Q.Discuss the continuity of the function $f$ defined by $f(x) = \begin{cases} x + 2, & \text{if } x \leq 1 \\ x - 2, & \text{if } x > 1 \end{cases}$.

We check continuity at the point where the definition changes, $x = 1$, because the function is clearly continuous elsewhere (each piece is a polynomial).

Step 1 — Left-hand limit:

$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x + 2) = 1 + 2 = 3$.

Step 2 — Right-hand limit:

$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x - 2) = 1 - 2 = -1$.

Step 3 — Compare with function value:

$f(1) = 1 + 2 = 3$. …

Concept First: Continuity at a Point

A function is continuous at a point $x = a$ if three things hold:

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

The critical idea is that the function’s value and its limit must agree. For a piecewise function, the limit exists only when the left-hand limit and right-hand limit are equal. If they differ, the function “jumps” — and continuity fails.

Here, the two pieces are simple lines: $x+2$ for $x \le 1$ and $x-2$ for $x > 1$. The only possible trouble spot is the boundary $x=1$, because inside each piece the function is a polynomial (hence continuous). So we check $x=1$ carefully.

Step-by-Step Solution

1. Find $f(1)$.

Since $x=1$ falls in the first case ($x \le 1$), we use $f(x) = x+2$.

2. Compute the left-hand limit as $x \to 1^-$.

For $x < 1$, the function is $x+2$. As $x$ approaches $1$ from the left,

3. Compute the right-hand limit as $x \to 1^+$.

For $x > 1$, the function is $x-2$. As $x$ approaches $1$ from the right,

4. Compare the two one-sided limits.

Left-hand limit = $3$, right-hand limit = $-1$. They are not equal.

Therefore, $\lim_{x \to 1} f(x)$ does not exist.

Method: Detecting a Jump Discontinuity at a Piecewise Function's Transition Point

This method checks continuity at the single point where a piecewise-linear (or piecewise-polynomial) function switches formulas, and correctly identifies a jump discontinuity when the two sides don't meet.

Steps

Step 1: Locate the transition point and note which piece defines it

Identify the boundary value $a$ where the definition switches, and read off which of the two formulas actually applies AT $x=a$ itself (usually determined by whether the inequality is $\le$ or $<$).

Step 2: Compute $f(a)$ using the formula that applies exactly at the boundary

Step 3: Compute the left-hand and right-hand limits using each side's own formula

$\lim_{x \to a^-} f(x) \quad \text{using the left piece,} \qquad \lim_{x \to a^+} f(x) \quad \text{using the right piece}$ …

Common Mistakes

Mistake 1: Concluding continuity from $f(1)$ alone without checking both one-sided limits

Why it's wrong: $f(1)=3$ being a well-defined number says nothing about whether the function approaches the same value from both sides — here the right-hand limit is $-1$, completely different from $f(1)$. Correct approach: always compute both $\lim_{x\to 1^-} f(x)$ and $\lim_{x\to 1^+} f(x)$ separately using the correct piece on each side before drawing any conclusion.

Mistake 2: Using the wrong piece of the definition to evaluate a one-sided limit …