4.9 note p9. added 2/24/17, asking question about problem in book

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?