Limits, Continuity & Differentiability
Left and Right Hand Limits — Fractional Part and Inverse Trig
DAILY_CHALLENGE
Grade 12
Question:
Let $\{x\}$ denote the fractional part of $x$ and $f(x)=\dfrac{\cos^{-1}(1-\{x\}^2)\sin^{-1}(1-\{x\})}{\{x\}-\{x\}^3}$, $x\neq0$. If $L$ and $R$ respectively denote the left hand limit and the right hand limit of $f(x)$ at $x=0$, then $\dfrac{32}{\pi^2}(L^2+R^2)$ is equal to
Step-by-Step Solution
Key Concept: For RHL ($x\to0^+$): $\{x\}\to h$. Simplify $f(h)=\frac{\cos^{-1}(1-h^2)\sin^{-1}(1-h)}{h(1-h^2)}$. As $h\to0$, $\sin^{-1}(1-h)\to\pi/2$ and for $\cos^{-1}(1-h^2)$, let $\cos^{-1}(1-h^2)=\theta\Rightarrow\cos\theta=1-h^2$: limit gives $R=\pi/\sqrt{2}$. For LHL ($x\to0^-$): $\{x\}\to1-h$ (since $\{-h\}=1-h$ for small $h>0$). After simplification, $L=\pi/4$.
$R=\pi/\sqrt{2}$, $L=\pi/4$. $\frac{32}{\pi^2}\left(\frac{\pi^2}{16}+\frac{\pi^2}{2}\right)=\frac{32}{\pi^2}\cdot\frac{9\pi^2}{16}=18$.
Correct Answer: 18