There are multiple mistakes on both sides.
I said $0$ multiplied by anything should equal to $0$ even when $x$ approaches infinity.
This is not true! Indeed, $\lim_{x \to \infty} \frac 1x = 0$, but $\lim_{x \to \infty} \frac 1x \cdot x^2 = \infty$. Therefore, you were wrong on this front, but it turns out your answer was correct, and you acknowledged it by writing $\left[\frac 6x\right] = 0$ as $x$ approaches infinity. This is a stronger condition than the limit existing and equalling zero.
$\left[\frac{6}{x}\right] = \frac 6x - k$ for some $0 \leq k < 1$.
Note that we know what $k$ is, because $\frac 6x < 1$ for $x > 6$, so in fact $\left[\frac 6x \right] = 0$ for $x > 6$ , which simply makes $k = \frac 6x$! This allows us to work out what happens when $x \to \infty$, so you can conclude the answer here, and it is not $-\infty$, but in fact $0$.
Then my friend said $0 \times \infty$ can't be zero.
Once again not true, take $\frac 1{x^2}$ and the sequence $x$ whose product goes to zero although one goes to infinity.
For the answers:
Zero times infinity is not zero. However, the first sequence $\left[\frac 6x\right ]\frac x3$ is zero after $x>6$ because the first fraction is $0$, so the limit is $0$. Note that terms being equal to zero after some time, is stronger than the limit being zero. This absorbs the sequence $\frac x3$ regardless of what properties it may have.
The second you know.
For the third, we have to be more careful : we have an infinity i.e. $\left[\frac 6x\right]$ and a zero i.e. $\frac x3$ convergent sequence. Now, here $\infty \times 0$ confusion comes in, which is sorted by setting appropriate bounds on $\left[\frac 6x\right]$.
What bounds? Obviously, $\frac 6x -1 \leq \left[\frac 6x\right] \leq \frac 6x $.
Setting these in, we get :
$$
\left(\frac 6x -1 \right)\frac x3\leq \left[\frac{6}{x}\right]\frac x3 \leq \frac 6x\frac{x}{3}
$$
Now all we are left to notice is that the left and right hand side have limit $2$ as $x$ converges to $0$, hence the middle also has the same limit by squeeze theorem.
One textbook proceeds like this.
(1) Bernoulli's inequality:
$$
(1+x)^n \ge 1+nx\qquad\text{when } x > 0, n \in \mathbb N
$$
Hint: induction.
(2) Use (1) to show
$$
t^n \to \infty\quad\text{as } n \to \infty, \text{when } t > 1
$$
Hint: use $1+x=t$.
(3) Use (2) to show
$$
s^n \to 0\quad\text{as } n \to \infty, \text{when } 0<s<1
$$
Hint: use $t=1/s$.
Best Answer
[Note] : Notice that $n\cdot n! = (n+1)!-n!$, $\forall n\in\mathbb{N}$ go ahead and try to prove that using induction. It follows that : \begin{align*} \lim _{n \to \infty} \left[\displaystyle\frac{1 \cdot 1 !+2 \cdot 2 !+\ldots+n \cdot n !}{(n+1) !}\right]^{(n+1)!}&=\lim_{n\to\infty}\left[\frac{(n+1)!-1}{(n+1)!}\right]^{(n+1)!}\\ &=\lim_{n\to\infty}\left(1-\frac{1}{(n+1)!}\right)^{(n+1)!}\\ &=\boxed{\frac{1}{e}} \end{align*}