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Limit of two real variables function $f(x,y)=|x|^$ at the origin
help with limit and lines passing through the originLimit of a function of two variablesTwo variables limit questionlimit of a function of three variablesHow to find the limit of a function involving two variables.How to find the limit of a $2$ variables function.Difficult limit of two variablesLimit at origin of function of two variablesRational limit on two variables approaching the originCompute the limit $lim_x to 0^+ (sin x)^sin x$ using L'Hopital's
$begingroup$
Any help in computing the limit of the function $|x|^y$ at the origin will be appreciated.
limits multivariable-calculus
asked Mar 29 at 20:30
M.R. YeganM.R. Yegan
436312
$endgroup$
add a comment |
$begingroup$
Any help in computing the limit of the function $|x|^y$ at the origin will be appreciated.
limits multivariable-calculus
asked Mar 29 at 20:30
M.R. YeganM.R. Yegan
436312
$endgroup$
1
$begingroup$
When $0<y<1$ we have $1/y>1$ So if $0<x<1$ we have $x^1/y < x$. Is that enough? Also: is $|x|^$ defined on the $x$-axis?
$endgroup$
– GEdgar
Mar 29 at 20:36
$begingroup$
It's OK, Dear GEdgar
$endgroup$
– M.R. Yegan
Mar 29 at 20:42
add a comment |
$begingroup$
Any help in computing the limit of the function $|x|^y$ at the origin will be appreciated.
limits multivariable-calculus
asked Mar 29 at 20:30
M.R. YeganM.R. Yegan
436312
$endgroup$
Any help in computing the limit of the function $|x|^y$ at the origin will be appreciated.
limits multivariable-calculus
limits multivariable-calculus
asked Mar 29 at 20:30
M.R. YeganM.R. Yegan
436312
asked Mar 29 at 20:30
M.R. YeganM.R. Yegan
436312
asked Mar 29 at 20:30
M.R. YeganM.R. Yegan
436312
asked Mar 29 at 20:30
M.R. YeganM.R. Yegan
436312
asked Mar 29 at 20:30
M.R. YeganM.R. Yegan
436312
436312
1
$begingroup$
When $0<y<1$ we have $1/y>1$ So if $0<x<1$ we have $x^1/y < x$. Is that enough? Also: is $|x|^$ defined on the $x$-axis?
$endgroup$
– GEdgar
Mar 29 at 20:36
$begingroup$
It's OK, Dear GEdgar
$endgroup$
– M.R. Yegan
Mar 29 at 20:42
add a comment |
1
$begingroup$
When $0<y<1$ we have $1/y>1$ So if $0<x<1$ we have $x^1/y < x$. Is that enough? Also: is $|x|^$ defined on the $x$-axis?
$endgroup$
– GEdgar
Mar 29 at 20:36
$begingroup$
It's OK, Dear GEdgar
$endgroup$
– M.R. Yegan
Mar 29 at 20:42
1
1
$begingroup$
When $0<y<1$ we have $1/y>1$ So if $0<x<1$ we have $x^1/y < x$. Is that enough? Also: is $|x|^$ defined on the $x$-axis?
$endgroup$
– GEdgar
Mar 29 at 20:36
$begingroup$
When $0<y<1$ we have $1/y>1$ So if $0<x<1$ we have $x^1/y < x$. Is that enough? Also: is $|x|^$ defined on the $x$-axis?
$endgroup$
– GEdgar
Mar 29 at 20:36
$begingroup$
It's OK, Dear GEdgar
$endgroup$
– M.R. Yegan
Mar 29 at 20:42
$begingroup$
It's OK, Dear GEdgar
$endgroup$
– M.R. Yegan
Mar 29 at 20:42
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Assume $x>0$, $y>0$, then
$$
x^1/y = e^(ln x)cdot(1/y).
$$
As $xto 0$ and $yto 0$, we have
$$
ln xto-infty, quad 1/yto + infty,
$$
therefore,
$$
(ln x)cdot(1/y)to-infty,
$$
which implies that
$$
x^1/y = e^(ln x)cdot(1/y) to 0.
$$
answered Mar 29 at 20:57
EagleToLearnEagleToLearn
484
$endgroup$
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Assume $x>0$, $y>0$, then
$$
x^1/y = e^(ln x)cdot(1/y).
$$
As $xto 0$ and $yto 0$, we have
$$
ln xto-infty, quad 1/yto + infty,
$$
therefore,
$$
(ln x)cdot(1/y)to-infty,
$$
which implies that
$$
x^1/y = e^(ln x)cdot(1/y) to 0.
$$
answered Mar 29 at 20:57
EagleToLearnEagleToLearn
484
$endgroup$
add a comment |
$begingroup$
Assume $x>0$, $y>0$, then
$$
x^1/y = e^(ln x)cdot(1/y).
$$
As $xto 0$ and $yto 0$, we have
$$
ln xto-infty, quad 1/yto + infty,
$$
therefore,
$$
(ln x)cdot(1/y)to-infty,
$$
which implies that
$$
x^1/y = e^(ln x)cdot(1/y) to 0.
$$
answered Mar 29 at 20:57
EagleToLearnEagleToLearn
484
$endgroup$
add a comment |
$begingroup$
Assume $x>0$, $y>0$, then
$$
x^1/y = e^(ln x)cdot(1/y).
$$
As $xto 0$ and $yto 0$, we have
$$
ln xto-infty, quad 1/yto + infty,
$$
therefore,
$$
(ln x)cdot(1/y)to-infty,
$$
which implies that
$$
x^1/y = e^(ln x)cdot(1/y) to 0.
$$
answered Mar 29 at 20:57
EagleToLearnEagleToLearn
484
$endgroup$
Assume $x>0$, $y>0$, then
$$
x^1/y = e^(ln x)cdot(1/y).
$$
As $xto 0$ and $yto 0$, we have
$$
ln xto-infty, quad 1/yto + infty,
$$
therefore,
$$
(ln x)cdot(1/y)to-infty,
$$
which implies that
$$
x^1/y = e^(ln x)cdot(1/y) to 0.
$$
answered Mar 29 at 20:57
EagleToLearnEagleToLearn
484
answered Mar 29 at 20:57
EagleToLearnEagleToLearn
484
answered Mar 29 at 20:57
EagleToLearnEagleToLearn
484
answered Mar 29 at 20:57
EagleToLearnEagleToLearn
484
484
add a comment |
add a comment |
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1
$begingroup$
When $0<y<1$ we have $1/y>1$ So if $0<x<1$ we have $x^1/y < x$. Is that enough? Also: is $|x|^$ defined on the $x$-axis?
$endgroup$
– GEdgar
Mar 29 at 20:36
$begingroup$
It's OK, Dear GEdgar
$endgroup$
– M.R. Yegan
Mar 29 at 20:42