Verifying a limit with Lambert W function
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Is the following limit computation correct:$$a = limlimits_{xrightarrow 1} expleft{frac{W_{-1}left(xln(x)right)}{x}right} = expleft{frac{W_{-1}left(1cdot 0right)}{1}right} = exp(-infty) = 0$$
More generally, when can we write:$$limlimits_{xrightarrow x_0} W(x) = Wleft(limlimits_{xrightarrow x_0} xright) = W(x_0)$$
limits proof-verification lambert-w
asked Dec 4 '18 at 9:51
jlandercyjlandercy
256213
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add a comment |
$begingroup$
Is the following limit computation correct:$$a = limlimits_{xrightarrow 1} expleft{frac{W_{-1}left(xln(x)right)}{x}right} = expleft{frac{W_{-1}left(1cdot 0right)}{1}right} = exp(-infty) = 0$$
More generally, when can we write:$$limlimits_{xrightarrow x_0} W(x) = Wleft(limlimits_{xrightarrow x_0} xright) = W(x_0)$$
limits proof-verification lambert-w
asked Dec 4 '18 at 9:51
jlandercyjlandercy
256213
$endgroup$
$begingroup$
The limit is $-infty$ since $xlog(x)to 0$ when $xto 1$. Look at the expansion in the Wikipedia page.
$endgroup$
– Claude Leibovici
Dec 4 '18 at 10:48
add a comment |
$begingroup$
Is the following limit computation correct:$$a = limlimits_{xrightarrow 1} expleft{frac{W_{-1}left(xln(x)right)}{x}right} = expleft{frac{W_{-1}left(1cdot 0right)}{1}right} = exp(-infty) = 0$$
More generally, when can we write:$$limlimits_{xrightarrow x_0} W(x) = Wleft(limlimits_{xrightarrow x_0} xright) = W(x_0)$$
limits proof-verification lambert-w
asked Dec 4 '18 at 9:51
jlandercyjlandercy
256213
$endgroup$
Is the following limit computation correct:$$a = limlimits_{xrightarrow 1} expleft{frac{W_{-1}left(xln(x)right)}{x}right} = expleft{frac{W_{-1}left(1cdot 0right)}{1}right} = exp(-infty) = 0$$
More generally, when can we write:$$limlimits_{xrightarrow x_0} W(x) = Wleft(limlimits_{xrightarrow x_0} xright) = W(x_0)$$
limits proof-verification lambert-w
limits proof-verification lambert-w
asked Dec 4 '18 at 9:51
jlandercyjlandercy
256213
asked Dec 4 '18 at 9:51
jlandercyjlandercy
256213
asked Dec 4 '18 at 9:51
jlandercyjlandercy
256213
asked Dec 4 '18 at 9:51
jlandercyjlandercy
256213
asked Dec 4 '18 at 9:51
jlandercyjlandercy
256213
256213
$begingroup$
The limit is $-infty$ since $xlog(x)to 0$ when $xto 1$. Look at the expansion in the Wikipedia page.
$endgroup$
– Claude Leibovici
Dec 4 '18 at 10:48
add a comment |
$begingroup$
The limit is $-infty$ since $xlog(x)to 0$ when $xto 1$. Look at the expansion in the Wikipedia page.
$endgroup$
– Claude Leibovici
Dec 4 '18 at 10:48
$begingroup$
The limit is $-infty$ since $xlog(x)to 0$ when $xto 1$. Look at the expansion in the Wikipedia page.
$endgroup$
– Claude Leibovici
Dec 4 '18 at 10:48
$begingroup$
The limit is $-infty$ since $xlog(x)to 0$ when $xto 1$. Look at the expansion in the Wikipedia page.
$endgroup$
– Claude Leibovici
Dec 4 '18 at 10:48
add a comment |
1 Answer
1
active
oldest
votes
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The principal branch of the Lambert function (considered as function of a real variable, which seems to be the case in the question) is continuous on $[-1/e,infty)$, so the answer is yes if $x_0in[-1/e,infty)$ (limit from the right if $x_0=-1/e$.)
answered Dec 4 '18 at 10:16
Julián AguirreJulián Aguirre
68.1k24094
$endgroup$
$begingroup$
Thank you for answering, I am using the second real branch in my computation where it is defined and seems to be continuous. So continuity is an enough criterion to compute the limit as this?
$endgroup$
– jlandercy
Dec 4 '18 at 10:23
1
$begingroup$
Yes, that is the definition of continuity.
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– Julián Aguirre
Dec 4 '18 at 10:34
$begingroup$
I was expecting a subtlety and I missed this obvious fact. I feel a bit fool about it, thank you for the refresh.
$endgroup$
– jlandercy
Dec 4 '18 at 10:40
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
$begingroup$
The principal branch of the Lambert function (considered as function of a real variable, which seems to be the case in the question) is continuous on $[-1/e,infty)$, so the answer is yes if $x_0in[-1/e,infty)$ (limit from the right if $x_0=-1/e$.)
answered Dec 4 '18 at 10:16
Julián AguirreJulián Aguirre
68.1k24094
$endgroup$
$begingroup$
Thank you for answering, I am using the second real branch in my computation where it is defined and seems to be continuous. So continuity is an enough criterion to compute the limit as this?
$endgroup$
– jlandercy
Dec 4 '18 at 10:23
1
$begingroup$
Yes, that is the definition of continuity.
$endgroup$
– Julián Aguirre
Dec 4 '18 at 10:34
$begingroup$
I was expecting a subtlety and I missed this obvious fact. I feel a bit fool about it, thank you for the refresh.
$endgroup$
– jlandercy
Dec 4 '18 at 10:40
add a comment |
$begingroup$
The principal branch of the Lambert function (considered as function of a real variable, which seems to be the case in the question) is continuous on $[-1/e,infty)$, so the answer is yes if $x_0in[-1/e,infty)$ (limit from the right if $x_0=-1/e$.)
answered Dec 4 '18 at 10:16
Julián AguirreJulián Aguirre
68.1k24094
$endgroup$
$begingroup$
Thank you for answering, I am using the second real branch in my computation where it is defined and seems to be continuous. So continuity is an enough criterion to compute the limit as this?
$endgroup$
– jlandercy
Dec 4 '18 at 10:23
1
$begingroup$
Yes, that is the definition of continuity.
$endgroup$
– Julián Aguirre
Dec 4 '18 at 10:34
$begingroup$
I was expecting a subtlety and I missed this obvious fact. I feel a bit fool about it, thank you for the refresh.
$endgroup$
– jlandercy
Dec 4 '18 at 10:40
add a comment |
$begingroup$
The principal branch of the Lambert function (considered as function of a real variable, which seems to be the case in the question) is continuous on $[-1/e,infty)$, so the answer is yes if $x_0in[-1/e,infty)$ (limit from the right if $x_0=-1/e$.)
answered Dec 4 '18 at 10:16
Julián AguirreJulián Aguirre
68.1k24094
$endgroup$
The principal branch of the Lambert function (considered as function of a real variable, which seems to be the case in the question) is continuous on $[-1/e,infty)$, so the answer is yes if $x_0in[-1/e,infty)$ (limit from the right if $x_0=-1/e$.)
answered Dec 4 '18 at 10:16
Julián AguirreJulián Aguirre
68.1k24094
answered Dec 4 '18 at 10:16
Julián AguirreJulián Aguirre
68.1k24094
answered Dec 4 '18 at 10:16
Julián AguirreJulián Aguirre
68.1k24094
answered Dec 4 '18 at 10:16
Julián AguirreJulián Aguirre
68.1k24094
68.1k24094
$begingroup$
Thank you for answering, I am using the second real branch in my computation where it is defined and seems to be continuous. So continuity is an enough criterion to compute the limit as this?
$endgroup$
– jlandercy
Dec 4 '18 at 10:23
1
$begingroup$
Yes, that is the definition of continuity.
$endgroup$
– Julián Aguirre
Dec 4 '18 at 10:34
$begingroup$
I was expecting a subtlety and I missed this obvious fact. I feel a bit fool about it, thank you for the refresh.
$endgroup$
– jlandercy
Dec 4 '18 at 10:40
add a comment |
$begingroup$
Thank you for answering, I am using the second real branch in my computation where it is defined and seems to be continuous. So continuity is an enough criterion to compute the limit as this?
$endgroup$
– jlandercy
Dec 4 '18 at 10:23
1
$begingroup$
Yes, that is the definition of continuity.
$endgroup$
– Julián Aguirre
Dec 4 '18 at 10:34
$begingroup$
I was expecting a subtlety and I missed this obvious fact. I feel a bit fool about it, thank you for the refresh.
$endgroup$
– jlandercy
Dec 4 '18 at 10:40
$begingroup$
Thank you for answering, I am using the second real branch in my computation where it is defined and seems to be continuous. So continuity is an enough criterion to compute the limit as this?
$endgroup$
– jlandercy
Dec 4 '18 at 10:23
$begingroup$
Thank you for answering, I am using the second real branch in my computation where it is defined and seems to be continuous. So continuity is an enough criterion to compute the limit as this?
$endgroup$
– jlandercy
Dec 4 '18 at 10:23
1
1
$begingroup$
Yes, that is the definition of continuity.
$endgroup$
– Julián Aguirre
Dec 4 '18 at 10:34
$begingroup$
Yes, that is the definition of continuity.
$endgroup$
– Julián Aguirre
Dec 4 '18 at 10:34
$begingroup$
I was expecting a subtlety and I missed this obvious fact. I feel a bit fool about it, thank you for the refresh.
$endgroup$
– jlandercy
Dec 4 '18 at 10:40
$begingroup$
I was expecting a subtlety and I missed this obvious fact. I feel a bit fool about it, thank you for the refresh.
$endgroup$
– jlandercy
Dec 4 '18 at 10:40
add a comment |
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$begingroup$
The limit is $-infty$ since $xlog(x)to 0$ when $xto 1$. Look at the expansion in the Wikipedia page.
$endgroup$
– Claude Leibovici
Dec 4 '18 at 10:48