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.
In general the growth of $u$ is not related to the growth of $n$ (the summation argument). The correct answer to this question is that the $\sum_{n=1}^{\infty}{\lim_{u\rightarrow\infty}{\frac{(-1)^nu^{n+s}}{(n-1)!(n+s)}}}$ has no definite value.
In contrast,
$$\lim_{u\rightarrow\infty}\sum_{n=1}^{\infty}{{\frac{(-1)^nu^{n+s}}{(n-1)!(n+s)}}}$$ exists as the summation parameter vanishing.
If we denote:
$$f(u,n)=\frac{(-1)^nu^{n+s}}{(n-1)!(n+s)}$$
then
$$g(u)=\sum_{n=1}^{\infty}f(u,n)$$
$$g'(u)=f(u,n)_u'=\frac{(-1)^nu^{n+s-1}}{(n-1)!}$$
$$\sum_{n=1}^{\infty}f(u,n)_u'=-e^{s-u}$$
So after integration:
$$\sum_{n=1}^{\infty}f(u,n)=e^{s-u} + C(s)$$
Now
$$\lim_{u\rightarrow\infty}\sum_{n=1}^{\infty}f(u,n)=\lim_{u\rightarrow\infty}e^{s-u}+C(s)=C(s)$$
$C(s)$ here is $-\Gamma(s+1,0)$. Incomplete $\Gamma$ function https://en.wikipedia.org/wiki/Incomplete_gamma_function.
Best Answer
Your "problem" already occurs with $n^{1/n}$. If you think "of $n$ first", then the limit would be infinite. If you think "of $1/n$ first", the limit would be zero. But it's neither! (as Chris mentioned above, this already happens with products, and even with sums).
There is no justification in isolating one part from the other: each new value of $n$ obviously implies a new value of $1/n$, so there's no ground in thinking that they should behave "independently".
This mistake is probably induced by the fact that there is actually "independence" in several situations, like "the limit of the sum is the sum of the limit", or "the limit of the product is the product of the limits"; but even these situations require both limits to exist. The power situation is no different, in the sense that you cannot freely "distribute" the limit.