P (x) = x³- 2x + 6 is a polynomial.
Hence, \(lim _{x\rightarrow 3}p(x)=p(3)=3^3-2\times3+6=27\) .

Limit of f(x) as x approaches 2 is the nature of (x) on both sides of 2.
It is independent ofthe nature of f(x) at x = 2.
Therefore we can not conclude anything about \(lim_{x \rightarrow 2}f(x) from f(2)=4\)

tan x can be made arbitrarily large when x is chosen suficiently close to \(\frac{\pi}{2}\) in the left side.
Similarly it can be made arbitrarily small when x is chosen sufficiently closer to \(\frac{\pi}{2}\) in the right side.
x is chosen sufficiently closer \(\frac{\pi}{2}\) to in the right side.
tan x does not approach any value when x approaches \(\frac{\pi}{2}\)
\(Indeed \lim _{x \rightarrow \frac{\pi^{-}}{2}} \tan x= and \lim _{x \rightarrow \frac{\pi^{+}}{2}} \tan x=-\infty .\)
Hence the limit does not exist.

\(lim_{x\rightarrow{1}}sin \pi x\)
At x = 1, the curve meets the x-axis.
\(\therefore lim_{x\rightarrow{1}}sin \pi x=0\)

Let \( f(x)=\frac{\cos x-1}{x} \)
\(\therefore \lim _{x \rightarrow 0} \frac{\cos x-1}{x}=0\)

Let \( f(x)=\frac{\sin x}{x} \)
\(\therefore \lim _{x \rightarrow 0} \frac{\sin x}{x}=0.999 \ldots=1 \)

Let \( f(x)=\frac{x-2}{x^2-x-2}=\frac{x-2}{(x-2)(x+1)}=\frac{1}{x+1} \)
\( \therefore \lim _{x \rightarrow 2} \frac{x-2}{\dot{x}^2-x-2}=\lim _{x \rightarrow 2} \frac{1}{x+1}=\frac{1}{3}=0 . \overline{3} \)

No. f(x) = \(\sqrt{x}\) is not even defined for x < 0.

Therefore as x \(\rightarrow 0^-,lim_{x\rightarrow0^-}\sqrt{x}\) does not exist.
However, \(lim_{x\rightarrow0^+}\sqrt{x}=0.\) Therefore \(lim_{x\rightarrow0}\sqrt{x}\) does not exist.

\(|x|= \begin{cases}-x & \text { if } x<0 \\ 0 & \text { if } x=0 \\ x & \text { if } x>0\end{cases}\)
If x > 0,then |x| =  x, which tends to 0 as
\(x \rightarrow 0\) from the right of 0. That is, \(\lim _{ x\rightarrow0^+}{|x| } =0\)
If x < 0, then |x| = - x which again tends to 0 as x\(\rightarrow\)0. from the left of 0. That is, \(\lim _{ x\rightarrow0^-}{|x| } =0\).
Thus, \(\lim _{ x\rightarrow0^-}{|x| } =0=\lim _{ x\rightarrow0^+}{|x| }.\)
Hence \(\lim _{ x\rightarrow0}{|x| } =0\).

The greatest integer function f(x) =\(\left\lfloor x \right\rfloor \) is defined as the greatest integer lesser than or equal to x.
  \(lim_{x\rightarrow 2^-}\left\lfloor x \right\rfloor =1\)  and   \(lim_{x\rightarrow 2^+}\left\lfloor x \right\rfloor =2\) .