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

Concept: Continuity At A Point — A function is continuous at $x = a$ if $\lim_{x \to a} f(x) = f(a)$. For a piecewise function, the critical point is the boundary where the definition changes.

Step 1: Check at $x = 1$

Left-hand limit: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x+5) = 1+5 = 6$

Right-hand limit: $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x-5) = 1-5 = -4$

Step 2: Compare limits

Since $6 \neq -4$, the two-sided limit $\lim_{x \to 1} f(x)$ does not exist.

Step 3: Conclusion …

Why This Question Is About "The Gap"

Continuity at a point asks a simple question: if you walk up to that point from the left, and then from the right, do you arrive at the same height? If the function jumps, there's a break — a discontinuity.

For a piecewise function like this one, the only place where a break could happen is at the boundary where the rule changes: $x=1$. Everywhere else, the function is just a straight line (either $x+5$ or $x-5$), and straight lines are continuous. So the entire problem reduces to checking what happens at $x=1$.

Step-by-Step Check at $x=1$

1. Is $f(1)$ defined?

Yes. The definition says: if $x \leq 1$, use $x+5$. So $f(1) = 1+5 = 6$.

2. What is the left-hand limit as $x \to 1^{-}$?

When $x$ is just less than 1, we are in the first piece: $f(x) = x+5$.

So $\displaystyle \lim_{x \to 1^{-}} f(x) = \lim_{x \to 1^{-}} (x+5) = 1+5 = 6$.

This matches $f(1)$. So far so good.

3. What is the right-hand limit as $x \to 1^{+}$?

When $x$ is just greater than 1, we switch to the second piece: $f(x) = x-5$.

So $\displaystyle \lim_{x \to 1^{+}} f(x) = \lim_{x \to 1^{+}} (x-5) = 1-5 = -4$.

4. Do the two limits agree?

No. The left-hand limit is $6$, the right-hand limit is $-4$. They are not equal. Therefore, $\displaystyle \lim_{x \to 1} f(x)$ does not exist. …

Method: Checking All Three Continuity Conditions, Not Just One

A function can be "defined" at a boundary point and still fail to be continuous — this method emphasizes checking every required condition, not stopping early.

Steps

Step 1: Confirm $f(a)$ is defined

Substitute the boundary value into the piece whose inequality includes it. This confirms Condition 1 of the continuity test, but is not sufficient on its own.

Step 2: Compute the left-hand limit

Substitute $a$ into the formula valid just to the left, giving $\lim_{x\to a^-} f(x)$.

Step 3: Compute the right-hand limit

Substitute $a$ into the formula valid just to the right, giving $\lim_{x\to a^+} f(x)$.

Step 4: Require all three to match …

Common Mistakes

Mistake 1: Checking only whether $f(1)$ is defined and concluding continuity from that alone

Why it's wrong: an existing function value is only one of the three required conditions — here $f(1)=6$ is perfectly well-defined, yet the function is still discontinuous because the two one-sided limits ($6$ and $-4$) disagree. Correct approach: always compute both one-sided limits independently, even when $f(a)$ is clearly defined.

Mistake 2: Assuming that combining two straight lines piecewise must always give a continuous function …