How  Binomial and Normal distributions approximate Poisson distribution respectively? From Wikipedia:


*

*In some cases, the cdf of the Poisson distribution is the limit of
the cdf of the binomial distribution： 

The Poisson distribution can be derived as a limiting case to the
  binomial distribution $\text{bin}(n,p)$ as the number $n$ of trials
  goes to infinity and the expected number $np$ of successes remains
  fixed — see law of rare events below. Therefore it can be used as an
  approximation of the binomial distribution if $n$ is sufficiently
  large and $p$ is sufficiently small.


*In some cases, the cdf of the Poisson distribution is the limit of
the cdf of the normal distribution:

For sufficiently large values of $λ$, (say $λ>1000$), the normal
  distribution with mean $λ$ and variance $λ$ (standard deviation
  $\sqrt{\lambda}$), is an excellent approximation to the Poisson
  distribution. If $λ$ is greater than about $10$, then the normal
  distribution is a good approximation if an appropriate continuity
  correction is performed, i.e., $P(X ≤ x)$, where (lower-case) $x$ is a
  non-negative integer, is replaced by $P(X ≤ x + 0.5)$. $$
             F_\mathrm{Poisson}(x;\lambda) \approx F_\mathrm{normal}(x;\mu=\lambda,\sigma^2=\lambda)\, $$

I wonder in what best senses these functional approximations are? Pointwise, uniformly, $L_2$, ...?
Thanks and regards!
 A: In comments above, Tim asked why it must be that if $X\sim\mathrm{Poisson}(\lambda)$ and $Y\sim\mathrm{Poisson}(\mu)$ and $X$ and $Y$ are independent, then we must have $X+Y\sim\mathrm{Poisson}(\lambda+\mu)$.
Here's one way to show that.
\begin{align}
& \Pr(X+Y= w) \\[8pt]
= {} & \Pr\Big( (X=0\ \& \ Y=w)\text{ or }(X=1\ \&\ Y=w-1) \\
& {}\qquad\qquad\text{ or } (X=2\ \&\ Y=w-2)\text{ or } \ldots \text{ or }(X=0\ \&\ Y=w)\Big) \\[8pt]
= {} & \sum_{u=0}^w \Pr(X=u)\Pr(Y=w-u)\qquad(\text{independence was used here}) \\[8pt]
= {} & \sum_{u=0}^w \frac{\lambda^u e^{-\lambda}}{u!} \cdot \frac{\mu^{w-u} e^{-\mu}}{(w-u)!} \\[8pt]
= {} & e^{-(\lambda+\mu)} \sum_{u=0}^w \frac{1}{u!(w-u)!} \mu^u\lambda^{w-u} \\[8pt]
= {} & \frac{e^{-(\lambda+\mu)}}{w!} \sum_{u=0}^w \frac{w!}{u!(w-u)!} \mu^u\lambda^{w-u} \\[8pt]
= {} & \frac{e^{-(\lambda+\mu)}}{w!} (\lambda+\mu)^w
\end{align}
and that is what was to be shown.
A: "In some cases, the cdf of the Poisson distribution is the limit of the cdf of the normal distribution:"
That is false.  The limit of Poisson distributions is a normal distribution; the limit of normal distributions is not a Poisson distribution.
"Best senses" is a term I've never come across and I don't know what you mean by it.
Suppose $X\sim\mathrm{Poisson}(\lambda)$.  Then the cdf of $(X-\lambda)/\sqrt{\lambda}$ converges pointwise to the cdf of the standard normal distribution as $\lambda\to\infty$.
Now suppose $X\sim\mathrm{Bin}(n,\lambda/n)$.  Then the cdf of $X$ converges pointwise to the cdf of the Poisson distribution with expectation $\lambda$ as $n\to\infty$.
Generally "convergence in distribution" means a sequence of cdf's converges to a cdf pointwise except that it need not converge at points where the limiting cdf is not continuous.  The reason for the exception is things like this: Concentrate probability $1$ at $1/n$.  Then the value of the cdf at $0$ is $0$.  But the limiting cdf has value $1$ at $0$; that's where it concentrates all the probability.
But don't write about normal distributions approaching a Poisson distribution.  It's the other way around.
