\(f(-1) =\lim _{x \rightarrow-1} f(x) \)
\(=\lim _{x \rightarrow-1} \frac{x^{2}-1}{x^{3}+1} \)
\(=\lim _{x \rightarrow-1} \frac{x^{2}-(-1)^{2}}{x^{3}-(-1)^{3}}=\frac{2(-1)^{2-1}}{3(-1)^{3-1}}=\frac{-2}{3} \)

\(\lim _{x \rightarrow 0^{-}} \frac{\sin x}{\sqrt{x^{2}}} =\lim _{x \rightarrow 0^{-}} \frac{\sin x}{|x|} \)
\(=\lim _{x \rightarrow 0^{-}} \frac{\sin x}{-x}(\because x<0) \)
= -1

\(\lim _{x \rightarrow 1^{-}} f(x)=3 \text { and } \lim _{x \rightarrow 1^{+}} f(x)=2\)
\(\therefore \lim _{x \rightarrow 1} f(x) \text { does not exist }\)

\(f(x)=x(-1)^{\left\lfloor\frac{1}{x}\right\rfloor}\)
\(=x(\pm 1) \left(\because\left[\frac{1}{x}\right\rfloor \text { is an integer }\right) \)
\(=\pm x \)
\(\therefore \lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0}(\pm x)=0\)

\(\text { Let } f(x)=\frac{\sqrt{1-\cos 2 x}}{x}\)
\(=\frac{\sqrt{2 \sin ^{2} x}}{x} \)
\(=\frac{\sqrt{2} \sqrt{\sin ^{2} x}}{x} \)
\(=\frac{\sqrt{2}|\sin x|}{x}\left(\therefore \sqrt{x^{2}}=|x|\right) \)
\(= \begin{cases}\frac{\sqrt{2}(-\sin x)}{x}, & \sin x<0 \\ \sqrt{2}\left(\frac{\sin x}{x}\right), & \sin x>0\end{cases} \)
\(= \begin{cases}-\sqrt{2} \frac{\sin x}{x}, x \in\left(-\frac{\pi}{2}, 0\right) \\ \sqrt{2} \frac{\sin x}{x}, x \in\left(0, \frac{\pi}{2}\right)\end{cases} \)
\(\therefore \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \frac{-\sqrt{2} \sin x}{x}=-\sqrt{2}\)
\(\text { and } \lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} \frac{\sqrt{2} \sin x}{x}=\sqrt{2}\)
\(\text { From (1) and (2) }\)
\(\lim _{x \rightarrow 0^{-}} f(x) \neq \lim _{x \rightarrow 0^{+}} f(x)\)
\(\therefore \lim _{x \rightarrow 0} f(x) \text { does not exist }\)