Is the function defined by $f(x) = \begin{cases} x + 5, & \text{if } x \le 1 \\ x - 5, & \text{if } x > 1 \end{cases}$ a continuous function?

(B) The given function is $f(x) = \begin{cases} x + 5, & \text{if } x \le 1 \\ x - 5, & \text{if } x > 1 \end{cases}$.
The function $f$ is defined for all real numbers.
Let $c$ be any real number.
Case $I$: If $c < 1$,then $f(c) = c + 5$ and $\lim_{x \to c} f(x) = \lim_{x \to c} (x + 5) = c + 5$. Since $\lim_{x \to c} f(x) = f(c)$,$f$ is continuous for all $x < 1$.
Case $II$: If $c = 1$,then $f(1) = 1 + 5 = 6$. The left-hand limit is $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x + 5) = 6$. The right-hand limit is $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x - 5) = -4$. Since the left-hand limit $\neq$ right-hand limit,$f$ is not continuous at $x = 1$.
Case $III$: If $c > 1$,then $f(c) = c - 5$ and $\lim_{x \to c} f(x) = \lim_{x \to c} (x - 5) = c - 5$. Since $\lim_{x \to c} f(x) = f(c)$,$f$ is continuous for all $x > 1$.
Conclusion: The function $f$ is not continuous at $x = 1$.