Normal approximation of Poisson using asymptotics on mass function I have an excercise in probability and I need your help:

Let $X \sim Poi(\lambda )$. Use asymptotics on the mass function directly to prove that, as $\lambda \to \infty$,
$\sqrt{\lambda}\cdot\Pr \{ X=\lfloor\lambda +x\sqrt\lambda\rfloor\}=\frac{1}{\sqrt{2\pi}}\exp (-x^2 /2)+\mathcal{O}(\lambda^{-1/2})$
(HINT: First argue
  that the floor is not important, then apply Stirling’s formula)

I did it up to a certain point where I'm stuck now. Here is the last developement I found (by using Stirling from the Poisson mass function):
$\sqrt{\lambda}\cdot\Pr \{ X=\lfloor\lambda +x\sqrt\lambda\rfloor\}=\sqrt\lambda\frac{\lambda^{\lambda+x\sqrt\lambda}\exp(-\lambda)}{(\lambda+x\sqrt\lambda)!}
=\frac{1}{\sqrt{2\pi}}\exp(x\sqrt{\lambda})\frac{1}{(\lambda ^2+x\lambda\sqrt{\lambda})^{\frac{1}{2}+\lambda+x\sqrt{\lambda}}}$
I tried to develop with Newton's Binomial then reuse Stirling but it's a mess.
Please I need help,
thank you

Let $z=\lambda + x \sqrt{\lambda}$, with $x$ fixed and $\lambda \to \infty$
Then, assumming $z$ is an integer, $$P(X=z)= e^{-\lambda}\frac{\lambda^z}{z!}=e^{-\lambda} \frac{1}{\sqrt{2\pi z}} \left(\frac{\lambda e}{z}\right)^z (1+O(1/z)) \tag{1}$$
Now consider
$$\begin{array}{rcl} \displaystyle\log \left(\frac{\lambda e}{z}\right)^z &=&\displaystyle(\lambda +x\sqrt{\lambda})\left(1-\log\left(1+\frac{x}{\sqrt{\lambda}}\right)\right)\\
&=&\displaystyle(\lambda +x\sqrt{\lambda})\left(1-\frac{x}{\sqrt{\lambda}}+\frac{x^2}{2\lambda} + O(\lambda^{-3/2})\right) \\
&=&\displaystyle\lambda -\frac{x^2}{2} + O(1/\sqrt{\lambda}) \tag{2}
\end{array} 
$$
(Notice that a cancelation of terms occurs here - that's why you need to take the second order Taylor expansion of the logarithm)
Hence $$e^{-\lambda}  \left(\frac{\lambda e}{z}\right)^z = e^{-x^2/2+ O(1/\sqrt{\lambda}) } \tag{3}$$
Can you go on from here?
