Use the Central Limit Theorem to deduce that if $λ$ is large, then $X$ approximately has a normal distribution. 
The time instants of incoming requests at a data server can be modelled with a Poisson process. 
  Let $X$ be the number of requests in one hour and let $λ$ be the intensity (requests per hour) of the Poisson process. 
  - Use the Central Limit Theorem to deduce that if $λ$ is large, then $X$ approximately has a normal distribution. Also specify its parameters.

Can I assume that, if $X\sim\operatorname{Poisson}(\lambda)$ then:
$$
\frac{X-\lambda}{\sqrt\lambda} \overset{\text{distribution}}\longrightarrow N(0,1) \text{ as } \lambda\to\infty \\  $$
It is a correct way to prove it? The parameters for the normal distribution are $\mu=0$ and $\sigma^2=1$ right?
 A: What you could say is that if $Y_i\sim\operatorname{Poisson}(\lambda_0)$ and you take $n$ i.i.d. samples 
then $\sqrt{n}\left(\frac1n\sum\limits_{i=1}^n Y_i - \lambda_0\right)  \overset{\text{d}}\longrightarrow N(0,\lambda_0)$ in distribution as $n$ increases by the central limit theorem
Meanwhile $S_n = \sum\limits_{i=1}^n Y_i \sim\operatorname{Poisson}(n\lambda_0)$ so $\sqrt{n}\left(\frac{S_n}n - \lambda_0\right) \overset{\text{d}}\longrightarrow N(0,\lambda_0)$ i.e. $\dfrac{S_n -{n\lambda_0} }{\sqrt{n\lambda_0}} \overset{\text{d}}\longrightarrow N(0,1)$ as $n$ increases
This is close to your $\dfrac{X -{\lambda} }{\sqrt{\lambda}} \overset{\text{d}}\longrightarrow N(0,1)$ as $\lambda$ increases, when $\lambda = n\lambda_0$ and $X=S_n$.  For a better proof you could use characteristic functions instead of the central limit theorem
A: $\def\l{\lambda}$
Let $F_{\lambda}(x)=P(\frac{X_\lambda-\lambda}{\sqrt{\lambda}}\le x),$ where $X_\lambda\sim \text{Poi}(\lambda)$. You need to show that 
$
\lim_{\lambda\to\infty} F_\lambda(x)=\Phi(x)
$
holds for all $x$. From Yves Duvast's answer, you know that the sequence
$$
F_1(x),F_2(x),F_3(x),\dots\to \Phi(x)
$$
 By the same logic, for any fixed $\mu>0$, the sequence
$$
F_\mu(x),F_{2\mu}(x),F_{3\mu}(x),\dots\to \Phi(x)\tag 1
$$
 Now, consider the function $f:\mathbb R^+\to \mathbb R$ defined by $$f(t)= F_{1/t}(x)-\Phi(x).$$ You can show that $f$ is continuous in $t$. Furthermore, for all fixed $t>0$,  letting $\mu=1/t$ in $(1)$ implies that $\lim_{n\to\infty}f(t/n)=0$. Appealing to this problem, we can then conclude that $\lim_{t\to 0} f(t)=0$, implying that $\lim_{\lambda\to \infty}F_\lambda(x)=\Phi(x)$.

This is not the proof I wanted to write; here is more along the lines of what I tried to do but could not finish. You know that $X_{\lambda}+X_{\mu}\stackrel{d}=X_{\lambda+\mu}$, provided $X_\lambda$ and $X_\mu$ are independent Poisson random variables. This should allow you to approximate $\frac{X_\lambda-\l}{\sqrt\l}$ by $\frac{X_{\lfloor\l\rfloor}-{\lfloor\l\rfloor}}{\sqrt{\lfloor\l\rfloor}}$, so that the probability the latter was less than $x$ is close to the former. This allows you to carry Yves's result about convergence to the integers to the reals. I think this could be salvaged.
A: Consider two Poisson iid variables of parameter $\lambda=1$.
Their pdf is $$p_1(k)=\frac1{ek!}.$$
The pdf of the sum of these variables is given by
$$p_2(k)=\sum_{i+j=k}\frac1{e^2i!j!}=\frac1{e^2k!}\sum\binom ki=\frac{2^k}{e^2k!}$$ which is simply a Poisson law of parameter $\lambda=2$.
More generally, you could show that the sum of $\lambda$ Poisson iid variables of parameter $1$ follows a Poisson law of parameter $\lambda$.
Then by the CLT,
$$\frac{p_\lambda-\lambda}{\sqrt\lambda}\to N(0,1)$$
and 
$$p_\lambda\to N(\lambda,\sqrt\lambda).$$
