Limits, Continuity & Differentiability
Evaluation of Limits
Grade 12
Question:
<p>Value of \(\lim_{x \to 0}\left[\dfrac{m\sin x}{x}\right]\) where \(m \in I\) and [.] is GIF, is</p>
<p>(a) \(m\) if \(m \leq 0\)</p>
<p>(b) \(m-1\) if \(m > 0\)</p>
<p>(c) \(m-1\) if \(m < 0\)</p>
<p>(d) \(m\) if \(m > 0\)</p>
Step-by-Step Solution
Key Concept: As x→0, sin(x)/x→1, so m·sin(x)/x→m. The GIF (Greatest Integer Function) of a number approaching m depends on whether m is an integer and how the expression approaches it from which side.
<p><strong>Step 1:</strong> Recognize that $\lim_{x \to 0} \frac{\sin x}{x} = 1$</p><p><strong>Step 2:</strong> Therefore, $\lim_{x \to 0} \frac{m\sin x}{x} = m \cdot 1 = m$ (where $m \in \mathbb{I}$)</p><p><strong>Step 3:</strong> For small positive x: $\sin x < x$, so $\frac{\sin x}{x} < 1$, which means $\frac{m\sin x}{x} < m$</p><p><strong>Step 4:</strong> For small negative x: $\frac{\sin x}{x} < 1$, so $\frac{m\sin x}{x} < m$ (approaching from below)</p><p><strong>Step 5:</strong> Since $\frac{m\sin x}{x}$ approaches m from below as $x \to 0$, we have $\left[\frac{m\sin x}{x}\right] = m-1$ for all sufficiently small $x \neq 0$</p><p>∴ Answer: <strong>m - 1</strong></p>
Correct Answer: B