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I was solving limits and it said $$\lim_{x\to 0} \left[\frac{x^2}{\tan(x)\sin(x)}\right]=0$$

I tried thinking about it but since I am not really good at math I couldn't proceed.

I know that $\lim_{x\to 0} \left[\frac{\sin(x)}{x}\right] = 0$ and $\lim_{x\to 0} \left[\frac{\tan(x)}{x}\right] = 1$

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    Please read an introduction to MathJax and use it to format our question. At the moment the poor formatting means it's not clear if the function you're looking at the limit of is $\frac{x^2}{\tan x\sin x}$ or $\frac{x^2}{\tan x}\sin x$. – 2017-01-30
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    Do you understand my answer? – 2017-03-14
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    umm not really, no – 2017-08-15
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    Do you understand [Big-O notation](https://en.wikipedia.org/wiki/Landau_notation)? I've written a bit about the motivation for asymptotic expansion [here](https://math.stackexchange.com/a/2131837/21820). After that, things like "$\cos(x) \in 1 - \frac12 x^2 + O(x^4)$ as $x \to 0$" follow directly from the Taylor expansion. If you wish, I can expand my answer a bit when I get back later; just let me know and tell me what you already know about asymptotic expansion so that I can tailor my explanation. – 2017-08-16

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Always use asymptotic expansion to analyze limits if possible, because you can fully understand what exactly is going on. $ \def\wi{\subseteq} \def\lfrac#1#2{{\large\frac{#1}{#2}}} $

As $x \to 0$:

  $\tan(x)\sin(x) = \lfrac{\sin(x)^2}{\cos(x)} \in \lfrac{x^2(1-\lfrac16x^2+O(x^4))^2}{1-\lfrac12x^2+O(x^4)} \wi x^2 \lfrac{1-\lfrac13x^2+O(x^4)}{1-\lfrac12x^2+O(x^4)} > x^2$ [eventually].

    [In case it is not clear, it is because $1-\lfrac13x^2+O(x^4) > 1-\lfrac12x^2+O(x^4) > 0$ eventually.]

If you really dislike asymptotic notation, you can use the above reasoning first and then extract whatever hard constants you need and replace all the Big-O notation with appropriate constants.

The strength of this method is its generality and systematicity. For instance, exactly the same reasoning as above (but with just one more term of the expansions) will easily give you that $\lim_{x \to 0} \Big\lfloor \lfrac{x^3}{\tan(x)\sin(x)^2} \Big\rfloor = 0$.

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Assuming "GIF" is meant to be an abbreviation of "Greatest Integer Function", the answer to the question in the title is no, a counterexample: $$ \lfloor 2.5\rfloor\cdot\lfloor 2.5\rfloor=2\cdot 2=4\neq 6 =\lfloor 6.25\rfloor=\lfloor 2.5\cdot 2.5\rfloor $$

It's not clear that will help you with the problem in the body of your question.

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    thank you.. i should probably ask a different question to solve my problem – 2017-01-30