How did the book, on page 429, near the end, arrive at $$A_{1}=-e$$ ?  I am not able to see it. This is what I tried. The book does things little diﬀerent than what we did in class. The book does not do $\lim _{\xi \rightarrow \infty }y^{in}\thicksim \lim _{x\rightarrow 0}y^{out}$ But instead, book replaces $$x$$ in the $$y^{out}\left ( x\right )$$ solution already obtained, with $$\xi \varepsilon$$, and rewrites $$y^{out}\left ( x\right )$$, which is what equation (9.2.14) is. So following this, I am trying to verify the book result for $$A_{1}=-e$$, but do not see how.     Using the book notation, of using $$X$$ in place of $$\xi$$, we have$Y_{1}\left ( X\right ) =\left ( A_{1}+A_{0}\right ) \left ( 1-e^{-X}\right ) -eX$ Which is the equation in the book just below 9.2.14. The goal now is to ﬁnd $$A_{1}$$. Book already found $$A_{0}=e$$ earlier. So we write\begin{align*} \lim _{X\rightarrow \infty }\overset{Y_{1}\left ( X\right ) }{\overbrace{\left ( A_{1}+A_{0}\right ) \left ( 1-e^{-X}\right ) -eX}} & \thicksim y^{out}\left ( x\right ) \\ \lim _{X\rightarrow \infty }\left ( A_{1}+A_{0}\right ) \left ( 1-e^{-X}\right ) -eX & \thicksim e\left ( 1-\varepsilon X+\frac{\varepsilon ^{2}X^{2}}{2!}-\cdots \right ) \end{align*}

So far so good. But now the book says ”comparing $$Y_{1}\left ( x\right )$$ when $$X\rightarrow \infty$$ with the second term in 9.2.14 gives $$A_{1}=-e$$”. But how? If we take $$X\rightarrow \infty$$ on the LHS above, we get$\lim _{X\rightarrow \infty }\left ( A_{1}+A_{0}\right ) -eX\thicksim e\left ( 1-\varepsilon X+\frac{\varepsilon X^{2}}{2!}-\cdots \right )$ But $$A_{0}=e$$, so\begin{align*} \lim _{X\rightarrow \infty }A_{1}+e-eX & \thicksim e-e\varepsilon X+e\frac{\varepsilon X^{2}}{2!}-\cdots \\ \lim _{X\rightarrow \infty }A_{1}-eX & \thicksim -e\varepsilon X+e\frac{\varepsilon X^{2}}{2!}-\cdots \end{align*}

How does the above says that $$A_{1}=-e$$ ? If we move $$-eX$$ to the right sides, it becomes\begin{align*} A_{1} & \thicksim eX-e\varepsilon X+e\frac{\varepsilon X^{2}}{2!}-\cdots \\ A_{1} & \thicksim e\left ( X-\varepsilon X\right ) +e\frac{\varepsilon X^{2}}{2!}-\cdots \end{align*}

I do not see how $$A_{1}=-e$$. Does any one see how to get $$A_{1}=-e$$?

Let redo this using the class method

\begin{align*} \lim _{X\rightarrow \infty }\overset{Y_{1}\left ( X\right ) }{\overbrace{\left ( A_{1}+A_{0}\right ) \left ( 1-e^{-X}\right ) -eX}} & \thicksim \lim _{x\rightarrow 0}y^{out}\left ( x\right ) \\ \lim _{X\rightarrow \infty }\left ( A_{1}+A_{0}\right ) \left ( 1-e^{-X}\right ) -eX & \thicksim \lim _{x\rightarrow 0}e\left ( 1-x+\frac{x^{2}}{2!}-\cdots \right ) \\ \lim _{X\rightarrow \infty }A_{1}+e-eX & \thicksim e\\ \lim _{X\rightarrow \infty }A_{1} & \thicksim eX \end{align*}

How does the above says that $$A_{1}=-e?$$ and what happend to the $$\lim _{X\rightarrow \infty }$$ of $$X$$ which remains there?