Proof. Because the limit \(c\) is a constant, \(Y_n \underset {P}{\longrightarrow } c\) gives joint convergence \((X_n,Y_n)\underset {D}{\longrightarrow }(X,c)\) — this is the step that fails for a non-constant limit, since the marginal behaviour of two sequences does not determine their joint behaviour. Each of the three maps \[(x,y)\mapsto x+y,\qquad (x,y)\mapsto xy,\qquad (x,y)\mapsto x/y\] is continuous at every point of the form \((x,c)\), the last provided \(c\neq 0\). The continuous mapping theorem applied to the pair then gives the three conclusions. □

Solution. By the central limit theorem, \[\frac {\sqrt {n}\left (\overline {X}-\mu \right )}{\sigma } \underset {D}{\longrightarrow } Z\sim N(0,1).\] By the weak law applied to \(X_i^{2}\) and to \(X_i\), together with the continuous mapping theorem, \(S_n^{2}\underset {P}{\longrightarrow }\sigma ^{2}\) and hence \(\sigma /S_n \underset {P}{\longrightarrow } 1\), the square root and the reciprocal both being continuous at \(\sigma ^{2}>0\). Writing \[\frac {\sqrt {n}\left (\overline {X}-\mu \right )}{S_n} = \underbrace {\frac {\sqrt {n}\left (\overline {X}-\mu \right )}{\sigma }}_{\underset {D}{\longrightarrow }\,Z} \ \cdot \ \underbrace {\frac {\sigma }{S_n}}_{\underset {P}{\longrightarrow }\,1},\] part (b) of Slutsky’s theorem gives the product converging in distribution to \(1\cdot Z = Z\).