Proving independence when Gaussian noise is added

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Consider r.v.'s $X$ and $Y$ which are not independent, i.e. $E[X\mid Y]\neq E[X]$. Can we show that if we add a variable $\epsilon$ following $\mathcal N(0,\sigma_{\epsilon})$, for which it holds that $\epsilon \perp X$ and $\epsilon \perp Y$, the following limit holds:

$$\lim_{\sigma_{\epsilon} \to \infty } E[X\mid Y+\epsilon ]=E[X] $$

Any help would be greatly appreciated!

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Supppose $X=Y \sim N(0,1)$ and $\epsilon=\sqrt {\sigma_{\epsilon}} X$. Then your conditions are satisfied but $E(X|Y+\epsilon)=E(X|(1+\sqrt {\sigma_{\epsilon}})X)=X $ which does not tend to $EX=0$.