<s>[INST] Evaluation of $ \lim_{x\to 0}\left\lfloor \frac{x^2}{\sin x\cdot \tan x}\right\rfloor$ Evaluation of $\displaystyle \lim_{x\to 0}\left\lfloor \frac{x^2}{\sin x\cdot \tan x}\right\rfloor$ where $\lfloor x \rfloor $ represent floor function of $x$.
My Try:: Here $\displaystyle f(x) = \frac{x^2}{\sin x\cdot \tan x}$ is an even function.
So we will calculate for $\displaystyle \lim_{x\to 0^{+}}\left\lfloor \frac{x^2}{\sin x\cdot \tan x}\right\rfloor$
Put $x=0+h$ where $h$ is a small positive quantity, and using series expansion
So limit convert into $\displaystyle \lim_{h\to 0}\left\lfloor \frac{h^2}{\sin h\cdot \tan h}\right\rfloor = \lim_{h\to 0}\left\lfloor \dfrac{h^2}{\left(h-\dfrac{h^3}{3!}+\dfrac{h^5}{5!}- \cdots\right)\cdot \left(h+\dfrac{h^3}{3}+\dfrac{2}{15}h^5+ \cdot\right)}\right\rfloor$
Now how can i solve after that, Help me 
Thanks
 [/INST] Note that the function under consideration is even and hence it is sufficient to consider $x \to 0^{+}$. Now we need to compare $x^{2}$ and $\sin x \tan x$ for $x > 0$. Let $$f(x) = x^{2} - \sin x\tan x$$ Then $f(0) = 0$ and $$f'(x) = 2x - \sin x\sec^{2}x - \sin x$$ Then we have $f'(0) = 0$. Further $$f''(x) = 2 - \sec x - 2\tan^{2}x - \cos x$$ and $f''(0) = 0$. Next we have $$f'''(x) = -\sec x\tan x - 4\tan x\sec^{2}x + \sin x = \sin x(1 - \sec^{2}x - 4\sec^{3}x)$$ As $x \to 0^{+}$ we can clearly see that $f'''(x) < 0$. This means that $f''(x)$ is decreasing and considering that $f''(0) = 0$ we have $f''(x) < 0$ when $x \to 0^{+}$. Continuing further in the same manner we see that $f(x) < 0$ when $x \to 0^{+}$. It follows that $\sin x\tan x > x^{2} > 0$ when $x \to 0^{+}$ so that $$0 < \dfrac{x^{2}}{\sin x\tan x} < 1$$ when $x \to 0^{+}$. Hence we have $$\left\lfloor\dfrac{x^{2}}{\sin x\tan x}\right\rfloor = 0$$ when $x \to 0^{+}$. Now it is obvious that the desired limit is $0$.
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