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Put ${\lambda }_i=z|V_i|$. Thenand ${\lambda }_i\to \infty$. The Poisson central limit theorem givesSince $\sqrt{{\lambda }_i}=\sqrt{z}\sqrt{|V_i|}$,

Equivalently, the characteristic function of the left-hand side iswhich is the characteristic function of that centered normal distribution.

The Poisson central limit theorem, or the ordinary central limit theorem applied to the $X_i$, gives $Y_n\stackrel{d}{\to }Y$ for a standard normal random variable $Y$. The negative-part map $x\mapsto x^{-}:=\max (-x,0)$ is continuous, so the continuous mapping theorem gives

For fixed real $t$, Taylor's theorem for the exponential function givesSubstituting into the previous moment-generating function,The continuity theorem for moment-generating functions states that, if the moment-generating functions of $Z_n$ and $Z$ are finite on a common open interval containing zero and $M_{Z_n}(t)\to M_Z(t)$ throughout that interval, then $Z_n\stackrel{\mathrm{d}}{\to }Z$. Here finiteness holds for all real $t$, and the limit is the moment-generating function of a standard normal variable. ThereforeThis proves the required instance of the Poisson central limit theorem directly from moment-generating functions; no appeal to a general central limit theorem is needed.

If $X_{\lambda }$ has a Poisson distribution, then $\mathbb{P}(X_{\lambda }\le \lfloor \lambda \rfloor )\to 1/2$ as $\lambda \to \infty$. The Poisson central limit theorem standardizes by mean $\lambda$ and variance $\lambda$; the cutoff $(\lfloor \lambda \rfloor -\lambda )/\sqrt{\lambda }$ tends to zero. Squeezing between fixed nearby cutoffs and using continuity of the standard normal distribution function gives the limit.