Solution
= Solution
At the origin, combine the rational coefficients before taking the limit:
$$
y(0)=\frac1{(1-a^2)(4-a^2)}-\frac a{3(1-a^2)}+\frac a{6(4-a^2)}
=\frac{a+3}{6(a+1)(a+2)}.
$$
Equivalently $y(0)=L(3L+1)/[6(L+1)(2L+1)]$. The apparent poles at $L=1$ cancel, giving
$$
\boxed{\lim_{L\to1^+}y(0)=\frac19.}
$$
The entire limiting profile is also finite:
$$
\lim_{L\to1^+}y(x)=\frac{(1+3|x|)e^{-|x|}+e^{-2|x|}}{18},
$$
which independently gives the same central value.