One-Sided, Greatest-Integer & Modulus Limits

LHL $=\lim_{x\to a^-}f(x)$, RHL $=\lim_{x\to a^+}f(x)$. The limit exists iff LHL $=$ RHL. For a piecewise $f$, use the piece valid on each side; for a product/quotient of one-sided-sensitive parts, evaluate each side end-to-end. The squeeze (sandwich) theorem is the tool for a bounded factor times something that dies: if $g\le f\le h$ near the point and $g,h\to L$, then $f\to L$. Standard uses: $|\cos x|\le1$ gives $\dfrac{\cos x}{x}\to0$ as $x\to\infty$ and $x\sin\tfrac1x\to0$ as $x\to0$; $x-1<\lfloor x\rfloor\le x$ gives $\dfrac{\lfloor x\rfloor}{x}\to1$ as $x\to\infty$. Note $\sin\tfrac1x$ alone has NO limit at 0 — the bounded factor must be multiplied by something tending to 0.

At an integer $n$: $\lim_{x\to n^-}\lfloor x\rfloor=n-1$, $\lim_{x\to n^+}\lfloor x\rfloor=n$ — so $\lim_{x\to n}\lfloor x\rfloor$ does not exist. Between integers $\lfloor x\rfloor$ is constant. For $\lfloor g(x)\rfloor$, track which integers $g$ crosses near the point (e.g. $\lfloor x^2\rfloor$ near $x=0$). Negatives round the other way: $\lfloor -y\rfloor=-\lfloor y\rfloor-1$ for non-integer $y$ (and $=-\lfloor y\rfloor$ when $y$ is an integer). So $\lfloor\pi^2\rfloor=9$ but $\lfloor-\pi^2\rfloor=-10$, and $\lfloor x\rfloor+\lfloor-x\rfloor=-1$ for every non-integer $x$. A limit like $\lim_{x\to\pi}\lfloor -x^2\rfloor$ (constant $-10$ near $\pi$, since $\pi^2\approx9.87$ is not an integer) is decided entirely by this rule; trig values at $\tfrac{9\pi}{2}$, $5\pi$ etc. reduce by period first.

Replace $|g(x)|$ by $+g$ on the side where $g>0$ and $-g$ where $g<0$, then take each one-sided limit. $\dfrac{x}{|x|}$ is $+1$ for $x>0$ and $-1$ for $x<0$. A surd hides a modulus: $\sqrt{A^2}=|A|$, which is sign-sensitive.