Limits, Continuity & Differentiability
Continuity using limits
Grade 12
Question:
<p>If <span class="math">\(f(x) = \begin{cases} \frac{\sin((p+1)x) + \sin x}{x}, & x < 0 \\ q, & x = 0 \\ \frac{x + x^2 - x^{3/2}}{x}, & x > 0 \end{cases}\)</span> is continuous at <span class="math">\(x = 0\)</span>, then the ordered pair <span class="math">\((p, q)\)</span> is equal to</p>
<p>(a) <span class="math">\(\left(-\frac{3}{2}, -\frac{1}{2}\right)\)</span></p>
<p>(b) <span class="math">\(\left(-\frac{1}{2}, \frac{3}{2}\right)\)</span></p>
<p>(c) <span class="math">\(\left(\frac{5}{2}, \frac{1}{2}\right)\)</span></p>
<p>(d) <span class="math">\(\left(-\frac{3}{2}, \frac{1}{2}\right)\)</span></p>
Step-by-Step Solution
Key Concept: For continuity at x=0, the left limit, right limit, and function value must all be equal. Use standard limit formulas for sin(u)/u.
<p><strong>Solution:</strong> For <span class="math">$f(x)$</span> to be continuous at <span class="math">$x = 0$</span>, we require <span class="math">$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = q$</span>.</p><p><strong>Left-hand limit:</strong> <span class="math">$\lim_{x \to 0^-} \frac{\sin((p+1)x) + \sin x}{x}$</span></p><p>Using <span class="math">$\lim_{u \to 0} \frac{\sin u}{u} = 1$</span>:</p><p><span class="math">$= \lim_{x \to 0^-} \left[\frac{\sin((p+1)x)}{x} + \frac{\sin x}{x}\right] = (p+1) + 1 = p + 2$</span></p><p><strong>Right-hand limit:</strong> <span class="math">$\lim_{x \to 0^+} \frac{x + x^2 - x^{3/2}}{x} = \lim_{x \to 0^+} \left(1 + x - x^{1/2}\right) = 1$</span></p><p>For continuity: <span class="math">$p + 2 = 1$</span>, so <span class="math">$p = -\frac{3}{2}$</span></p><p>And <span class="math">$q = 1 - 0 - 0 = \frac{1}{2}$</span></p><p>Therefore, <span class="math">$(p, q) = \left(-\frac{3}{2}, \frac{1}{2}\right)$</span></p>
Correct Answer: d