= Solution
A <rational function> with no poles away from $P_\infty$ is regular on the smooth affine curve. Its ring of <regular functions> is the coordinate ring
$$
k[x,y]/(y^2-\textstyle\prod_j(x-\lambda_j)),
$$
so it has a unique expression $A(x)+yB(x)$ with <polynomials> $A,B$. This regular-function assertion uses smoothness, hence normality, of the affine curve, not <Riemann-Roch theorem>.
The nonconstant leading term of $A$ has even pole order $2\deg A$ at infinity, while $yB$ has odd pole order $3+2\deg B$. These different parities prevent cancellation. Thus no nonconstant such function has a pole of order at most one. Therefore
$$
\boxed{L(P_\infty)=k.}
$$
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