Proof that a sequence decreases past a certain point 
I would like to try to prove that the sequence
  $$
\frac{\sqrt{6}^{x}}{x!}
$$
  for $x=0, 1, 2, \cdots$, is strictly decreasing when $x\geq 2$. 

It appears to be the case after listing out a few terms. Here's my attempt at a proof:
To the contrary, suppose $\frac{\sqrt{6}^{x}}{x!} \leq \frac{\sqrt{6}^{x+1}}{(x+1)!}$ for $x \geq 2$. Then we have
\begin{align*}
\begin{split}
&\frac{x\text{ln}\sqrt{6}}{x!} \leq \frac{(x+1)\text{ln}\sqrt{6}}{(x+1)!}\\
&\Rightarrow \frac{1}{(x-1)!} \leq \frac{1}{x!}\\
&\Rightarrow \frac{x!}{(x-1)!} \leq 1\\
&\Rightarrow x \leq 1.
\end{split}
\end{align*}
But this contradicts our assumption on $x$, so the inequality above must in fact be ">". Thus the sequence is strictly decreasing for $x \geq 2$.
Any corrects/input appreciated.

The given sequence IS strictly decreasing for $x\geq 2$. If we follow your approach in a correct way, we find that the inequality 
$$\frac{\sqrt{6}^{x}}{x!} > \frac{\sqrt{6}^{x+1}}{(x+1)!},$$ is equivalent to
$$\frac{(x+1)!}{x!} > \frac{\sqrt{6}^{x+1}}{\sqrt{6}^{x}}$$
which reduces to
$$x+1> \sqrt{6}$$
which is satisfied when $x> \sqrt{6}-1\approx 1.449$.
P.S. Note that $\sqrt{6}^{x}=\exp(x\ln(\sqrt{6}))$ (not $x\ln(\sqrt{6})$).
