Proper way to solve $\lim\limits_{x\rightarrow\infty}\left[\frac{6}{x}\right]\frac{x}{3}$ I made a problem for my friends, but not everyone agreed with my answer.
$$\lim_{x\rightarrow\infty}\left[\frac{6}{x}\right]\frac{x}{3}+\lim_{x\rightarrow\infty}\frac{6}{x}\left[\frac{x}{3}\right]+\lim_{x\rightarrow0}\left[\frac{6}{x}\right]\frac{x}{3}$$
$[x]$ is the biggest integer less than or equal to $x$
My answer is $0+2+2=4$.
Everyone agreed that the second limit should be $2$.
However, I said the first limit should be $0$ because $\left[\frac{6}{x}\right]=0$ as $x$ approaches infinity.
so I said $0$ multiplied by anything should equal to $0$ even when $x$ approaches infinity.
Then my friend said $\left[\frac{6}{x}\right]$ can be expressed as $\frac{6}{x}-k$ where $0\leq k<1$.
Therefore, $$\lim_{x\rightarrow\infty}\left(\frac{6}{x}-k\right)\frac{x}{3}=\lim_{x\rightarrow\infty}2-\frac{kx}{3}=-\infty$$
Then I said we don't know what $k$ approaches to, so we can't conclude the answer here.
Then my friend said $0\times\infty$ can't be 0.
When I put it on WolframAlpha, $$\displaystyle\lim_{x\rightarrow\infty}\left[\frac{6}{x}\right]\frac{x}{3}=0$$
But I am not sure which answer is correct.
Can someone explain which is correct and why?
I didn't learn L'Hôpital's rule yet.
 A: 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.
A: Note that when $x=6,\left \lfloor \frac{6}{x} \right \rfloor\frac{x}{3} = \left \lfloor \frac{6}{6} \right \rfloor\frac{6}{3} =2$.
While when $x>6$, say $x=8, \left \lfloor \frac{6}{x} \right \rfloor\frac{x}{3} = \left \lfloor \frac{6}{8} \right \rfloor\frac{8}{3} =0\cdot \frac{8}{3}=0$.
Similarly for any value of $x \in (6,\infty)$.
What your friend said: "$0×\infty$  can't be 0." is not true.
However, $0 \times \infty$ is indeterminate, it can be $0$, it can be $\infty$, and it can be any other value.
A: Notice $[6/x]$ is equal to $0$ for all $x>6$, so that
$$
\lim_{x\rightarrow\infty}\left[\frac{6}{x}\right]\frac{x}{3} = \lim_{x\rightarrow\infty}0 \times \frac{x}{3}=0
$$
It is not a form of $0 \cdot \infty$, for that the value of $\left [\dfrac{6}{x}\right]$ don't change as $x \to \infty$
A: The limit $\displaystyle\lim_{x\to\infty}\left\lfloor\frac6x\right\rfloor\frac x3$ is $0$, but not for the reason you stated. It is perfectly possible for $\lim_{x\to\infty}f(x)g(x)$ to be nonzero even when $\lim_{x\to\infty}f(x)=0$, for example we can take $f(x)=1/x$ and $g(x)=x$. The important thing is that we do not even need $x$ to tend to infinity for $\lfloor 6/x\rfloor$ to be equal to $0$--we can just pick any $x>6$. Indeed, let $x=6+y$ for some $y>0$, then the limit becomes
$$\lim_{y\to\infty}\left(\left\lfloor\frac{6}{6+y}\right\rfloor\frac{y+6}{3}\right)=\lim_{y\to\infty}\left(\left\lfloor 1-\frac{y}{y+6} \right\rfloor\frac{y+6}{3}\right)=\lim_{y\to\infty}\left(0\cdot\frac{y+6}{3}\right)=0.$$
In other words, if we pick a sufficiently large $x$ (in particular $x>6$), then the function we are taking the limit of becomes $0$. This argument does not work if it simply approaches $0$, but when it is in fact $0$ like in this case, then we are safe.
A: $ [\frac{6}{X}]=\frac{6}{X}\,-\,${${\frac{6}{X}}$} $\\$
Where {x} is fractional part of X .$\\$
$\lim_{x \rightarrow\infty}((\frac{6}{x})(\frac{x}{3})$-{ $(6/x)$}$\frac{x}{3} $.   $\\$ 
=2
