Question

The function $\phi(x) = [|x| - \sin|x|]$ (where [.] denotes greater integer function) is -

• A. derivable at x = 0
• B. continuous at x = 0
• C. $\lim_{x\to 0}\phi(x)$ does not exists
• D. continuous and derivable at x = 0

Answer 1

Answer: (D)

(D) continuous and derivable at x = 0

Answer 2

The function $\phi(x) = [|x| - \sin|x|]$. For $x$ in a small neighborhood of $0$, let $g(x) = |x| - \sin|x|$.

For $x > 0$, $g(x) = x - \sin x$. Using Taylor expansion, $x - \sin x = x - (x - x^3/6 + \dots) = x^3/6 - x^5/120 + \dots$. For sufficiently small $x > 0$, this value is positive and less than $1$.

For $x < 0$, $g(x) = -x - \sin(-x) = -x + \sin x$. Let $x = -h$ for $h > 0$. Then $g(x) = h - \sin h$, which is the same form as for $x > 0$. For sufficiently small $h > 0$, this value is positive and less than $1$.

At $x=0$, $g(0) = 0 - \sin 0 = 0$.

Therefore, for $x$ in a small open interval around $0$ (e.g., $(-\delta, \delta)$ for $\delta$ small enough), $0 \le |x| - \sin|x| < 1$.

This implies $\phi(x) = [|x| - \sin|x|] = 0$ for all $x \in (-\delta, \delta)$.

Since $\phi(x)$ is identically zero in an open interval containing $x=0$, it is a constant function in that interval. A constant function is both continuous and derivable.