{x^2 y^2}{sin x cos y}right| leq 2left|frac{x}{sin x} right| left| y^2 right| left| x right| leq
|xy^2| =o(1)
$$
as $(x,y) to (0,0)$.
It is actually easier:
$$
left|frac{x^2 y^2}{sin x cos y}right| leq 2left|frac{x}{sin x} right| left| y^2 right| left| x right| leq
|xy^2| =o(1)
$$
as $(x,y) to (0,0)$.
answered Feb 7 '16 at 9:02
Siminore
30.2k23368
30.2k23368
add a comment |
add a comment |
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