Asymptotic expansion of $x\exp(1/x)=\exp(\lambda)$ as $\lambda\to\infty$ I am struggling to find a three-term expansion of the following equation $x\exp(1/x)=\exp(\lambda)$ as $\lambda\to\infty$ for each of the solutions for $x$.
I graphed the functions $1/x$ and $\log(x)$ after taking the logarithm on both sides of the equation. I know, correct me if I am wrong, that for small root I get the $x\sim 1/\lambda$ balance, while for the large root the balance is $\log(x)\sim\lambda$.
Please guide me on how to proceed from here.
 A: We first can solve the equation exactly as follows:
$$
xe^{1/x}=z\implies \frac{1}{x}e^{-1/x}=\frac{1}{z}\implies -\frac{1}{x}e^{-1/x}=-\frac{1}{z}.
$$
It follows that
$$
x=-\frac{1}{W_0(-1/z)}\ \text{or}\ x=-\frac{1}{W_{-1}(-1/z)},
$$
where $W_0(\cdot)$ is the principal branch of the Lambert $W$-function.  In your question you have $z=e^\lambda$, where $\lambda\to\infty$ and so we want the solution for $x$ in terms of $W_0(\cdot)$. As such, we can use the Taylor series for $W_0(\cdot)$ to obtain the desired asymptotic expansion.  First note
$$
W_0(s)=s\sum_{n=1}^\infty\frac{(-(n+1))^n}{(n+1)!}s^n.
$$
Using the formula for the multiplicative inverse of formal power series we write
$$
\frac{1}{W_0(s)}=\frac{1}{s}\left(1+s-\frac{1}{2}s^2+\mathcal O(s^3)\right).
$$
Substituting $s=-1/z$ we then obtain as $z\to\infty$:
$$
x=-\frac{1}{W_0(-1/z)}=z-1-\frac{1}{2z}+\mathcal O(z^{-2}).
$$
Substituting $z=e^\lambda$ then gives us the final asymptotic result for $\lambda\to\infty$:
$$
x\sim e^\lambda-1-\frac{1}{2}e^{-\lambda}.
$$
Edit:
If we want the second solution $x\ll 1$ then the asymptotic series for $z\to\infty$ can be derived using the same process as above and an appropriate series expansion for $W_{-1}(\cdot)$.
