Which one of the following functions is continuous at x = 3? A. \[{\text{f}}\left( {\text{x}} \right) = \left\{ {\begin{array}{*{20}{c}} {2,}&{{\text{if}}}&{{\text{x}} = 3} \\ {{\text{x}} – 1,}&{{\text{if}}}&{{\text{x}} > 3} \\ {\frac{{{\text{x}} + 3}}{3},}&{{\text{if}}}&{{\text{x}} < 3} \end{array}} \right.\] B. \[{\text{f}}\left( {\text{x}} \right) = \left\{ {\begin{array}{*{20}{c}} {4,}&{{\text{if}}}&{{\text{x}} = 3} \\ {8 - {\text{x,}}}&{{\text{if}}}&{{\text{x}} \ne 3} \end{array}} \right.\] C. \[{\text{f}}\left( {\text{x}} \right) = \left\{ {\begin{array}{*{20}{c}} {{\text{x}} + 3,}&{{\text{if}}}&{{\text{x}} \leqslant 3} \\ {{\text{x}} - 4,}&{{\text{if}}}&{{\text{x}} > 3} \end{array}} \right.\] D. $${\text{f}}\left( {\text{x}} \right) = \frac{1}{{{{\text{x}}^3} – 27}},\,{\text{if}}\,{\text{x}} \ne 3$$

A function is continuous at $x=3$ if the following three conditions hold:

1. $\lim_{x\to 3}f(x)$ exists.
2. $f(3)$ exists.
3. $\lim_{x\to 3}f(x)=f(3)$.

In option A, we have

$$\lim_{x\to 3^-}f(x)=\lim_{x\to 3^+}f(x)=2.$$

However, $f(3)=\frac{3+3}{3}=3$. Therefore, option A does not satisfy the third condition and is not continuous at $x=3$.

In option B, we have

$$\lim_{x\to 3^-}f(x)=4=\lim_{x\to 3^+}f(x)=f(3).$$

Therefore, option B satisfies all three conditions and is continuous at $x=3$.

In option C, we have

$$\lim_{x\to 3^-}f(x)=3=\lim_{x\to 3^+}f(x)=3.$$

Therefore, option C also satisfies all three conditions and is continuous at $x=3$.

In option D, we have

$$\lim_{x\to 3^-}f(x)=\lim_{x\to 3^+}f(x)=\frac{1}{(3)^3-27}=\frac{1}{0}.$$

However, the limit $\lim_{x\to 3}f(x)$ does not exist, since the function is undefined at $x=3$. Therefore, option D does not satisfy the first condition and is not continuous at $x=3$.

In conclusion, the only function that is continuous at $x=3$ is option B.