Limits, Continuity & Differentiability
Limit of Piecewise Functions
Grade 12
Question:
<p>Let <span class="latex">[ ]x</span> denote the greatest integer less than or equal to <span class="latex">x</span>. Then, <span class="latex">\lim_{x \to 0} \frac{\tan(\pi \sin^2 x) + (|x| - \sin(x[x]))^2}{x^2}</span></p>
<p>(a) equals <span class="latex">\pi</span></p>
<p>(b) equals <span class="latex">\pi + 1</span></p>
<p>(c) equals <span class="latex">0</span></p>
<p>(d) does not exist</p>
Step-by-Step Solution
Key Concept: Evaluate the right-hand and left-hand limits separately using the greatest integer function and properties of limits.
<p><strong>At x = 0, RHL:</strong></p><p><span class="latex">\lim_{x \to 0^+} \frac{\tan(\pi \sin^2 x) + (|x| - \sin(x[x]))^2}{x^2}</span></p><p><strong>Since</strong> <span class="latex">|x| = x</span> for <span class="latex">x > 0</span> and <span class="latex">[x] = 0</span> for <span class="latex">0 < x < 1</span>:</p><p><span class="latex">= \lim_{x \to 0^+} \frac{\tan(\pi \sin^2 x) + x^2}{x^2}</span></p><p><span class="latex">= \lim_{x \to 0^+} \left(\frac{\tan(\pi \sin^2 x)}{\pi \sin^2 x} \cdot \frac{\pi \sin^2 x}{x^2} + 1\right)</span></p><p>The limit does not exist because RHL and LHL give different results.</p><p>∴ Answer is (d).</p>
Correct Answer: D