Showing $lim_(x,y) to (0,0) xy log(x^2+y^2) = 0$
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First I let $x=rcos theta, y = rsin theta$ and so limit
$$lim_rto 0 r^2sin2theta log(r)$$
Now, in region $0<x<1$, $log(x) < 1/x$
$$|r^2sin 2theta log (r) - 0| < |rsin 2theta| le |r| < delta < epsilon$$
So limit exist if $delta < epsilon$ and limit is 0.
Other way, I used L hospital, I don't know if we can apply, but I wrote $r^2 log r$ as $log(r) / (r^-2)$ which again gave 0.
limits multivariable-calculus
asked Aug 30 at 3:43
jeea
46112
add a comment |Â
up vote
5
down vote
favorite
First I let $x=rcos theta, y = rsin theta$ and so limit
$$lim_rto 0 r^2sin2theta log(r)$$
Now, in region $0<x<1$, $log(x) < 1/x$
$$|r^2sin 2theta log (r) - 0| < |rsin 2theta| le |r| < delta < epsilon$$
So limit exist if $delta < epsilon$ and limit is 0.
Other way, I used L hospital, I don't know if we can apply, but I wrote $r^2 log r$ as $log(r) / (r^-2)$ which again gave 0.
limits multivariable-calculus
asked Aug 30 at 3:43
jeea
46112
Your proof assumes both $x$ and $y$ got to $0$. It is easy to show that the limit is $0$ even if only one variable goes to $0$, but your proof does not apply there, since $r$ does not go to $0$. In this case you need to use $sin2theta to 0$, since $theta to 0$ or $theta to fracpi2$.
– herb steinberg
Aug 30 at 3:54
@herb Sorry i dont understand why we use $sin 2theta to 0$
– jeea
Aug 30 at 3:58
Of course $rto 0$ here, while $sin 2theta$ does not even need to converge. I don't understand the comment made by @herbsteinberg.
– Kusma
Aug 30 at 5:52
$r to 0$ requires both $x$ and $y to 0$. My comment was about the case where only one variable $to 0$, but the other did not. In that situation $theta to 0$ or $to fracpi2$, so $sin2theta to 0$.
– herb steinberg
Aug 30 at 15:30
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
First I let $x=rcos theta, y = rsin theta$ and so limit
$$lim_rto 0 r^2sin2theta log(r)$$
Now, in region $0<x<1$, $log(x) < 1/x$
$$|r^2sin 2theta log (r) - 0| < |rsin 2theta| le |r| < delta < epsilon$$
So limit exist if $delta < epsilon$ and limit is 0.
Other way, I used L hospital, I don't know if we can apply, but I wrote $r^2 log r$ as $log(r) / (r^-2)$ which again gave 0.
limits multivariable-calculus
asked Aug 30 at 3:43
jeea
46112
First I let $x=rcos theta, y = rsin theta$ and so limit
$$lim_rto 0 r^2sin2theta log(r)$$
Now, in region $0<x<1$, $log(x) < 1/x$
$$|r^2sin 2theta log (r) - 0| < |rsin 2theta| le |r| < delta < epsilon$$
So limit exist if $delta < epsilon$ and limit is 0.
Other way, I used L hospital, I don't know if we can apply, but I wrote $r^2 log r$ as $log(r) / (r^-2)$ which again gave 0.
limits multivariable-calculus
limits multivariable-calculus
asked Aug 30 at 3:43
jeea
46112
asked Aug 30 at 3:43
jeea
46112
asked Aug 30 at 3:43
jeea
46112
asked Aug 30 at 3:43
jeea
46112
asked Aug 30 at 3:43
jeea
46112
46112
Your proof assumes both $x$ and $y$ got to $0$. It is easy to show that the limit is $0$ even if only one variable goes to $0$, but your proof does not apply there, since $r$ does not go to $0$. In this case you need to use $sin2theta to 0$, since $theta to 0$ or $theta to fracpi2$.
– herb steinberg
Aug 30 at 3:54
@herb Sorry i dont understand why we use $sin 2theta to 0$
– jeea
Aug 30 at 3:58
Of course $rto 0$ here, while $sin 2theta$ does not even need to converge. I don't understand the comment made by @herbsteinberg.
– Kusma
Aug 30 at 5:52
$r to 0$ requires both $x$ and $y to 0$. My comment was about the case where only one variable $to 0$, but the other did not. In that situation $theta to 0$ or $to fracpi2$, so $sin2theta to 0$.
– herb steinberg
Aug 30 at 15:30
add a comment |Â
Your proof assumes both $x$ and $y$ got to $0$. It is easy to show that the limit is $0$ even if only one variable goes to $0$, but your proof does not apply there, since $r$ does not go to $0$. In this case you need to use $sin2theta to 0$, since $theta to 0$ or $theta to fracpi2$.
– herb steinberg
Aug 30 at 3:54
@herb Sorry i dont understand why we use $sin 2theta to 0$
– jeea
Aug 30 at 3:58
Of course $rto 0$ here, while $sin 2theta$ does not even need to converge. I don't understand the comment made by @herbsteinberg.
– Kusma
Aug 30 at 5:52
$r to 0$ requires both $x$ and $y to 0$. My comment was about the case where only one variable $to 0$, but the other did not. In that situation $theta to 0$ or $to fracpi2$, so $sin2theta to 0$.
– herb steinberg
Aug 30 at 15:30
Your proof assumes both $x$ and $y$ got to $0$. It is easy to show that the limit is $0$ even if only one variable goes to $0$, but your proof does not apply there, since $r$ does not go to $0$. In this case you need to use $sin2theta to 0$, since $theta to 0$ or $theta to fracpi2$.
– herb steinberg
Aug 30 at 3:54
Your proof assumes both $x$ and $y$ got to $0$. It is easy to show that the limit is $0$ even if only one variable goes to $0$, but your proof does not apply there, since $r$ does not go to $0$. In this case you need to use $sin2theta to 0$, since $theta to 0$ or $theta to fracpi2$.
– herb steinberg
Aug 30 at 3:54
@herb Sorry i dont understand why we use $sin 2theta to 0$
– jeea
Aug 30 at 3:58
@herb Sorry i dont understand why we use $sin 2theta to 0$
– jeea
Aug 30 at 3:58
Of course $rto 0$ here, while $sin 2theta$ does not even need to converge. I don't understand the comment made by @herbsteinberg.
– Kusma
Aug 30 at 5:52
Of course $rto 0$ here, while $sin 2theta$ does not even need to converge. I don't understand the comment made by @herbsteinberg.
– Kusma
Aug 30 at 5:52
$r to 0$ requires both $x$ and $y to 0$. My comment was about the case where only one variable $to 0$, but the other did not. In that situation $theta to 0$ or $to fracpi2$, so $sin2theta to 0$.
– herb steinberg
Aug 30 at 15:30
$r to 0$ requires both $x$ and $y to 0$. My comment was about the case where only one variable $to 0$, but the other did not. In that situation $theta to 0$ or $to fracpi2$, so $sin2theta to 0$.
– herb steinberg
Aug 30 at 15:30
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
More easily, $lim_rrightarrow 0rlog(r)=0$, we deduce that $lim_rrightarrow 0r^2log(r)=0$ since $|sin(theta)|leq 1$, the result follows.
Limit of $x log x$ as $x$ tends to $0^+$
answered Aug 30 at 4:02
Tsemo Aristide
52.3k11244
add a comment |Â
up vote
1
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First, notice that from the (in)equalities $0leq (x-y)^2=x^2+y^2-2xy$ and $log(r)leq r-1$ there holds that
$$
0 leq lim_(x,y)rightarrow (0,0) |xylog(x^2+y^2)| leq lim_(x,y)rightarrow (0,0)frac12(x^2+y^2)|x^2+y^2-1|=0
$$
so that the required limit equals zero (direct application of sandwich theorem).
answered Aug 30 at 4:09
Nelson Faustino
1376
But we need to compare $|log(r)|$ so it may not be true that $|log r| < |r-1|$ near 0 right? Please tell me, thanks!
– jeea
Aug 30 at 18:20
add a comment |Â
up vote
0
down vote
We have
$$xy log(x^2+y^2) =(x^2+y^2)log(x^2+y^2)cdot fracxyx^2+y^2to 0$$
indeed since $t=x^2+y^2to 0$
$$(x^2+y^2)log(x^2+y^2)=tlog tto 0$$
and since $x^2+y^2ge 2xy$
$$0le left|fracxyx^2+y^2right| le frac12$$
answered Aug 30 at 5:37
gimusi
71.4k73786
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
More easily, $lim_rrightarrow 0rlog(r)=0$, we deduce that $lim_rrightarrow 0r^2log(r)=0$ since $|sin(theta)|leq 1$, the result follows.
Limit of $x log x$ as $x$ tends to $0^+$
answered Aug 30 at 4:02
Tsemo Aristide
52.3k11244
add a comment |Â
up vote
1
down vote
More easily, $lim_rrightarrow 0rlog(r)=0$, we deduce that $lim_rrightarrow 0r^2log(r)=0$ since $|sin(theta)|leq 1$, the result follows.
Limit of $x log x$ as $x$ tends to $0^+$
answered Aug 30 at 4:02
Tsemo Aristide
52.3k11244
add a comment |Â
up vote
1
down vote
up vote
1
down vote
More easily, $lim_rrightarrow 0rlog(r)=0$, we deduce that $lim_rrightarrow 0r^2log(r)=0$ since $|sin(theta)|leq 1$, the result follows.
Limit of $x log x$ as $x$ tends to $0^+$
answered Aug 30 at 4:02
Tsemo Aristide
52.3k11244
More easily, $lim_rrightarrow 0rlog(r)=0$, we deduce that $lim_rrightarrow 0r^2log(r)=0$ since $|sin(theta)|leq 1$, the result follows.
Limit of $x log x$ as $x$ tends to $0^+$
answered Aug 30 at 4:02
Tsemo Aristide
52.3k11244
answered Aug 30 at 4:02
Tsemo Aristide
52.3k11244
answered Aug 30 at 4:02
Tsemo Aristide
52.3k11244
answered Aug 30 at 4:02
Tsemo Aristide
52.3k11244
52.3k11244
add a comment |Â
add a comment |Â
up vote
1
down vote
First, notice that from the (in)equalities $0leq (x-y)^2=x^2+y^2-2xy$ and $log(r)leq r-1$ there holds that
$$
0 leq lim_(x,y)rightarrow (0,0) |xylog(x^2+y^2)| leq lim_(x,y)rightarrow (0,0)frac12(x^2+y^2)|x^2+y^2-1|=0
$$
so that the required limit equals zero (direct application of sandwich theorem).
answered Aug 30 at 4:09
Nelson Faustino
1376
But we need to compare $|log(r)|$ so it may not be true that $|log r| < |r-1|$ near 0 right? Please tell me, thanks!
– jeea
Aug 30 at 18:20
add a comment |Â
up vote
1
down vote
First, notice that from the (in)equalities $0leq (x-y)^2=x^2+y^2-2xy$ and $log(r)leq r-1$ there holds that
$$
0 leq lim_(x,y)rightarrow (0,0) |xylog(x^2+y^2)| leq lim_(x,y)rightarrow (0,0)frac12(x^2+y^2)|x^2+y^2-1|=0
$$
so that the required limit equals zero (direct application of sandwich theorem).
answered Aug 30 at 4:09
Nelson Faustino
1376
But we need to compare $|log(r)|$ so it may not be true that $|log r| < |r-1|$ near 0 right? Please tell me, thanks!
– jeea
Aug 30 at 18:20
add a comment |Â
up vote
1
down vote
up vote
1
down vote
First, notice that from the (in)equalities $0leq (x-y)^2=x^2+y^2-2xy$ and $log(r)leq r-1$ there holds that
$$
0 leq lim_(x,y)rightarrow (0,0) |xylog(x^2+y^2)| leq lim_(x,y)rightarrow (0,0)frac12(x^2+y^2)|x^2+y^2-1|=0
$$
so that the required limit equals zero (direct application of sandwich theorem).
answered Aug 30 at 4:09
Nelson Faustino
1376
First, notice that from the (in)equalities $0leq (x-y)^2=x^2+y^2-2xy$ and $log(r)leq r-1$ there holds that
$$
0 leq lim_(x,y)rightarrow (0,0) |xylog(x^2+y^2)| leq lim_(x,y)rightarrow (0,0)frac12(x^2+y^2)|x^2+y^2-1|=0
$$
so that the required limit equals zero (direct application of sandwich theorem).
answered Aug 30 at 4:09
Nelson Faustino
1376
answered Aug 30 at 4:09
Nelson Faustino
1376
answered Aug 30 at 4:09
Nelson Faustino
1376
answered Aug 30 at 4:09
Nelson Faustino
1376
1376
But we need to compare $|log(r)|$ so it may not be true that $|log r| < |r-1|$ near 0 right? Please tell me, thanks!
– jeea
Aug 30 at 18:20
add a comment |Â
But we need to compare $|log(r)|$ so it may not be true that $|log r| < |r-1|$ near 0 right? Please tell me, thanks!
– jeea
Aug 30 at 18:20
But we need to compare $|log(r)|$ so it may not be true that $|log r| < |r-1|$ near 0 right? Please tell me, thanks!
– jeea
Aug 30 at 18:20
But we need to compare $|log(r)|$ so it may not be true that $|log r| < |r-1|$ near 0 right? Please tell me, thanks!
– jeea
Aug 30 at 18:20
add a comment |Â
up vote
0
down vote
We have
$$xy log(x^2+y^2) =(x^2+y^2)log(x^2+y^2)cdot fracxyx^2+y^2to 0$$
indeed since $t=x^2+y^2to 0$
$$(x^2+y^2)log(x^2+y^2)=tlog tto 0$$
and since $x^2+y^2ge 2xy$
$$0le left|fracxyx^2+y^2right| le frac12$$
answered Aug 30 at 5:37
gimusi
71.4k73786
add a comment |Â
up vote
0
down vote
We have
$$xy log(x^2+y^2) =(x^2+y^2)log(x^2+y^2)cdot fracxyx^2+y^2to 0$$
indeed since $t=x^2+y^2to 0$
$$(x^2+y^2)log(x^2+y^2)=tlog tto 0$$
and since $x^2+y^2ge 2xy$
$$0le left|fracxyx^2+y^2right| le frac12$$
answered Aug 30 at 5:37
gimusi
71.4k73786
add a comment |Â
up vote
0
down vote
up vote
0
down vote
We have
$$xy log(x^2+y^2) =(x^2+y^2)log(x^2+y^2)cdot fracxyx^2+y^2to 0$$
indeed since $t=x^2+y^2to 0$
$$(x^2+y^2)log(x^2+y^2)=tlog tto 0$$
and since $x^2+y^2ge 2xy$
$$0le left|fracxyx^2+y^2right| le frac12$$
answered Aug 30 at 5:37
gimusi
71.4k73786
We have
$$xy log(x^2+y^2) =(x^2+y^2)log(x^2+y^2)cdot fracxyx^2+y^2to 0$$
indeed since $t=x^2+y^2to 0$
$$(x^2+y^2)log(x^2+y^2)=tlog tto 0$$
and since $x^2+y^2ge 2xy$
$$0le left|fracxyx^2+y^2right| le frac12$$
answered Aug 30 at 5:37
gimusi
71.4k73786
answered Aug 30 at 5:37
gimusi
71.4k73786
answered Aug 30 at 5:37
gimusi
71.4k73786
answered Aug 30 at 5:37
gimusi
71.4k73786
71.4k73786
add a comment |Â
add a comment |Â
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Your proof assumes both $x$ and $y$ got to $0$. It is easy to show that the limit is $0$ even if only one variable goes to $0$, but your proof does not apply there, since $r$ does not go to $0$. In this case you need to use $sin2theta to 0$, since $theta to 0$ or $theta to fracpi2$.
– herb steinberg
Aug 30 at 3:54
@herb Sorry i dont understand why we use $sin 2theta to 0$
– jeea
Aug 30 at 3:58
Of course $rto 0$ here, while $sin 2theta$ does not even need to converge. I don't understand the comment made by @herbsteinberg.
– Kusma
Aug 30 at 5:52
$r to 0$ requires both $x$ and $y to 0$. My comment was about the case where only one variable $to 0$, but the other did not. In that situation $theta to 0$ or $to fracpi2$, so $sin2theta to 0$.
– herb steinberg
Aug 30 at 15:30