Q.Discuss the continuity of the function $f$, where $f$ is defined by $f(x) = \begin{cases} 3, & \text{if } 0 \le x \le 1 \\ 4, & \text{if } 1 < x < 3 \\ 5, & \text{if } 3 \le x \le 10 \end{cases}$

The key idea is that a piecewise constant function is continuous at a point only if the left-hand limit, right-hand limit, and function value all match — which fails at the boundaries where the constant jumps.

Step 1: Check continuity at $x = 1$.

Left limit: $\lim_{x \to 1^-} f(x) = 3$ (since $f(x)=3$ for $0 \le x \le 1$).

Right limit: $\lim_{x \to 1^+} f(x) = 4$ (since $f(x)=4$ for $1 < x < 3$).

The limits are unequal, so $f$ is discontinuous at $x=1$.

Step 2: Check continuity at $x = 3$.

Left limit: $\lim_{x \to 3^-} f(x) = 4$ (since $f(x)=4$ for $1 < x < 3$). …

Why this approach works

A piecewise function is like a set of separate rules for different intervals. For continuity at a point inside an interval, we just check that the rule there is well-behaved (constant functions always are). The trouble comes at the boundary points where the rule changes. At those points, we must check three things:

1. Does the function have a value there?
2. Do the left-hand and right-hand limits exist and agree?
3. Does that common limit equal the function’s value?

If any of these fails, the function is discontinuous at that point. Since $f$ is constant on each piece, the only possible discontinuities are at $x=1$ and $x=3$.

Step-by-step reasoning

1. Check continuity on the open intervals

• On $0 \le x \le 1$, $f(x)=3$ — a constant, so it’s continuous on $[0,1]$, but we’ll check the endpoint $x=1$ separately.
• On $1 < x < 3$, $f(x)=4$ — constant, so continuous on $(1,3)$.
• On $3 \le x \le 10$, $f(x)=5$ — constant, so continuous on $[3,10]$.

So the only points that need scrutiny are $x=1$ and $x=3$.

2. Examine $x=1$

• Function value: $f(1)=3$ (from the first piece).
• Left-hand limit as $x \to 1^-$: For $x$ just less than 1, we are still in $[0,1]$, so $f(x)=3$. Hence $\lim_{x \to 1^-} f(x) = 3$.
• Right-hand limit as $x \to 1^+$: For $x$ just greater than 1, we are in $(1,3)$, so $f(x)=4$. Hence $\lim_{x \to 1^+} f(x) = 4$.

Since the left-hand limit (3) and right-hand limit (4) are not equal, the two-sided limit does not exist. Therefore $f$ is discontinuous at $x=1$.

3. Examine $x=3$

• Function value: $f(3)=5$ (from the third piece). …

Method: Testing Continuity of a Multi-Piece Function at Every Boundary Independently

When a function has more than two pieces, it has more than one candidate point for discontinuity — this method shows how to handle each one systematically.

Steps

Step 1: List every boundary point

A function built from $n$ pieces has exactly $n-1$ points where the definition switches. Write down all of them before starting — skipping one is a common source of an incomplete answer.

Step 2: Test each boundary using the standard three-condition check

For each boundary point $a$ in turn:

using the appropriate piece for $f(a)$ (whichever inequality includes equality) and the neighbouring pieces for each one-sided limit.

Step 3: Record the individual verdict at each boundary …

Common Mistakes

Mistake 1: Checking only one of the two boundary points and forgetting the other

Why it's wrong: this function has two switch points, $x=1$ and $x=3$ — checking only one and assuming the rest of the domain is automatically fine misses a genuine second discontinuity. Correct approach: list every point where the piecewise rule changes before starting, and test each one independently.

Mistake 2: At $x=1$, seeing $f(1)=3$ match the left-hand limit and stopping there …