Q.Discuss the continuity of the function defined by $f(x) = \begin{cases} x + 2, & \text{if } x < 0 \\ -x + 2, & \text{if } x > 0 \end{cases}$.

For a piecewise function, check the change-point $x=0$: the value $f(0)$ and the two one-sided limits.

• Left limit: $\lim_{x\to 0^-}(x+2) = 2$
• Right limit: $\lim_{x\to 0^+}(-x+2) = 2$

So the two-sided limit exists and equals $2$. But the rule only covers $x<0$ and $x>0$ — it never assigns a value at $x=0$, so $f(0)$ is undefined. Continuity requires $f(0)$ to exist, so $f$ is not continuous at $x=0$. Every other point lies inside a li …

What we must check

$f$ is continuous at $x=a$ only if all three hold: (1) $f(a)$ is defined, (2) $\lim_{x\to a}f(x)$ exists, (3) the limit equals $f(a)$. Each piece here is linear, so the only point that can cause trouble is the boundary $x=0$.

The one-sided limits

For $x<0$, $f(x)=x+2$:

For $x>0$, $f(x)=-x+2$:

Both equal $2$, so $\lim_{x\to 0} f(x) = 2$ exists.

The value at the point

The definition gives a formula only for $x<0$ and for $x>0$. It says nothing about $x=0$, so $f(0)$ is undefined — condition (1) fails.

Conclusion …

Method: Testing a Piecewise Function When the Boundary Point Might Not Be in the Domain

This method applies to piecewise functions where the pieces are given only for $x<a$ and $x>a$, with no formula stated for $x=a$ itself.

Steps

Step 1: Recognise that the boundary point needs a domain check, not just a limit check.

Before computing any limit, look at the definition and ask: is $f(a)$ actually assigned a value? If the rule only covers $x<a$ and $x>a$ and is silent about $x=a$, then $f(a)$ is undefined, and Condition 1 of the continuity test already fails — no further limit work can rescue continuity at that point.

Step 2: Compute the one-sided limits anyway (they still matter for classifying the discontinuity).

using the appropriate piece on each side.

Step 3: Interpret the result correctly. …

Common Mistakes

Mistake 1: Concluding "continuous" because the left-hand and right-hand limits are equal.

Why it's wrong: equal one-sided limits only establish that $\displaystyle\lim_{x\to a} f(x)$ exists — Condition 2 of the three-condition test. It says nothing about Condition 1 (whether $f(a)$ is even defined) or Condition 3 (whether that limit equals $f(a)$). Correct approach: always separately check that $f(a)$ is defined and that it matches the common one-sided-limit value before declaring continuity.

Mistake 2: Assuming a piecewise definition that skips $x=a$ secretly assigns some default value there. …