 Confidence interval from a relative normal pivot

Suppose $\theta >0$ and a positive estimator $T_n$ satisfies $\sqrt{n}(T_n-\theta )/\theta \stackrel{d}{\to }N(0,1)$. Choose the standard normal quantile $c={\Phi }^{-1}(1-\alpha /2)$. For $n>c^2$, an asymptotic confidence interval of level $1-\alpha$ isProof. Convergence in distribution gives $\mathbb{P}(-c<\sqrt{n}(T_n-\theta )/\theta <c)\to 1-\alpha$. Multiplication by the positive $\theta$ and rearrangement makes this event equivalent to $\theta (1-c/\sqrt{n})<T_n<\theta (1+c/\sqrt{n})$. Both factors are positive when $n>c^2$, so division yields exactly the displayed interval. Thus its coverage probability tends to $1-\alpha$. For $95\%$ coverage, $c\approx 1.96$. This inversion retains the unknown scale $\theta$ in the normalized error, rather than replacing it by the observed estimator before inversion.