Q.The greatest integer function $f(x) = [x]$, defined for $0 < x < 2$, is not differentiable at $x = \_\_\_\_\_\_\_\_\_\_.$

The key idea is that the greatest integer function has jump discontinuities at integer points, which makes it non-differentiable there.

1. For $0 < x < 2$, the function $f(x) = [x]$ takes the value $0$ on $(0,1)$ and $1$ on $[1,2)$.
2. At $x = 1$, the left-hand derivative is $0$ (since the function is constant $0$ just left of $1$), but the right-hand derivative does not exist because the function jumps from $0$ to $1$ — the slope is effectively infinite. …

The greatest integer function $f(x) = [x]$ returns the largest integer less than or equal to $x$. On the open interval $(0,2)$, the only integer inside is $1$. At every other point, the function is locally constant — flat horizontal segments — so the derivative exists and equals $0$. The trouble is only at the jump.

For differentiability at a point, the function must be continuous there first. But $[x]$ has a jump discontinuity at every integer: the left-hand limit and right-hand limit differ by $1$. At $x=1$, the left limit is $0$ and the right limit is $1$, so the function isn't even continuous — and therefore cannot be differentiable.

Even if we ignored continuity and tried to compute the derivative from the definition, the left and right difference quotients would give different results. Let's check that explicitly.

1. Left-hand derivative at $x=1$ For $x$ just less than $1$, say $x = 1 - h$ with $h > 0$ small, $[x] = 0$. The difference quotient is

As $h \to 0^+$, this blows up to $+\infty$. So the left-hand derivative does not exist as a finite number.

2. Right-hand derivative at $x=1$ For $x$ just greater than $1$, say $x = 1 + h$ with $h > 0$, $[x] = 1$. The difference quotient is $\frac{f(1+h) - f(1)}{h} = \frac{1 - 1}{h} = 0.$ …