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I am trying to evaluate the following limit $$ L=\lim_{x \rightarrow 0^+}\frac{2 \operatorname{W}\left( -{{ e}^{-x-1}}\right) \left( {{\operatorname{W}\left( -{{e}^{-x-1}}\right) }^{2}}+2 \operatorname{W}\left( -{{ e}^{-x-1}}\right) -2 x+1\right) }{{{\left( \operatorname{W}\left( -{{ e}^{-x-1}}\right) +1\right) }^{3}}}$$ where $W(z)$ is the principal branch of Lambert's function. The numerical experiments show that it is $\sqrt{2}$ but the l'Hopital's rule does not produce anything useful.

Here is the numerical experiment computed with Maxima: $L(x) - \sqrt{2}$ L-sqrt(2)

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    $\begingroup$ What's "%"? Is it a typo? $\endgroup$
    – user127032
    Commented Jun 10, 2020 at 23:59
  • $\begingroup$ I use Maxima, so that came from the Latex output. $\endgroup$
    – user48672
    Commented Jun 11, 2020 at 7:59
  • $\begingroup$ It's undefined: For $x=0$, $W(-\exp(-x-1))$ becomes $W(-1/e)=-1$, so your numerator becomes 4, while the denominator is 0. L'Hospital can't help you here. $\endgroup$
    – user127032
    Commented Jun 11, 2020 at 10:20
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    $\begingroup$ @YiannisGalidakis. The numerator and denominator tend to $0$ when $x\to 0^+$ $\endgroup$ Commented Jun 11, 2020 at 11:31
  • $\begingroup$ Could you check the numerical experiments ? $\endgroup$ Commented Jun 11, 2020 at 11:33

4 Answers 4

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L'Hopital's rule works.

Note that $$\lim_{x\to 0^{+}} W(-\mathrm{e}^{-x-1}) = -1 \tag{1}$$ and $$\frac{\mathrm{d}}{\mathrm{d} x}W(-\mathrm{e}^{-x-1}) = -\frac{W(-\mathrm{e}^{-x-1})}{W(-\mathrm{e}^{-x-1}) + 1}, \quad x > 0 \tag{2}$$ where we have used $W'(y) = \frac{W(y)}{y(1+W(y))}$ and the chain rule. See: https://en.wikipedia.org/wiki/Lambert_W_function

Let \begin{align} f(x) &= W(-\mathrm{e}^{-x-1})^2 + 2W(-\mathrm{e}^{-x-1}) - 2x + 1, \\ g(x) &= (W(-\mathrm{e}^{-x-1}) + 1)^3. \end{align} We have (noting (2)) \begin{align} f'(x) &= -2 W(-\mathrm{e}^{-x-1}) - 2, \quad x > 0\\ g'(x) &= -3 (W(-\mathrm{e}^{-x-1}) + 1)W(-\mathrm{e}^{-x-1}), \quad x > 0. \end{align}

Clearly, $\lim_{x\to 0^{+}} f(x) = 0$ and $\lim_{x\to 0^{+}} g(x) = 0$. Also, we have (noting (1)) $$\lim_{x\to 0^{+}} \frac{f'}{g'} = \lim_{x\to 0^{+}} \frac{2}{3 W(-\mathrm{e}^{-x-1}) } = -\frac{2}{3}.$$ By L'Hopital's rule, we have $\lim_{x\to 0^{+}} \frac{f}{g} = - \frac{2}{3}$. Thus, we have (noting (1)) \begin{align} \lim_{x\to 0^{+}} L &= 2 \cdot \lim_{x\to 0^{+}} W(-\mathrm{e}^{-x-1}) \cdot \lim_{x\to 0^{+}} \frac{f}{g}\\ &= \frac{4}{3}. \end{align}

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Compose Taylor expansion starting with $$e^{-x-1}=\frac{1}{e}-\frac{x}{e}+\frac{x^2}{2 e}-\frac{x^3}{6 e}+O\left(x^4\right)$$ $$W\left(-e^{-x-1}\right)=-1+\sqrt{2}\, x^{1/2}-\frac{2 x}{3}+\frac{x^{3/2}}{9 \sqrt{2}}+\frac{2 x^2}{135}-\frac{17 x^{5/2}}{48 \sqrt{2}}+O\left(x^{3}\right)$$ Then, the expression becomes $$\frac{4}{3}-\frac{\sqrt{2} }{3}x^{1/2}-\frac{8 }{45}x+\frac{27 }{40 \sqrt{2}}x^{3/2}+O\left(x^2\right)$$ which shows the limit and how it is approached.

Moreover, this gives you a shortcut method for the evaluation of the expression. For checking, let $\color{red}{x=10^{-k}}$ and compute $$\left( \begin{array}{ccc} k & \text{approximation} & \text{exact} \\ 1 & 1.181577816 & 1.165840097 \\ 2 & 1.284892401 & 1.284390451 \\ 3 & 1.318263529 & 1.318247622 \\ 4 & 1.328601988 & 1.328601484 \\ 5 & 1.331840859 & 1.331840843 \\ 6 & 1.332861752 & 1.332861751 \\ 7 & 1.333184244 & 1.333184244 \\ 8 & 1.333286191 & 1.333286191 \\ 9 & 1.333318426 & 1.333318426 \\ 10 & 1.333328619 & 1.333328619 \end{array} \right)$$

Edit (just for your curiosity)

Sooner or later, you will learn that for function approximations, Padé approximants are much better than Taylor series (even if the look similar). A simple one for your function is $$\frac{\frac{4}{3}-\frac{2179 \sqrt{2} }{2457}\sqrt{x}+\frac{11798 }{36855}x}{1-\frac{340 \sqrt{2} }{819}\sqrt{x}+\frac{815}{4914}x }$$ For $k=1$ this would give $1.165842516$ (!!).

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  • $\begingroup$ I wanted to evaluate an asymptotic integral where $(W(-exp(x))+1)$ was in the denominator. This is why I was looking for a fractional Taylor expansion around the branch point. $\endgroup$
    – user48672
    Commented Jun 11, 2020 at 15:10
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Computing the Series

First note that if $x=we^w$, then at $w=-1$, $$\newcommand{\W}{\operatorname{W}} \begin{align} x^{(k)} &=(w+k)e^w\\ &=(k-1)e^{-1}\tag1 \end{align} $$ So that at $w=-1$, we have the Taylor series $$ x=\sum_{k=0}^\infty\frac{k-1}{k!}e^{-1}(w+1)^k\tag2 $$ Multiplying by $e$, adding $1$, then doubling gives $$ \begin{align} 2(ex+1) &=\sum_{k=2}^\infty\frac{2(k-1)}{k!}(w+1)^k\\ &=(w+1)^2+\frac23(w+1)^3+\frac14(w+1)^4+\frac1{15}(w+1)^5+O(w+1)^6\tag3 \end{align} $$ Take the positive square root; i.e. $w\gt-1$: $$ \scriptsize\sqrt{2(ex+1)}=(w+1)+\frac13(w+1)^2+\frac5{72}(w+1)^3+\frac{11}{1080}(w+1)^4+O(w+1)^5\tag4 $$ Revert the series $$ \scriptsize\W(x)+1=\sqrt{2(ex+1)}-\frac23(ex+1)+\frac{11\sqrt2}{36}(ex+1)^{3/2}-\frac{43}{135}(ex+1)^2+O(ex+1)^{5/2}\tag5 $$ Substitute $x\mapsto-e^{-x-1}$, then apply the series $1-e^{-x}=x-\frac12x^2+\frac16x^3+O(x)^4$: $$ \begin{align} &\W\left(-e^{-x-1}\right)+1\\ &\scriptsize=\sqrt{2\left(1-e^{-x}\right)}-\frac23\left(1-e^{-x}\right)+\frac{11\sqrt2}{36}\left(1-e^{-x}\right)^{3/2}-\frac{43}{135}\left(1-e^{-x}\right)^2+O\left(1-e^{-x}\right)^{5/2}\\ &=\sqrt{2x}-\frac23x+\frac{\sqrt2}{18}x^{3/2}+\frac2{135}x^2+O(x)^{5/2}\tag6 \end{align} $$ Square and cube the series $$ \begin{align} \left(\W\left(-e^{-x-1}\right)+1\right)^2&=2x-\frac{4\sqrt2}3x^{3/2}+\frac23x^2-\frac{2\sqrt2}{45}x^{5/2}+O(x)^3\tag7\\ \left(\W\left(-e^{-x-1}\right)+1\right)^3&=2\sqrt2\,x^{3/2}-4x^2+\frac{5\sqrt2}3x^{5/2}-\frac{88}{135}x^3+O(x)^{7/2}\tag8 \end{align} $$ Note that $(6)$ says that $\W\left(-e^{-x-1}\right)=-1+\sqrt{2x}-\frac23x+\frac{\sqrt2}{18}x^{3/2}+\frac2{135}x^2+O(x)^{5/2}$, which combined with $(7)$ and $(8)$ gives $$ \bbox[5px,border:2px solid #C0A000]{2\W\left(-e^{-x-1}\right)\frac{\left(\W\left(-e^{-x-1}\right)+1\right)^2-2x}{\left(\W\left(-e^{-x-1}\right)+1\right)^3}=\frac43-\frac{\sqrt2}3\sqrt{x}-\frac8{45}x+O(x)^{3/2}}\tag9 $$


Graphing the Function

The graph in the question spans too large a domain to see clearly what the limit is. The limit is $\frac43=1.3333333$, not $\sqrt2=1.4142136$; however, the large domain, $[0,5]$ makes it hard to see what the limit is at $0$. In the graph below, the domain is $[0,0.1]$, and the line $y=\frac43$ is plotted for reference.

I have also graphed $\frac43-\frac{\sqrt2}3\sqrt{x}-\frac8{45}x$ to compare it with the actual function, but the line thickness covers the difference for most of the graph. I made that curve red so that it would be more noticeable where the difference can be seen.

enter image description here


L'Hôpital Approach

Taking the derivative of $x=we^w$ gives $1=(w+1)e^ww'=(w+1)\frac xww'$, so that $$ \W'(x)=\frac{\W(x)}{x(\W(x)+1)}\tag{10} $$ Suppose $u=-e^{-x-1}$, then $x=-1-\log(-u)$ and $$ \begin{align} &\lim_{x\to0^+}2\W\left(-e^{-x-1}\right)\frac{\left(\W\left(-e^{-x-1}\right)+1\right)^2-2x}{\left(\W\left(-e^{-x-1}\right)+1\right)^3}\tag{11}\\ &=\lim_{u\to-1/e}2\W(u)\frac{(\W(u)+1)^2+2+2\log(-u)}{(\W(u)+1)^3}\tag{12}\\ &=-2\lim_{u\to-1/e}\frac{2(\W(u)+1)\W'(u)+\frac2u}{3(\W(u)+1)^2\W'(u)}\tag{13}\\ &=-\frac43\lim_{u\to-1/e}\frac{\W(u)+1+\frac{\W(u)+1}{\W(u)}}{(\W(u)+1)^2}\tag{14}\\ &=-\frac43\lim_{u\to-1/e}\frac1{\W(u)}\tag{15}\\[3pt] &=\frac43\tag{16} \end{align} $$ Explanation:
$(11)$: the limit we seek
$(12)$: substitute $x=-1-\log(-u)$
$(13)$: pull $-2$ outside and apply L'Hôpital
$(14)$: pull $\frac23$ outside and divide numerator and denominator by $\W'(u)$
$(15)$: divide numerator and denominator by $(\W(u)+1)^2$
$(16)$: $\W(-1/e)=-1$

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  • $\begingroup$ Do you treat x as a composite function? I haven't thought about this. $\endgroup$
    – user48672
    Commented Jun 11, 2020 at 18:23
  • $\begingroup$ Yes. At the beginning, I look at $x$ as a function of $w\approx-1$; i.e. $x=we^w$. Then in $(5)$, we use series reversion to get $w$ as a function of $x\ge-e^{-1}$. $\endgroup$
    – robjohn
    Commented Jun 11, 2020 at 18:35
  • $\begingroup$ Now, understand. I managed to derive it algorithmically as well. $\endgroup$
    – user48672
    Commented Jun 12, 2020 at 21:14
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$\require{begingroup} \begingroup$ $\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}$

\begin{align} L &=\lim_{x \rightarrow 0^+} \frac{2 \W(-e^{-x-1}) \left(\W(-e^{-x-1})^2+2 \W(-e^{-x-1}) -2 x+1\right)} {(\W(-e^{-x-1}) +1)^3} \tag{1}\label{1} \\ &= 2 \lim_{x \rightarrow 0^+}\W(-e^{-x-1}) \cdot \lim_{x \rightarrow 0^+} \frac{(\W(-e^{-x-1})+1)^2-2 x} {(\W(-e^{-x-1}) +1)^3} \tag{2}\label{2} =-2\cdot L_1 ,\\ L_1&= \lim_{x \rightarrow 0^+} \frac{(\W(-e^{-x-1})+1)^2-2 x} {(\W(-e^{-x-1}) +1)^3} \tag{3}\label{3} . \end{align}

Let

\begin{align} y&=\W(-e^{-x-1})+1 \tag{4}\label{4} ,\\ x&=-(y+\ln(1-y)) \tag{5}\label{5} , \end{align}

$y\rightarrow 0^+$ when $x\rightarrow 0^+$, so

\begin{align} L_1&= \lim_{y \rightarrow 0^+} \frac{y^2+2 (y+\ln(1-y))} {y^3} \tag{6}\label{6} \end{align}

Now we can apply the L'Hopital's rule just once:

\begin{align} L_1&= \lim_{y \rightarrow 0^+} \frac{2y+2-\frac2{1-y}} {3y^2} \tag{7}\label{7} \\ &= \lim_{y \rightarrow 0^+} \frac23\cdot\frac1{y-1} =-\frac23 \tag{8}\label{8} , \end{align}

hence \begin{align} L&=-2\cdot L_1 =\frac43 \tag{9}\label{9} . \end{align}

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