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Q19:
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 $x$.

Solution :

Given Variables & Initial Setup

The function is defined as $g(x) = x - [x]$, where $[x]$ is the greatest integer function (floor function). We are required to investigate the continuity of $g(x)$ at any arbitrary integer $n$, where $n \in \mathbb{Z}$.

A function $g(x)$ is continuous at $x = n$ if and only if:

$\lim_{x \to n^-} g(x) = \lim_{x \to n^+} g(x) = g(n)$

Step 1: Evaluating the Function at $x = n$

By the definition of the greatest integer function, for any integer $n$, $[n] = n$ [By definition of the floor function for integers].

Substituting this into the function definition:

$g(n) = n - [n] = n - n = 0$

Step 2: Evaluating the Left-Hand Limit (LHL)

Let $x = n - h$, where $h > 0$ and $h \to 0$. As $x \to n^-$, $h \to 0^+$.

For $x$ in the interval $(n-1, n)$, the greatest integer $[x] = n - 1$.

$\lim_{x \to n^-} g(x) = \lim_{h \to 0} ((n - h) - [n - h])$

Since $n-h$ is slightly less than $n$, $[n - h] = n - 1$.

$\lim_{x \to n^-} g(x) = \lim_{h \to 0} (n - h - (n - 1)) = \lim_{h \to 0} (1 - h) = 1$

Step 3: Evaluating the Right-Hand Limit (RHL)

Let $x = n + h$, where $h > 0$ and $h \to 0$. As $x \to n^+$, $h \to 0^+$.

For $x$ in the interval $[n, n+1)$, the greatest integer $[x] = n$.

$\lim_{x \to n^+} g(x) = \lim_{h \to 0} ((n + h) - [n + h])$

Since $n+h$ is slightly greater than or equal to $n$, $[n + h] = n$.

$\lim_{x \to n^+} g(x) = \lim_{h \to 0} (n + h - n) = \lim_{h \to 0} (h) = 0$

Step 4: Comparison and Conclusion

Condition Value
LHL ($x \to n^-$) $1$
RHL ($x \to n^+$) $0$
Function Value $g(n)$ $0$

Since $\lim_{x \to n^-} g(x) \neq \lim_{x \to n^+} g(x)$, the limit $\lim_{x \to n} g(x)$ does not exist [By the Limit Existence Theorem]. Therefore, the function $g(x) = x - [x]$ is discontinuous at all integral points $n \in \mathbb{Z}$.

Visual Representation of $g(x) = x - [x]$

x y 0 1 2

Final Solution: The function $g(x) = x - [x]$ is discontinuous at every integer $n$ because the left-hand limit is $1$ and the right-hand limit is $0$, resulting in a jump discontinuity of magnitude $1$.


More Questions from Class 12 Mathematics Continuity and Differentiability EXERCISE 5.1


CBSE Solutions for Class 12 Mathematics Continuity and Differentiability


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