Distance between the noise and the corrupted signal

How can one formalize the fact that the law of $$X+Z$$ where $$X \in \mathbb{R}^d$$ is any vector-valued random variable and $$Z\sim \mathcal{N}(0, \sigma^2 \mathbf{I}_d)$$ closely resembles the law of $$Z$$ if $$\sigma^2$$ is sufficiently large ? $$X$$ and $$Z$$ are supposed independent. Ideally, I would like to prove that some distance/divergence between the law of $$X+Z$$ and $$Z$$ approaches zero as $$\sigma^2\to \infty$$.

If 𝑋 were Gaussian, for example, with a non-zero mean, $$𝜇_{𝑋+𝑍}=𝜇_𝑋+𝜇_𝑍≠𝜇_𝑍.$$
$$\displaystyle \lim_{\sigma\rightarrow \infty} (\mu_{X+Z} - \mu_{Z})=\mu_X \ne 0.$$
So the difference between the law of $$X+Z$$ and the law of $$Z$$ does not approach zero.
• $X$ is any fixed vector-valued random variable, i.e. the law of $X$ cannot change or depend on the law of $Z$. – Nocturne Mar 26 at 19:18
• $X$ is constant? or a random variable? – mjw Mar 26 at 19:39
• Okay, $X\sim \mathcal{N}(0, \sigma_0^2 \mathbf{I}_d)$, for example. A fixed distribution, whereas you want to show what happens as the distribution of $Z$ spreads out, yes? – mjw Mar 26 at 19:43
• Well, if $X$ were Gaussian with a non-zero mean, $\mu_{X+Z} = \mu_X + \mu_Z \ne \mu_Z$. – mjw Mar 26 at 19:45