<s>[INST] 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.
 [/INST] 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.
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