Clarification of Solution of PDE Laplace Transform Problem I have the following problem:

The transverse displacement $u(x, t)$ of a semi-infinite elastic string satisfies 
$$\dfrac{\partial^2{u}}{\partial{t}^2} = c^2 \dfrac{\partial^2{u}}{\partial{x}^2}$$
for $x > 0$, $t > 0$,
with initial conditions 
$$u(x, 0) = \dfrac{\partial{u(x, 0)}}{\partial{t}} = 0$$
for $x > 0$,
and boundary condition
$$-\beta \dfrac{\partial{u(0, t)}}{\partial{x}} = f(t)$$
for $t > 0$.
Show, using Laplace transforms, that the solution can be written 
$$u(x, t) = \dfrac{c}{\beta} H \left( t - \dfrac{x}{c} \right) \int^{t - x/c}_0 f(u) \ du,$$
where $H(x)$ is the Heaviside function. Can you interpret this result physically?

And I have the following solution:

Taking the Laplace transform of the PDE yields
$$s^2 \mathcal{L} \{u\} - su(x, 0) - u_t(x, 0) = c^2 \dfrac{d^2}{dx^2} \mathcal{L} \{ u(x, t) \}$$.
Substituting the initial conditions gets us
\begin{align} &s^2 \mathcal{L} \{ u \} = c^2 \dfrac{d^2}{dx^2} \mathcal{L} \{ u(x, t) \} \\ 
&\Rightarrow \dfrac{d^2}{dx^2} \mathcal{L} \{ u(x, t) \} = \dfrac{s^2}{c^2} \mathcal{L} \{ u(x, t) \} \\ 
&\Rightarrow \dfrac{d^2}{dx^2} \mathcal{L} \{ u(x, t) \} - \dfrac{s^2}{c^2} \mathcal{L} \{u(x, t) \} = 0 \end{align}
We solve this homogeneous second-order linear DE:
\begin{align} &m^2 - \dfrac{s^2}{c^2} = 0 \\
&\Rightarrow m = \pm \dfrac{s}{c} \end{align}
Therefore, the general solution to this DE is 
$$\mathcal{L} \{ u(x, t) \} = A(s) e^{x\frac{s}{c}} + B(s)e^{-x \frac{s}{c}},$$
where $A(s)$ and $B(s)$ are arbitrary functions.
To apply the boundary conditions, we differentiate the general solution with respect to $x$:
$$\dfrac{\partial}{\partial{x}} \mathcal{L} \{ u(x, t) \} = \dfrac{sA(s)}{c}e^{\frac{sx}{c}} - \dfrac{sB(s)}{c} e^{-\frac{sx}{c}}$$
And since we had the boundary condition
$$- \beta \dfrac{\partial}{\partial{x}} u(0, t) = f(t),$$
we have 
$$\dfrac{\partial}{\partial{x}} \mathcal{L} \{ u(0, t) \} = \dfrac{sA(s)}{c} - \dfrac{sB(s)}{c}$$
Taking the Laplace transform of the boundary condition, we get
$$- \beta \dfrac{\partial}{\partial{x}} \mathcal{L} \{ u(0, t) \} = \mathcal{L} \{ f(t) \}$$
So we get 
\begin{align} &-\dfrac{\mathcal{L} \{f(t) \}}{\beta} = \dfrac{sA(s)}{c} - \dfrac{sB(s)}{c} \\
& \Rightarrow -\dfrac{\mathcal{L} \{ f(t) \}}{\beta} = \dfrac{s}{c}(A(s) - B(s)) \\
& \Rightarrow \dfrac{c}{\beta} \dfrac{ \mathcal{L} \{ f(t) \}}{s} + A(s) = B(s) \end{align}
Using this to eliminate $B(s)$ from the general solution leaves
$$\dfrac{\partial}{\partial{x}} \mathcal{L} \{ u(x, t) \} = \dfrac{sA(s)}{c} e^{\frac{sx}{c}} - \left( \dfrac{c}{\beta} \dfrac{ \mathcal{L} \{ f(t) \}}{s} + A(s) \right) \dfrac{s}{c} e^{-\frac{sx}{c}}$$
We can now constrain $A(s)$ by using the initial value theorem with the initial conditions:
\begin{align} 0 &= u(x, 0) \\
&= \lim_{t \to 0} u(x, t) \\
&= \lim_{s \to \infty} s \mathcal{L} \{ u(x, t) \} \end{align} 
for all $x > 0$.
Using the general solution, we see that 
$$\lim_{s \to \infty} s \mathcal{L} \{ u(x, t) \} = \lim_{s \to \infty} A(s) e^{\frac{sx}{c}} s,$$ 
since $\lim_{t \to 0} f(t) = 0$ means that $\lim_{s \to \infty} s \mathcal{L} \{ f(t) \} = 0$.

I'm confused about this last part:

Using the general solution, we see that 
$$\lim_{s \to \infty} s \mathcal{L} \{ u(x, t) \} = \lim_{s \to \infty} A(s) e^{\frac{sx}{c}} s,$$ 
since $\lim_{t \to 0} f(t) = 0$ means that $\lim_{s \to \infty} s \mathcal{L} \{ f(t) \} = 0$.

I don't see how they got 
$$\lim_{s \to \infty} s \mathcal{L} \{ u(x, t) \} = \lim_{s \to \infty} A(s) e^{\frac{sx}{c}} s$$
?
I would greatly appreciate it if someone could please take the time to clarify this.
 A: First I want to mention that you forgot the $\mathcal{L}$ symbol at some places:

So we get 
\begin{align} &-\dfrac{\mathcal{L}\{f(t)\}}{\beta} = \dfrac{sA(s)}{c} - \dfrac{sB(s)}{c} \\
& \Rightarrow -\dfrac{\mathcal{L}\{f(t)\}}{\beta} = \dfrac{s}{c}(A(s) - B(s)) \\
& \Rightarrow \dfrac{c}{\beta} \dfrac{\mathcal{L}\{f(t)\}}{s} + A(s) = B(s) \end{align}
Using this to eliminate $B(s)$ from the general solution leaves
$$\dfrac{\partial}{\partial{x}} \mathcal{L} \{ u(x, t) \} = \dfrac{sA(s)}{c} e^{\frac{sx}{c}} - \left( \dfrac{c}{\beta} \dfrac{\mathcal{L}\{f(t)\}}{s} + A(s) \right) \dfrac{s}{c} e^{-\frac{sx}{c}}$$

Now we can show the equation. Substituting the expression for $B(s)$ in $\mathcal{L}\{u(x,t)\}$ and multiplying with $s$ gives
$$ s\mathcal{L} \{ u(x, t) \} = s A(s) \left(e^{\frac{sx}{c}}+e^{-\frac{sx}{c}}\right) + \dfrac{c}{\beta} \dfrac{s\mathcal{L}\{f(t)\}}{s} e^{-\frac{sx}{c}}.  $$
From the boundary condition $f(t) =-\beta \frac{\partial u(0,t)}{\partial x}$ for all $t>0$, and the initial condition $u(x,0)=0$ for all $x>0$, it follows that $\lim_{t\to 0}f(t)=0$. Then the initial value theorem implies $\lim_{s\to\infty}s\mathcal{L}\{f(t)\}=0$. This and the fact that $x$, $s$ and $c$ are positive implies that.
$$\lim_{s\to\infty} \frac{c}{\beta} \frac{s\mathcal{L}\{f(t)\}}{s} e^{-\frac{sx}{c}}=0.$$

Edit: As was pointed out by The Pointer in the comments (pun not intended), it suffices to note that $\lim_{s\to \infty} \mathcal{L}\{f(t)\}=0$ (asymptotic property of Laplace transforms) and that the constants are positive.

Since $\lim_{s\to\infty} \mathcal{L}\{u(x,t)\} = 0$ (asymptotic property of Laplace transforms), also $\lim_{s\to\infty} A(s)=0$, and thus $\lim_{s\to \infty} A(s) s e^{-\frac{sx}{c}}=0$. 
We conclude that
$$\lim_{s\to \infty} s \mathcal{L}\{u(x,t)\} = \lim_{s\to\infty }s A(s) e^{\frac{sx}{c}}.$$
I hope this answer is helpful to you. 
