Limit with Lambert-$W$ function

Question

I have asked a similar question about this one particular limit:

\begin{equation} A=\lim_{c\to 1}\exp\left[ -\left(\frac{1}{1-c}\right)\left(W_{0}\left[ B\left( 1+\frac{x}{rc}\right) \right]-W_{0}[B]\right)\right] \end{equation} where $r>0$, $x \in \mathbb{R}$, $W_0$ is the $k=0$ branch of the Lambert-$W$ function defined as: \begin{equation} x=W(x)e^{W(x)} \end{equation} and B is defined as: \begin{equation} B=\frac{(1-c)r}{1-(1-c)r}\exp\left[ \frac{(1-c)r}{1-(1-c)r} \right] \end{equation} Now, I tried with Mathematica and it is shown that $A \to e^{-x}$ when $c\to 1$ but I have trouble proving that analytically.

If anyone could assist me with this one I would be grateful. Thank you!

Answer 1

First note that as $c\rightarrow 1$, $(1-c)\rightarrow 0$, and so $$ B \sim (1-c)r e^{(1-c)r} \sim (1-c) r. $$ Moreover, for small $x$, $W_0(x) \sim x$. So $$ W_0\left[B\left(1+\frac{x}{rc}\right)\right]-W_0[B]\sim\frac{B x}{rc}\sim(1-c)x. $$ Hence $$ A=\lim_{c\rightarrow 1}\exp\left[-\left(\frac{1}{1-c}\right)\left((1-c)x\right)\right]=e^{-x}. $$

Answer 2

This is not an answer but it is too long for a comment.

mjqxxxx's answer contains all the required steps.

Concerning the approximation made for $B$, consider the definition $$B=\frac{(1-c)r}{1-(1-c)r}\exp\left[ \frac{(1-c)r}{1-(1-c)r} \right]$$ and let us define $a=(1-c)r$ which makes $$B=\frac a{1-a}\exp\left[ \frac{a}{1-a} \right]$$ and develop as a Taylor series built at $a=0$; this gives $$B=a+2 a^2+O\left(a^3\right)$$

Similarly, for small values of $y$, using Taylor again $$W(B(1+y))-W(B)=\frac{ W(B)}{1+W(B)}y+O\left(y^2\right)$$ and, for small $B$ $$\frac{ W(B)}{1+W(B)}=B-2 B^2+O\left(B^3\right)$$ Combining all the above leads to

$$W_{0}\left[ B\left( 1+\frac{x}{rc}\right) \right]-W_{0}[B]=(1-c) x+(1-c)^2 \left(x-x^2\right)+O\left((c-1)^3\right)$$ $$ -\left(\frac{1}{1-c}\right)\left(W_{0}\left[ B\left( 1+\frac{x}{rc}\right) \right]-W_{0}[B]\right)=-x-(1-c) \left(x-x^2\right)+O\left((c-1)^2\right) $$