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







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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
















0












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











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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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















1












$begingroup$

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 $x_0=-1/e$.)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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 $x_0=-1/e$.)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















1












$begingroup$

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 $x_0=-1/e$.)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

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 $x_0=-1/e$.)






share|cite|improve this answer









$endgroup$



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 $x_0=-1/e$.)







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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