Can someone help me do this limit? $ \lim_{n\to\infty} \frac{n!\times(2n)!}{(3n)!}$

In this case I'd just write it out recursively. Denote by $a_n$ the $n^{th}$ term of the sequence, so that $$ a_n = \frac{n!(2n)!}{(3n)!} = \frac{n(2n)(2n-1)}{(3n)(3n-1)(3n-2)}a_{n-1}$$ The basic idea behind understanding sequences is finding something simpler to compare it to. So let's take apart that factor: $$ \frac{n(2n)(2n-1)}{(3n)(3n-1)(3n-2)} = \frac{1}{3}\frac{2n-1}{3n-1}\frac{2n-1}{3n-2} < \frac{1}{3} $$ since each of those latter two fractions is less than $1$.

That means every new term in the sequence is less than a third the previous term: $a_n < a_{n-1}/3$. Your sequence starts with $1$, so $a_n < (1/3)^n$.

So now, ask a simpler question: what does $(1/3)^n$ do? Then use the answer to answer your original question.

This is going to be a horrible overkill, but a funny one. Since

$$ \sum_{n\geq 0}\frac{n!(2n)!}{(3n)!} = \sum_{n\geq 0}(3n+1) B(n+1,2n+1)=\int_{0}^{1}\sum_{n\geq 0}(3n+1)(1-x)^{n}x^{2n}\,dx $$ equals $\phantom{}_3 F_2\left(\frac{1}{2},1,1;\frac{1}{3},\frac{2}{3};\frac{4}{27}\right)$ or $\int_{0}^{1}\frac{1+2x^2-2x^3}{(1-x^2+x^3)^2}\,dx$, which is a convergent integral (and pretty close to $\sqrt{2}$), the main term of the original series goes to zero as $n\to +\infty$.

Hint: $$\frac{n! (2n)!}{(3n)!} = \prod_{i=1}^{n}\frac{i}{2n+i},$$ and $$\frac{i}{2n+i} \le \frac{1}{3}.$$