Let $X_\lambda$ have the Poisson distribution with parameter $\lambda$. Consider the limit distribution of $(X_\lambda-\lambda)/\lambda^{1/2}$ as $\lambda\rightarrow\infty$. Since $X_\lambda\sim\textit{Poi}\,(\lambda)$, we have $$\mathscr{E}(X_\lambda)=\lambda\mbox{ and }\sigma^2(X_\lambda)=\lambda.$$ $X_\lambda$ is a single random variable which of course be i.i.d. Thus by the Classical Central Limit Theorem, we have $$\frac{X_\lambda-\mathscr{E}(X_\lambda)}{\sigma(X_\lambda)\sqrt{1}} = \frac{X_\lambda-\lambda}{\lambda^{1/2}}\overset{\mathscr{L}}{\longrightarrow}\boldsymbol{\Phi},
$$where $\boldsymbol{\Phi}$ is normal distribution with mean 0 and variance 1.
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