Step 1: Picture the jump directly by evaluating the two branch formulas at $x=1$ itself, as a quick first check:
Branch 1 at $x=1$ gives $1+5=6$; branch 2, if it were evaluated at $x=1$ (even though it technically starts just after 1), gives $1-5=-4$. Such a mismatch between the two branch values at the meeting point is a strong hint of a jump discontinuity.
Step 2: Confirm rigorously with one-sided limits, since a hint is not a proof:
$\lim_{x\to1^-}(x+5)=6$ and $\lim_{x\to1^+}(x-5)=-4$. These two one-sided limits are different real numbers.
Step 3: Apply the continuity test:
Continuity at $x=1$ needs $\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=f(1)$. Here the two one-sided limits themselves disagree ($6\ne-4$), so the two-sided limit fails to exist, and continuity is impossible regardless of what $f(1)$ is.
Final Answer:
No, $f(x)$ has a jump discontinuity at $x=1$, since the left limit (6) and right limit (−4) do not match.
\[ \boxed{\text{Not continuous at } x=1} \]