Q.Find all the points of discontinuity of the greatest integer function defined by $f(x) = [x]$, where $[x]$ denotes the greatest integer less than or equal to $x$.

Concept: Greatest Integer Continuity — The floor function $[x]$ jumps at every integer because the left-hand limit and right-hand limit differ by exactly 1.

1. For any integer $n$, as $x \to n^-$, $[x] = n-1$; as $x \to n^+$, $[x] = n$.
2. The left-hand limit is $n-1$, the right-hand limit is $n$, and $f(n) = n$.
3. Since $\lim_{x \to n^-} f(x) \neq \lim_{x \to n^+} f(x)$, the function is discontinuous at every integer $n \in \mathbb{Z}$. …

Why this problem matters

The greatest integer function — often called the floor function — is one of the simplest piecewise-constant functions. It jumps at every integer. Understanding where and why it breaks continuity is the foundation for analyzing more complex piecewise functions in calculus.

The key insight: continuity at a point $x = a$ requires three things to be equal — the left-hand limit, the right-hand limit, and the function value at $a$. For $[x]$, the function is constant on intervals between integers, but at each integer, the value on the left is different from the value on the right.

Step-by-step reasoning

1. Understand the definition of $[x]$

For any real $x$, $[x]$ is the greatest integer that is less than or equal to $x$. So:

• If $x = 2.3$, then $[x] = 2$.
• If $x = 2$, then $[x] = 2$.
• If $x = 1.9$, then $[x] = 1$.

The function is constant on every interval $[n, n+1)$ for integer $n$, and at $x = n+1$ it jumps to the next integer.

2. Check continuity at a non-integer point

Take any $a$ that is not an integer. Then $a$ lies strictly between two consecutive integers, say $n < a < n+1$. Since $[x] = n$ for all $x$ in $(n, n+1)$, the function is constant in a neighbourhood of $a$. Therefore:

• $\lim_{x \to a^-} [x] = n$
• $\lim_{x \to a^+} [x] = n$
• $f(a) = n$

All three match. So $f$ is continuous at every non-integer.

3. Check continuity at an integer point

Let $a = n$, where $n$ is an integer. Now examine the left-hand and right-hand behaviour.

• Left-hand limit: As $x \to n^-$, $x$ is slightly less than $n$, so $[x] = n-1$. Hence:

• Right-hand limit: As $x \to n^+$, $x$ is slightly greater than $n$, so $[x] = n$. Hence:

• Function value: $f(n) = [n] = n$ …

Method: Proving (Dis)continuity at an Infinite Family of Points (e.g., the Greatest Integer Function)

This method applies when a function's rule changes at infinitely many points that all share the same structural pattern (every integer, every multiple of $\pi$, etc.), so you cannot check each one individually.

Steps

Step 1: Identify the general pattern of "special" points.

Rather than listing points one by one, describe them algebraically — here, "every integer $n$" rather than trying to check $0, 1, 2, \dots$ separately.

Step 2: Test an arbitrary representative point in each category.

Split the real line into two categories: an arbitrary non-integer point, and an arbitrary integer $n$. You only need to analyse one representative of each category, because the same argument structure applies to every point in that category.

Step 3: For a non-integer, show the function is locally constant (or otherwise continuous) nearby. …

Common Mistakes

Mistake 1: Concluding continuity at an integer because the right-hand limit matches $f(n)$.

Why it's wrong: continuity requires the two-sided limit to exist and match $f(n)$ — matching only the right-hand side, while the left-hand limit is different, still leaves a genuine jump. Correct approach: always check both one-sided limits before declaring continuity, not just the one that happens to agree.

Mistake 2: Trying to check every integer individually instead of arguing for a general integer $n$.

Why it's wrong: there are infinitely many integers, so a point-by-point check can never be completed — an answer that only checks $n=0,1,2$ and stops has not actually proven the general claim. Correct approach: state the left-hand and right-hand limits in terms of an arbitrary integer $n$ (namely $n-1$ and $n$), so the argument covers every integer at once. …