$$\lim_{x \to 0} \left[ \frac{100 \tan(x) \sin (x)}{x^2} \right],$$ where $\left[ \phantom{\frac{1}{1}} \right]$ denotes the greatest integer (floor) function.
I am having problem as $\frac{\sin(x)}{x}$ is always less than $1$ and $\frac{\tan(x)}{x}$ is always greater than $1.$
write it in the form $$100\frac{\sin(x)}{x}\frac{1}{\cos(x)}\frac{\sin(x)}{x}$$ can you proceed?
Hint
$$\frac{\tan x}{x}\cdot \frac{\sin x}{x}$$
Now use fundamental limit.
$\lim_{x\rightarrow 0} \frac{\tan x}{x}=1$ and $\lim_{x\rightarrow 0} \frac{\sin x}{x}=1$
Consider $$ f(x)=\tan x\sin x-x^2 $$ over $(-\pi/2,\pi/2)$. Then $$ f'(x)=(1+\tan^2x)\sin x+\sin x-2x=2\sin x+\tan^2x\sin x-2x $$ and \begin{align} f''(x) &=2\cos x+2\tan x(1+\tan^2x)\sin x+\tan^2x\cos x-2\\ &=2\cos x+2\frac{\sin^2x}{\cos^3x}+\frac{\sin^2x}{\cos x}-2\\ &=\frac{2\cos^4x+2\sin^2x+\sin^2x\cos^2x-2\cos^3x}{\cos^3x}\\ &=\frac{2\cos^4x+2-2\cos^2x+\cos^2x-\cos^4x-2\cos^3x}{\cos^3x}\\ &=\frac{\cos^4x-2\cos^3x-\cos^2x+2}{\cos^3x}\\ &=\frac{(\cos x-1)(\cos^3x-\cos^2x-2\cos x-2)}{\cos^3x} \end{align} and $$ \cos^3x-\cos^2x-2\cos x-2= \cos^2x(\cos x-1)-2(\cos x+1)<0 $$ Thus $f''(x)>0$ except at $0$ and therefore the function is convex.
In particular, $f(x)>0$ for $x\ne0$, so $$ \frac{\tan x\sin x}{x^2}>1 $$ (except, of course, at $0$).
First, write the limit without the floor in the form $100\frac{1}{\cos x}\frac{\sin x}{x}\frac{\sin x}{x}$ as Dr. Sonnhard Graubner suggested. This will lead you to believe the answer is $100$.
However, now you must worry that the limit may not exist as we are dealing with the floor $\lfloor100\frac{1}{\cos x}\frac{\sin x}{x}\frac{\sin x}{x}\rfloor$.
It is enough to show that $100$ is being approached from above in both $x$ directions.
One way to do this is by showing the derivative of $100\frac{1}{\cos x}\frac{\sin x}{x}\frac{\sin x}{x}$ is positive for $x>0$ and negative for $x<0$.