Vtyjkyuk

Evaluate $$lim _x to 0 left[fracx^2sin x tan x right]$$ where $[cdot]$ denotes the greatest integer function.

Can anyone give me a hint to proceed?

I know that $$frac sin xx < 1$$ for all $x in (-pi/2 ,pi/2) setminus 0$ and $$frac tan xx > 1$$ for all $x in (-pi/2 ,pi/2) setminus 0$. But will these two inequalities be helpful here?

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 $0leq frac x^2 sin x tan x <1$ 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 $sin x$.

Can you just use the known expansions for $sin x, tan x$ for small $x$? Then have $$fracx^2sin x tan x = fracx^2(x - fracx^36 + cdots)(x + fracx^33 + cdots).$$ The rhs only including terms up to $x^2$ can be written $$frac1(1 - x^2/6)(1 + x^2/3)$$ and the denominator to order $x^2$ is $1 + x^2/6$ so that finally have $$fracx^2sin x tan x = frac11 + fracx^26 + cdots = 1 - x^2/6$$ and the integer part is 0.

Alternatively write the original expression as $$fracx^2 cos xsin^2 x.$$