Limits, Continuity & Differentiability
Limits with Fractional Part
Grade 12
Question:
<p>The value of $\lim_{x \to 0} \sin^{-1}\{x\}$ (where $\{\cdot\}$ denotes fractional part of $x$) is</p>
<p>(a) $0$</p>
<p>(b) $\frac{\pi}{2}$</p>
<p>(c) Doesn't exist</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: The fractional part function has different behavior on the left and right of zero, causing discontinuity.
<p>The fractional part function $\{x\}$ is defined as $\{x\} = x - [x]$, where $0 \le \{x\} < 1$.</p><p>As $x \to 0^+$: $\{x\} = x$, so $\sin^{-1}\{x\} \to \sin^{-1}(0) = 0$</p><p>As $x \to 0^-$: $\{x\} = x - (-1) = x + 1 \to 1$, so $\sin^{-1}\{x\} \to \sin^{-1}(1) = \frac{\pi}{2}$</p><p>Since LHL $\neq$ RHL, the limit doesn't exist.</p>
Correct Answer: C