Limit probability of a complete bipartite random graph $G(n,n,p)$ is connected I need to calculate the following probability limit for a complete bipartite random graph $G(n,n,p)$ in the Erdos-Renyi model:
\begin{equation}
\lim_{n\rightarrow\infty}\mathbb{P}[G(n,n,p) \text{ is connected}]
\end{equation}
when $np=\ln n + w(n)$ where $w(n)\rightarrow -\infty$ when $n\rightarrow\infty$.
I know that the expected value of the number of isolated vertices in $G(n,n,p)$ is:
\begin{equation}
\mathbb{E}X=2n(1-p)^n
\end{equation}
Where X stands for the number of isolated vertices of course.
So the graph $G(n,n,p)$ is connected if it doesn't have any isolated vertices then by Markov inequality:
\begin{equation}
\mathbb{P}(X\ge 1)\le \mathbb{E}X
\end{equation}
and so substituting the value of $p$ i get:
\begin{eqnarray}
\lim_{n\rightarrow\infty}\mathbb{P}[G(n,n,p) \text{ is connected}]&=&\lim_{n\rightarrow\infty}\mathbb{P}(X\ge1)\leq\lim_{n\rightarrow\infty}\mathbb{E}X\\
&=&\lim_{n\rightarrow\infty}2n\bigg(1-\cfrac{\ln n + w(n)}{n}\bigg)^n
\end{eqnarray}
My questions are 2 actually:
1- Is correct what i did?
2- If the answer for 1 is YES. How can i calculate that limit?
Thanks in advance!
The variance for the number of isolated vertices is 
\begin{equation}
 \mathbb{V}X=4n^2(1-p)^n+2n(n-1)(1-p)^{2n}+n^2(1-p)^{2n-1}
\end{equation}
And by Chebyshev inequality:
\begin{eqnarray}
\mathbb{P}(X=0)&\leq&\cfrac{\mathbb{V}X}{(\mathbb{E}X)^2}
&\leq&\cfrac{4n^2(1-p)^n+2n(n-1)(1-p)^{2n}+n^2(1-p)^{2n-1}}{4n^2(1-p)^{2n}}
\end{eqnarray}
And sadly once again i got stuck with the limiting probability any hand on this?

You are right that if $X$ is the number of isolated vertices, then $P(X \geq 1) \leq E[X]$, because $X$ is a non-negative integer-valued random variable. 
Note that on the event that $\{X \geq 1\}$, the graph is not connected. Thus $\{X \geq 1\} \subseteq \{G(n,n,p) \text{ is not connected}\}$, and in fact,
$$
P(X \geq 1) \leq P(G(n,n,p) \text{ is not connected}).
$$
So having an upper bound on $P(X \geq 1)$ will not help us here. In fact, 
\begin{align*}
E[X] &= 2n \left( 1 - p_n \right)^n = 2 \exp \left( \ln n +  n \ln \left( 1- p_n \right)\right) \\&= 2 \exp \left( \ln n + n (-p_n + O(p_n^2)) \right) \\ &= 2 \exp \left( -\omega_n + O(\ln^2 n/ n) \right) \to \infty,
\end{align*}
as $n \to \infty$. 
Now it is possible to show that $P(X \geq 1) \to 1$ (and hence, $P(G(n,n,p) \text{ is not connected}) \to 1$) without so much more work. If you calculate the second moment of $X$, you can use Chebyshev's Inequality to get this fact. 
