Q.Show that the function defined by $g(x) = x - [x]$ is discontinuous at all integral points. Here $[x]$ denotes the greatest integer less than or equal to $x$.

Concept: Greatest Integer Continuity — the floor function $[x]$ jumps at integers, so $g(x) = x - [x]$ (the fractional part) also jumps there.

Reasoning:

1. At an integer $n$, $g(n) = n - [n] = n - n = 0$.
2. For $x \to n^-$, $[x] = n-1$, so $g(x) = x - (n-1) \to n - (n-1) = 1$.
3. For $x \to n^+$, $[x] = n$, so $g(x) = x - n \to n - n = 0$. …

The core idea: what $g(x)$ really is

The greatest integer function $[x]$ returns the largest integer not exceeding $x$. So $g(x) = x - [x]$ gives the fractional part of $x$ — the distance from $x$ down to the nearest integer below it.

For example:

• At $x = 2.3$, $[2.3] = 2$, so $g(2.3) = 0.3$.
• At $x = 5$, $[5] = 5$, so $g(5) = 0$.
• At $x = -1.2$, $[-1.2] = -2$, so $g(-1.2) = 0.8$.

The graph of $g$ is a “sawtooth”: it rises linearly from $0$ to just below $1$ on each interval $[n, n+1)$, then drops sharply back to $0$ at the next integer. That drop is the discontinuity.

Step-by-step reasoning

1. Pick an arbitrary integer $n$.

We need to check continuity at $x = n$. The definition of continuity requires:

If the two-sided limit does not exist (or exists but doesn’t match $g(n)$), then $g$ is discontinuous at $n$.

2. Compute $g(n)$ directly.

Since $[n] = n$,

3. Find the right-hand limit as $x \to n^+$.

For $x$ just greater than $n$, say $x = n + h$ with $h > 0$ small, the greatest integer less than or equal to $x$ is still $n$ (because $x$ hasn’t reached $n+1$ yet). So $[x] = n$, and

As $h \to 0^+$, $g(x) \to 0$. Hence

4. Find the left-hand limit as $x \to n^-$.

For $x$ just less than $n$, say $x = n - h$ with $h > 0$ small, the greatest integer less than or equal to $x$ is $n-1$ (because $x$ is between $n-1$ and $n$). So $[x] = n-1$, and

As $h \to 0^+$, $g(x) \to 1$. Hence

$\lim_{x \to n^-} g(x) = 1$ …

Method: Proving a Function Built From $[x]$ Is Discontinuous at Every Integer

Use this whenever you must show a function involving the greatest integer function $[x]$ (floor function) is discontinuous at integer points, by examining its behaviour on either side of a general integer $n$.

Steps

Step 1: Recall how $[x]$ behaves near an integer

For an integer $n$: $[n] = n$. Just below $n$ (i.e., $x = n-h$ for small $h>0$), $[x] = n-1$. Just above $n$ (i.e., $x=n+h$), $[x] = n$. This is the key fact the whole argument rests on — the floor value drops by exactly $1$ the instant $x$ crosses below an integer.

Step 2: Fix an arbitrary integer $n$ and write the function value $f(n)$

Substitute $[n] = n$ into the given expression to get $f(n)$ in terms of $n$.

Step 3: Compute the right-hand limit as $x \to n^+$

Substitute $[x] = n$ (valid for $x$ slightly above $n$) into the function's formula and simplify as $x \to n$. …

Common Mistakes

Mistake 1: Assuming $[x] = n$ for $x$ just below the integer $n$

Why it's wrong: the greatest integer function always rounds down; for $x$ slightly less than $n$ (like $2.999$), $[x] = n-1$, not $n$. Correct approach: always test with a concrete example (is $[1.9]$ equal to $1$ or $2$?) before writing the left-hand limit formula.

Mistake 2: Thinking the function is continuous because it's "defined at every real number"

Why it's wrong: $[x]$ (and functions built from it) being defined everywhere says nothing about the limit existing — continuity needs the one-sided limits to also agree with each other and with the function value, not just that a value exists. Correct approach: explicitly compute both one-sided limits at the integer and compare them, rather than assuming "defined = continuous." …