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Okay, so I read this somewhere that,
$$ \lim_{x \to 0^+} \left[ \frac{\sin x}{x} \right] = 0 $$ Where, [] denotes the greatest integer function.
But, on the other hand, this is also true,
$$ [0.9999...] = 1 $$
Aren't these two contradictory? I mean if,
$$ \lim_{x \to 0^+} \frac{\sin x}{x} = 1 $$
and $$ 0.9999... = 1 $$
Then why is the greatest Integer function behaving differently for these two functions?
$0.999\ldots$ isn't a sequence / function you take the limit of. It is a fixed number, and it's equal to $1$ in value (more strictly: If you allow it to represent a value, then any value except $1$ will get you into inconsistencies). Round it down all you like, that doesn't change. On the other hand, we do have $$ \lim_{n \to \infty}\left[0.\underbrace{999\ldots99}_{n\text{ times}}\right] = 0 $$ As for why $\lim_{x \to 0^+} \left[ \frac{\sin x}{x} \right] = 0$, that's simply because for all non-zero $x$, we have $\left[ \frac{\sin x}{x} \right] = 0$, so of course the limit is going to be $0$ as $x \to 0$.
There are too many red herrings in your question. Let's remove them entirely.
Consider the function $F$ such that $F(x)=0$ for all $x<1$ and $F(1)=1$. This function is not continuous at $1$. More specifically, $$\lim_{x\to 1^{-}} F(x)=0$$ because it is the limit of a constant function which always returns $0$. This is despite the obvious fact that $\lim_{x\to 1^{-}}x=1$.
This is the same situation in your question. You compose the $\frac{\sin x}x$ function with a non-continuous function which ensures that the value becomes a constant $0$; and then you compare it to a different limit.
$$ 0.9999... = 1 $$ is perfectly true, but when you take the limit, you never evaluate the function at exactly $0$, as it is undefined, so you never have infinitely many $9$s, and the above equality does not enter into play.
