Evaluate $ \lim _{x \to 0} \left[{\frac{x^2}{\sin x \tan x}} \right]$ where $[\cdot]$ denotes the greatest integer function. Evaluate $$\lim _{x \to 0} \left[{\frac{x^2}{\sin x \tan x}} \right]$$ where $[\cdot]$ denotes the greatest integer function.
Can anyone give me a hint to proceed?
I know that $$\frac {\sin x}{x} < 1$$ for all $x \in (-\pi/2 ,\pi/2) \setminus \{0\}$ and $$\frac {\tan x}{x} > 1$$ for all $x \in (-\pi/2 ,\pi/2) \setminus \{0\}$. But will these two inequalities be helpful here?
 A: Using the inequalities $\cos x < 1-\frac {x^{2}} {2} + \frac {x^{4}} {24}$ and $\sin x > x -\frac {x^{3}} {6}$ you can check that $0\leq \frac {x^{2}} {\sin x \tan x} <1$ for all $x>0$ sufficiently small. Hence the integer part of this fraction is $0$ for such $x$. The function is even so the limit from both sides are $0$. To obtain the stated inequality for $\cos x$ note that $\frac {x^{2n+2}} {(2n+2)!} <\frac {x^{2n}} {(2n)!}$ if $0<x<1$. Group the terms of the Taylor series two by two and use this inequality. A similar argument works for $\sin x$.
A: Can you just use the known expansions for $\sin x, \tan x$ for small $x$? Then have $$\frac{x^2}{\sin x \tan x} = \frac{x^2}{(x - \frac{x^3}{6} + \cdots)(x + \frac{x^3}{3} + \cdots)}.$$ The rhs only including terms up to $x^2$ can be written $$\frac{1}{(1 - x^2/6)(1 + x^2/3)}$$ and the denominator to order $x^2$ is $1 + x^2/6$ so that finally have $$\frac{x^2}{\sin x \tan x} = \frac{1}{1 + \frac{x^2}{6} + \cdots} = 1 - x^2/6$$ and the integer part is 0.
Alternatively write the original expression as $$\frac{x^2 \cos x}{\sin^2 x}.$$
