Find existence of limit
Show that $left(frac{1}{|x|} +frac{1}{|y |}right)$ tends to infinity as $(x,y)$ tends to $(0,0)$.
I have used $x=rcos alpha$ and $y=rsin alpha$, $r>0$.But I got stuck as $left(frac{1}{|cos alpha|} +frac{1}{ |rsin alpha|}right) left(frac{1}{r}right)$ lies between $frac{1}{r}$ and infinity.
calculus limits
edited Dec 16 at 11:54
Jam
4,94711431
asked Dec 13 at 14:32
Kashmira
463
add a comment |
Show that $left(frac{1}{|x|} +frac{1}{|y |}right)$ tends to infinity as $(x,y)$ tends to $(0,0)$.
I have used $x=rcos alpha$ and $y=rsin alpha$, $r>0$.But I got stuck as $left(frac{1}{|cos alpha|} +frac{1}{ |rsin alpha|}right) left(frac{1}{r}right)$ lies between $frac{1}{r}$ and infinity.
calculus limits
edited Dec 16 at 11:54
Jam
4,94711431
asked Dec 13 at 14:32
Kashmira
463
3
"I got stuck as" has a mistake with the $r$ in the denominator.
– Teepeemm
Dec 13 at 20:31
add a comment |
Show that $left(frac{1}{|x|} +frac{1}{|y |}right)$ tends to infinity as $(x,y)$ tends to $(0,0)$.
I have used $x=rcos alpha$ and $y=rsin alpha$, $r>0$.But I got stuck as $left(frac{1}{|cos alpha|} +frac{1}{ |rsin alpha|}right) left(frac{1}{r}right)$ lies between $frac{1}{r}$ and infinity.
calculus limits
edited Dec 16 at 11:54
Jam
4,94711431
asked Dec 13 at 14:32
Kashmira
463
Show that $left(frac{1}{|x|} +frac{1}{|y |}right)$ tends to infinity as $(x,y)$ tends to $(0,0)$.
I have used $x=rcos alpha$ and $y=rsin alpha$, $r>0$.But I got stuck as $left(frac{1}{|cos alpha|} +frac{1}{ |rsin alpha|}right) left(frac{1}{r}right)$ lies between $frac{1}{r}$ and infinity.
calculus limits
calculus limits
edited Dec 16 at 11:54
Jam
4,94711431
asked Dec 13 at 14:32
Kashmira
463
edited Dec 16 at 11:54
Jam
4,94711431
asked Dec 13 at 14:32
Kashmira
463
edited Dec 16 at 11:54
Jam
4,94711431
edited Dec 16 at 11:54
Jam
4,94711431
edited Dec 16 at 11:54
Jam
4,94711431
4,94711431
asked Dec 13 at 14:32
Kashmira
463
asked Dec 13 at 14:32
Kashmira
463
asked Dec 13 at 14:32
Kashmira
463
463
3
"I got stuck as" has a mistake with the $r$ in the denominator.
– Teepeemm
Dec 13 at 20:31
add a comment |
3
"I got stuck as" has a mistake with the $r$ in the denominator.
– Teepeemm
Dec 13 at 20:31
3
3
"I got stuck as" has a mistake with the $r$ in the denominator.
– Teepeemm
Dec 13 at 20:31
"I got stuck as" has a mistake with the $r$ in the denominator.
– Teepeemm
Dec 13 at 20:31
add a comment |
4 Answers
4
active
oldest
votes
Note that
$$frac1{|x|}+frac1{|y|}ge frac1{|x|}to infty$$
answered Dec 13 at 14:37
gimusi
1
add a comment |
Take the individual limits of each component. x and y are tending towards zero (the direction doesn't matter due to the magnitudes) so $frac{1}{|x|}$ and $frac{1}{|y|}$ both tend to infinity, and as a result so does their sum.
EDIT:
As suggested in the comments, this is an expansion as to why you can seperate the limits.
$$lim_{(x,y)to(0,0)}{bigg(frac{1}{|x|}+frac{1}{|y|}bigg)}=lim_{(x,y)to(0,0)}{bigg(frac{1}{|x|}bigg)}+lim_{(x,y)to(0,0)}{bigg(frac{1}{|y|}bigg)}$$
Examining the LHS we see that the first limit is independent of y, and the second independent of x. As a result we can rewrite the equation as follows:
$$=lim_{xto0}{bigg(frac{1}{|x|}bigg)}+lim_{yto0}{bigg(frac{1}{|y|}bigg)}$$
As stated above, the magnitudes mean that both limits tend to infinity as x and y tend to zero from either direction, so their sum is necessarily also infinity.
answered Dec 13 at 14:35
M.M.
637
2
You are right, but a few more words about why taking "individual limits of each component" can replace $(x,y)to (0,0)$ would be helpful in the present circumstance.
– hardmath
Dec 13 at 14:44
Thank you for the advice, I'll add an edit expanding on that.
– M.M.
Dec 14 at 10:06
add a comment |
Hint:
Use the fact that if $|(x,y)| < epsilon$, then $|x|<epsilon$ and $|y|<epsilon$
answered Dec 13 at 14:34
5xum
89.5k393161
add a comment |
We have:
$(x,y) rightarrow 0.$
Let $epsilon_n =1/n$, $n$ positive integer, be given.
Then $|(x^2+y^2)^{1/2}| < delta_n$ $(=epsilon_n)$ implies
$|(x^2+y^2)^{1/2}| < epsilon_n =1/n$, or
$n=1/epsilon_n lt dfrac{1}{(x^2+y^2)^{1/2}}.$
Note:
$(x^2+y^2)^{1/2} ge |x|$, and
$(x^2+y^2)^{1/2} ge |y|.$
Then
$dfrac{2}{(x^2+y^2)^{1/2}} le dfrac{1}{|x|}+dfrac{1}{|y|}.$
And finally :
$|(x^2+y^2)^{1/2}| lt delta_n$ implies
$2n =2/epsilon_n lt dfrac{2}{(x^2+y^2)^{1/2}} le $
$dfrac{1}{|x|} +dfrac{1}{|y|}.$
answered Dec 13 at 16:57
Peter Szilas
10.6k2720
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
Note that
$$frac1{|x|}+frac1{|y|}ge frac1{|x|}to infty$$
answered Dec 13 at 14:37
gimusi
1
add a comment |
Note that
$$frac1{|x|}+frac1{|y|}ge frac1{|x|}to infty$$
answered Dec 13 at 14:37
gimusi
1
add a comment |
Note that
$$frac1{|x|}+frac1{|y|}ge frac1{|x|}to infty$$
answered Dec 13 at 14:37
gimusi
1
Note that
$$frac1{|x|}+frac1{|y|}ge frac1{|x|}to infty$$
answered Dec 13 at 14:37
gimusi
1
answered Dec 13 at 14:37
gimusi
1
answered Dec 13 at 14:37
gimusi
1
answered Dec 13 at 14:37
gimusi
1
1
add a comment |
add a comment |
Take the individual limits of each component. x and y are tending towards zero (the direction doesn't matter due to the magnitudes) so $frac{1}{|x|}$ and $frac{1}{|y|}$ both tend to infinity, and as a result so does their sum.
EDIT:
As suggested in the comments, this is an expansion as to why you can seperate the limits.
$$lim_{(x,y)to(0,0)}{bigg(frac{1}{|x|}+frac{1}{|y|}bigg)}=lim_{(x,y)to(0,0)}{bigg(frac{1}{|x|}bigg)}+lim_{(x,y)to(0,0)}{bigg(frac{1}{|y|}bigg)}$$
Examining the LHS we see that the first limit is independent of y, and the second independent of x. As a result we can rewrite the equation as follows:
$$=lim_{xto0}{bigg(frac{1}{|x|}bigg)}+lim_{yto0}{bigg(frac{1}{|y|}bigg)}$$
As stated above, the magnitudes mean that both limits tend to infinity as x and y tend to zero from either direction, so their sum is necessarily also infinity.
answered Dec 13 at 14:35
M.M.
637
2
You are right, but a few more words about why taking "individual limits of each component" can replace $(x,y)to (0,0)$ would be helpful in the present circumstance.
– hardmath
Dec 13 at 14:44
Thank you for the advice, I'll add an edit expanding on that.
– M.M.
Dec 14 at 10:06
add a comment |
Take the individual limits of each component. x and y are tending towards zero (the direction doesn't matter due to the magnitudes) so $frac{1}{|x|}$ and $frac{1}{|y|}$ both tend to infinity, and as a result so does their sum.
EDIT:
As suggested in the comments, this is an expansion as to why you can seperate the limits.
$$lim_{(x,y)to(0,0)}{bigg(frac{1}{|x|}+frac{1}{|y|}bigg)}=lim_{(x,y)to(0,0)}{bigg(frac{1}{|x|}bigg)}+lim_{(x,y)to(0,0)}{bigg(frac{1}{|y|}bigg)}$$
Examining the LHS we see that the first limit is independent of y, and the second independent of x. As a result we can rewrite the equation as follows:
$$=lim_{xto0}{bigg(frac{1}{|x|}bigg)}+lim_{yto0}{bigg(frac{1}{|y|}bigg)}$$
As stated above, the magnitudes mean that both limits tend to infinity as x and y tend to zero from either direction, so their sum is necessarily also infinity.
answered Dec 13 at 14:35
M.M.
637
2
You are right, but a few more words about why taking "individual limits of each component" can replace $(x,y)to (0,0)$ would be helpful in the present circumstance.
– hardmath
Dec 13 at 14:44
Thank you for the advice, I'll add an edit expanding on that.
– M.M.
Dec 14 at 10:06
add a comment |
Take the individual limits of each component. x and y are tending towards zero (the direction doesn't matter due to the magnitudes) so $frac{1}{|x|}$ and $frac{1}{|y|}$ both tend to infinity, and as a result so does their sum.
EDIT:
As suggested in the comments, this is an expansion as to why you can seperate the limits.
$$lim_{(x,y)to(0,0)}{bigg(frac{1}{|x|}+frac{1}{|y|}bigg)}=lim_{(x,y)to(0,0)}{bigg(frac{1}{|x|}bigg)}+lim_{(x,y)to(0,0)}{bigg(frac{1}{|y|}bigg)}$$
Examining the LHS we see that the first limit is independent of y, and the second independent of x. As a result we can rewrite the equation as follows:
$$=lim_{xto0}{bigg(frac{1}{|x|}bigg)}+lim_{yto0}{bigg(frac{1}{|y|}bigg)}$$
As stated above, the magnitudes mean that both limits tend to infinity as x and y tend to zero from either direction, so their sum is necessarily also infinity.
answered Dec 13 at 14:35
M.M.
637
Take the individual limits of each component. x and y are tending towards zero (the direction doesn't matter due to the magnitudes) so $frac{1}{|x|}$ and $frac{1}{|y|}$ both tend to infinity, and as a result so does their sum.
EDIT:
As suggested in the comments, this is an expansion as to why you can seperate the limits.
$$lim_{(x,y)to(0,0)}{bigg(frac{1}{|x|}+frac{1}{|y|}bigg)}=lim_{(x,y)to(0,0)}{bigg(frac{1}{|x|}bigg)}+lim_{(x,y)to(0,0)}{bigg(frac{1}{|y|}bigg)}$$
Examining the LHS we see that the first limit is independent of y, and the second independent of x. As a result we can rewrite the equation as follows:
$$=lim_{xto0}{bigg(frac{1}{|x|}bigg)}+lim_{yto0}{bigg(frac{1}{|y|}bigg)}$$
As stated above, the magnitudes mean that both limits tend to infinity as x and y tend to zero from either direction, so their sum is necessarily also infinity.
answered Dec 13 at 14:35
M.M.
637
edited Dec 14 at 10:12
answered Dec 13 at 14:35
M.M.
637
answered Dec 13 at 14:35
M.M.
637
answered Dec 13 at 14:35
M.M.
637
637
2
You are right, but a few more words about why taking "individual limits of each component" can replace $(x,y)to (0,0)$ would be helpful in the present circumstance.
– hardmath
Dec 13 at 14:44
Thank you for the advice, I'll add an edit expanding on that.
– M.M.
Dec 14 at 10:06
add a comment |
2
You are right, but a few more words about why taking "individual limits of each component" can replace $(x,y)to (0,0)$ would be helpful in the present circumstance.
– hardmath
Dec 13 at 14:44
Thank you for the advice, I'll add an edit expanding on that.
– M.M.
Dec 14 at 10:06
2
2
You are right, but a few more words about why taking "individual limits of each component" can replace $(x,y)to (0,0)$ would be helpful in the present circumstance.
– hardmath
Dec 13 at 14:44
You are right, but a few more words about why taking "individual limits of each component" can replace $(x,y)to (0,0)$ would be helpful in the present circumstance.
– hardmath
Dec 13 at 14:44
Thank you for the advice, I'll add an edit expanding on that.
– M.M.
Dec 14 at 10:06
Thank you for the advice, I'll add an edit expanding on that.
– M.M.
Dec 14 at 10:06
add a comment |
Hint:
Use the fact that if $|(x,y)| < epsilon$, then $|x|<epsilon$ and $|y|<epsilon$
answered Dec 13 at 14:34
5xum
89.5k393161
add a comment |
Hint:
Use the fact that if $|(x,y)| < epsilon$, then $|x|<epsilon$ and $|y|<epsilon$
answered Dec 13 at 14:34
5xum
89.5k393161
add a comment |
Hint:
Use the fact that if $|(x,y)| < epsilon$, then $|x|<epsilon$ and $|y|<epsilon$
answered Dec 13 at 14:34
5xum
89.5k393161
Hint:
Use the fact that if $|(x,y)| < epsilon$, then $|x|<epsilon$ and $|y|<epsilon$
answered Dec 13 at 14:34
5xum
89.5k393161
answered Dec 13 at 14:34
5xum
89.5k393161
answered Dec 13 at 14:34
5xum
89.5k393161
answered Dec 13 at 14:34
5xum
89.5k393161
89.5k393161
add a comment |
add a comment |
We have:
$(x,y) rightarrow 0.$
Let $epsilon_n =1/n$, $n$ positive integer, be given.
Then $|(x^2+y^2)^{1/2}| < delta_n$ $(=epsilon_n)$ implies
$|(x^2+y^2)^{1/2}| < epsilon_n =1/n$, or
$n=1/epsilon_n lt dfrac{1}{(x^2+y^2)^{1/2}}.$
Note:
$(x^2+y^2)^{1/2} ge |x|$, and
$(x^2+y^2)^{1/2} ge |y|.$
Then
$dfrac{2}{(x^2+y^2)^{1/2}} le dfrac{1}{|x|}+dfrac{1}{|y|}.$
And finally :
$|(x^2+y^2)^{1/2}| lt delta_n$ implies
$2n =2/epsilon_n lt dfrac{2}{(x^2+y^2)^{1/2}} le $
$dfrac{1}{|x|} +dfrac{1}{|y|}.$
answered Dec 13 at 16:57
Peter Szilas
10.6k2720
add a comment |
We have:
$(x,y) rightarrow 0.$
Let $epsilon_n =1/n$, $n$ positive integer, be given.
Then $|(x^2+y^2)^{1/2}| < delta_n$ $(=epsilon_n)$ implies
$|(x^2+y^2)^{1/2}| < epsilon_n =1/n$, or
$n=1/epsilon_n lt dfrac{1}{(x^2+y^2)^{1/2}}.$
Note:
$(x^2+y^2)^{1/2} ge |x|$, and
$(x^2+y^2)^{1/2} ge |y|.$
Then
$dfrac{2}{(x^2+y^2)^{1/2}} le dfrac{1}{|x|}+dfrac{1}{|y|}.$
And finally :
$|(x^2+y^2)^{1/2}| lt delta_n$ implies
$2n =2/epsilon_n lt dfrac{2}{(x^2+y^2)^{1/2}} le $
$dfrac{1}{|x|} +dfrac{1}{|y|}.$
answered Dec 13 at 16:57
Peter Szilas
10.6k2720
add a comment |
We have:
$(x,y) rightarrow 0.$
Let $epsilon_n =1/n$, $n$ positive integer, be given.
Then $|(x^2+y^2)^{1/2}| < delta_n$ $(=epsilon_n)$ implies
$|(x^2+y^2)^{1/2}| < epsilon_n =1/n$, or
$n=1/epsilon_n lt dfrac{1}{(x^2+y^2)^{1/2}}.$
Note:
$(x^2+y^2)^{1/2} ge |x|$, and
$(x^2+y^2)^{1/2} ge |y|.$
Then
$dfrac{2}{(x^2+y^2)^{1/2}} le dfrac{1}{|x|}+dfrac{1}{|y|}.$
And finally :
$|(x^2+y^2)^{1/2}| lt delta_n$ implies
$2n =2/epsilon_n lt dfrac{2}{(x^2+y^2)^{1/2}} le $
$dfrac{1}{|x|} +dfrac{1}{|y|}.$
answered Dec 13 at 16:57
Peter Szilas
10.6k2720
We have:
$(x,y) rightarrow 0.$
Let $epsilon_n =1/n$, $n$ positive integer, be given.
Then $|(x^2+y^2)^{1/2}| < delta_n$ $(=epsilon_n)$ implies
$|(x^2+y^2)^{1/2}| < epsilon_n =1/n$, or
$n=1/epsilon_n lt dfrac{1}{(x^2+y^2)^{1/2}}.$
Note:
$(x^2+y^2)^{1/2} ge |x|$, and
$(x^2+y^2)^{1/2} ge |y|.$
Then
$dfrac{2}{(x^2+y^2)^{1/2}} le dfrac{1}{|x|}+dfrac{1}{|y|}.$
And finally :
$|(x^2+y^2)^{1/2}| lt delta_n$ implies
$2n =2/epsilon_n lt dfrac{2}{(x^2+y^2)^{1/2}} le $
$dfrac{1}{|x|} +dfrac{1}{|y|}.$
answered Dec 13 at 16:57
Peter Szilas
10.6k2720
edited Dec 13 at 17:05
answered Dec 13 at 16:57
Peter Szilas
10.6k2720
answered Dec 13 at 16:57
Peter Szilas
10.6k2720
answered Dec 13 at 16:57
Peter Szilas
10.6k2720
10.6k2720
add a comment |
add a comment |
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3
"I got stuck as" has a mistake with the $r$ in the denominator.
– Teepeemm
Dec 13 at 20:31