$lim_{(x,y)to(0,0)}frac{sin^p(xy)log(x^2+y^2)}{x}$ when $p>0$
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It seems the limit exists (I can't find a subset for which it doesn't) and it is $0$, but I don't know how to prove it.
EDIT: the hint helped me a lot. However, I solved it in a little different way, I hope it's alright. We have:
$$lim_{(x,y)to(0,0)}frac{sin^p(xy)log(x^2+y^2)}{x} = lim_{(x,y)to(0,0)}frac{sin^p(xy)}{xy} lim_{(x,y)to(0,0)}ylog(x^2+y^2)$$
The first limit goes to $0$ by L'Hôpital (supposing $p>1$) and the second goes to $0$ by polar coordinates, so the solution is $0$. Is it OK?
calculus limits
asked Nov 23 at 15:29
Seven
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up vote
0
down vote
favorite
It seems the limit exists (I can't find a subset for which it doesn't) and it is $0$, but I don't know how to prove it.
EDIT: the hint helped me a lot. However, I solved it in a little different way, I hope it's alright. We have:
$$lim_{(x,y)to(0,0)}frac{sin^p(xy)log(x^2+y^2)}{x} = lim_{(x,y)to(0,0)}frac{sin^p(xy)}{xy} lim_{(x,y)to(0,0)}ylog(x^2+y^2)$$
The first limit goes to $0$ by L'Hôpital (supposing $p>1$) and the second goes to $0$ by polar coordinates, so the solution is $0$. Is it OK?
calculus limits
asked Nov 23 at 15:29
Seven
829
What have you deduced from the given hint? It should be fine if you could add your thoughts about that editing your answer.
– gimusi
Nov 23 at 18:41
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
It seems the limit exists (I can't find a subset for which it doesn't) and it is $0$, but I don't know how to prove it.
EDIT: the hint helped me a lot. However, I solved it in a little different way, I hope it's alright. We have:
$$lim_{(x,y)to(0,0)}frac{sin^p(xy)log(x^2+y^2)}{x} = lim_{(x,y)to(0,0)}frac{sin^p(xy)}{xy} lim_{(x,y)to(0,0)}ylog(x^2+y^2)$$
The first limit goes to $0$ by L'Hôpital (supposing $p>1$) and the second goes to $0$ by polar coordinates, so the solution is $0$. Is it OK?
calculus limits
asked Nov 23 at 15:29
Seven
829
It seems the limit exists (I can't find a subset for which it doesn't) and it is $0$, but I don't know how to prove it.
EDIT: the hint helped me a lot. However, I solved it in a little different way, I hope it's alright. We have:
$$lim_{(x,y)to(0,0)}frac{sin^p(xy)log(x^2+y^2)}{x} = lim_{(x,y)to(0,0)}frac{sin^p(xy)}{xy} lim_{(x,y)to(0,0)}ylog(x^2+y^2)$$
The first limit goes to $0$ by L'Hôpital (supposing $p>1$) and the second goes to $0$ by polar coordinates, so the solution is $0$. Is it OK?
calculus limits
calculus limits
asked Nov 23 at 15:29
Seven
829
asked Nov 23 at 15:29
Seven
829
edited Nov 23 at 20:13
asked Nov 23 at 15:29
Seven
829
asked Nov 23 at 15:29
Seven
829
asked Nov 23 at 15:29
Seven
829
829
What have you deduced from the given hint? It should be fine if you could add your thoughts about that editing your answer.
– gimusi
Nov 23 at 18:41
add a comment |
What have you deduced from the given hint? It should be fine if you could add your thoughts about that editing your answer.
– gimusi
Nov 23 at 18:41
What have you deduced from the given hint? It should be fine if you could add your thoughts about that editing your answer.
– gimusi
Nov 23 at 18:41
What have you deduced from the given hint? It should be fine if you could add your thoughts about that editing your answer.
– gimusi
Nov 23 at 18:41
add a comment |
1 Answer
1
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2
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accepted
HINT
We have that
$$frac{sin^p(xy) log(x^2+y^2)}{x}=frac{sin^p(xy)}{(xy)^p}, (x^{p-1}y^p)log(x^2+y^2)$$
and by polar coordinates
$$(x^{p-1}y^p)log(x^2+y^2)=2r^{2p-1}log r cdot(cos^{p-1}theta cdot sin^p theta)$$
answered Nov 23 at 15:36
gimusi
88.5k74394
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
HINT
We have that
$$frac{sin^p(xy) log(x^2+y^2)}{x}=frac{sin^p(xy)}{(xy)^p}, (x^{p-1}y^p)log(x^2+y^2)$$
and by polar coordinates
$$(x^{p-1}y^p)log(x^2+y^2)=2r^{2p-1}log r cdot(cos^{p-1}theta cdot sin^p theta)$$
answered Nov 23 at 15:36
gimusi
88.5k74394
add a comment |
up vote
2
down vote
accepted
HINT
We have that
$$frac{sin^p(xy) log(x^2+y^2)}{x}=frac{sin^p(xy)}{(xy)^p}, (x^{p-1}y^p)log(x^2+y^2)$$
and by polar coordinates
$$(x^{p-1}y^p)log(x^2+y^2)=2r^{2p-1}log r cdot(cos^{p-1}theta cdot sin^p theta)$$
answered Nov 23 at 15:36
gimusi
88.5k74394
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
HINT
We have that
$$frac{sin^p(xy) log(x^2+y^2)}{x}=frac{sin^p(xy)}{(xy)^p}, (x^{p-1}y^p)log(x^2+y^2)$$
and by polar coordinates
$$(x^{p-1}y^p)log(x^2+y^2)=2r^{2p-1}log r cdot(cos^{p-1}theta cdot sin^p theta)$$
answered Nov 23 at 15:36
gimusi
88.5k74394
HINT
We have that
$$frac{sin^p(xy) log(x^2+y^2)}{x}=frac{sin^p(xy)}{(xy)^p}, (x^{p-1}y^p)log(x^2+y^2)$$
and by polar coordinates
$$(x^{p-1}y^p)log(x^2+y^2)=2r^{2p-1}log r cdot(cos^{p-1}theta cdot sin^p theta)$$
answered Nov 23 at 15:36
gimusi
88.5k74394
edited Nov 23 at 15:42
answered Nov 23 at 15:36
gimusi
88.5k74394
answered Nov 23 at 15:36
gimusi
88.5k74394
answered Nov 23 at 15:36
gimusi
88.5k74394
88.5k74394
add a comment |
add a comment |
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What have you deduced from the given hint? It should be fine if you could add your thoughts about that editing your answer.
– gimusi
Nov 23 at 18:41