Animated Solution for Mathematics - Limits, Continuity and Differentiability: The function $f(x)=[x{]}^2-[x^2]$ (where $[y]$ is the greatest integer less than or equal to $y$), is discontinuous at

Welcome, fellow traveler on the road to JEE excellence. Today, we are going to dissect a function that seems simple on the surface but hides a beautiful, rhythmic complexity beneath its skin: $f(x)=[x{]}^2-[x^2]$.

Many students see the Greatest Integer Function (GIF) and immediately panic, thinking of jagged, broken lines. You are right to be cautious—the GIF is indeed a staircase that jumps at every integer.

But in this problem, we aren't just dealing with one staircase; we are dealing with two, dancing in a way that might just lead to a moment of perfect continuity. Let's peel back the layers.

The Anatomy of the Function

Our function is $f(x)=[x{]}^2-[x^2]$. To understand where this function breaks, we must look at the points where the GIF itself breaks: the integers.

Let $n$ be any integer. For the function to be continuous at $x=n$, the Left Hand Limit (LHL), the Right Hand Limit (RHL), and the function value $f(n)$ must all be equal.

Let's test the function value first. At any integer $n$, $[n]=n$ and $[n^2]=n^2$. Therefore, $f(n)=n^2-n^2=0$.

This is our baseline. If the function is to be continuous, both the LHL and RHL must also equal $0$.

The Right Hand Limit (RHL)

Let's approach $n$ from the right, $x\to n^{+}$. We can write $x=n+h$, where $h$ is a tiny positive number.

The first term is $[n+h]=n$. Squaring it gives $n^2$.

The second term is $[(n+h{)}^2]=[n^2+2nh+h^2]$. Since $h$ is tiny, $n^2+2nh+h^2$ is just a hair larger than $n^2$. The greatest integer of a number slightly larger than $n^2$ is simply $n^2$.

So, the RHL is $n^2-n^2=0$. The right side of our function is perfectly well-behaved; it stays at zero.

The Left Hand Limit (LHL) - The Moment of Truth

Now, let's approach $n$ from the left, $x\to n^{-}$. We write $x=n-h$. This is where the magic—or the disaster—happens.

The first term is $[n-h]$. Since $n-h$ is just below $n$, its greatest integer is $n-1$. Squaring this gives $(n-1{)}^2=n^2-2n+1$.

The second term is $[(n-h{)}^2]=[n^2-2nh+h^2]$. Because $h$ is positive, $n^2-2nh+h^2$ is slightly less than $n^2$. The greatest integer of a value just shy of $n^2$ is $n^2-1$.

Now, let's combine them for the LHL:

Watch the algebra unfold. The $n^2$ terms cancel out, leaving us with:

The Verdict

We have our condition for continuity: $\text{LHL}=\text{RHL}$. We know $\text{RHL}=0$, so we set our LHL expression to zero:

This is the revelation! The function is continuous only when $n=1$.

At every other integer, the LHL will be $2-2n$, which is not zero, meaning the function will have a jump discontinuity.

Isn't that fascinating? At $x=1$, the two components of the function conspire to cancel out their jumps, creating a smooth transition. At every other integer, they fail to align.