Let $C$ be a curve in $\mathbb{P}^3$, possibly nonreduced. Assume, there exists a smooth surface in $\mathbb{P}^3$ containing $C$. Is it true that for $d \gg 0$, a generic element of $I_d(C)$ defines a smooth surface in $\mathbb{P}^3$?

I hope this is not a homework exercise, but I do not recall this from the standard textbooks. This is already false for planar double lines. Let $\mathbb{P}^3$ have homogeneous coordinates $[T_0,T_1,T_2,T_3]$. Let $S$ be the plane $Z(T_3)$. Let $C$ be the curve $Z(T_2^2,T_3)$ with induced reduced curve $L=Z(T_2,T_3)$. Let $R$ be a smooth hypersurface of degree $d$ in $\mathbb{P}^3$ that contains $L$. Since $C$ is a local complete intersection scheme, if $R$ contains $C$, then $C$ is a Cartier divisor on $R$. Thus $[C]$ equals $2[L]$ as an effective Cartier divisor on $R$. In particular, the kernel $\mathcal{J}$ of the reduction homomorphism, $$ \mathcal{O}_C \to \mathcal{O}_L, $$ equals $\mathcal{O}_R(L)_L$, i.e., the dual of the normal sheaf of $L$ in $R$ (the conormal sheaf). Considering the special case when $R$ equals $S$, we compute that $\mathcal{J}$ is isomorphic to $\mathcal{O}_L(1)$. On the other hand, by adjunction, Chern class computations, etc., for general $R$ of degree $d$ that contains $L$, $\mathcal{O}_R(L)_L$ is isomorphic to $\mathcal{O}_L(2d)$. Thus, if $R$ contains $C$, then $d$ must equal $1$. 


EDIT. The first version of this answer claimed that the result was true for any $C$, but it was uncorrect as remarked by J. Starr. In fact, Jason's answer shows that the result may be false for a nonreduced $C$. However, the claim is true when $C$ is reduced; let me give a sketch of the proof. Since $\mathcal{O}_{\mathbb{P}^3}(1)$ is ample, there exists $d_0 \in \mathbb{N}$ such that for any $d \geq d_0$ the sheaf $\mathcal{I}_C(d)$ is generated by its global sections. Hence the base locus of $\mathcal{I}_C(d)$ consists of the curve $C$ only. In particular, the linear system $\mathcal{I}_C(d)$ has not fixed components, so Bertini theorem implies that the general element of the linear system $\mathcal{I}_C(d)$ is smooth outside $C$. $(*)$ Let now $S$ be a smooth surface of degree $n$ containing $C$ (we may assume $d_0 > n$) and consider the elements in $\mathcal{I}_C(d)$ of the form $$X=S +H_1+H_2+ \cdots + H_{dn},$$ where the $H_i$ are hyperplanes. If $p$ is any point of $C$ and we choose as $H_i$ hyperplanes not containing $p$, then $X$ is a surface of degree $d$ which is smooth at $p$. Since $p \in C$ is arbitrary, by using $(*)$ it follows that the general element of $\mathcal{I}_C(d)$ is smooth everywhere for $d \geq d_0$. 

