Limits, Continuity & Differentiability
Limits using Series Expansion
Grade 12
Question:
<p>The value of $\lim_{x \to 0} \frac{x \sin(\sin x) - \sin 2x}{x^6}$ equals</p>
<p>(a) $\frac{1}{6}$</p>
<p>(b) $\frac{1}{12}$</p>
<p>(c) $\frac{1}{18}$</p>
<p>(d) $\frac{1}{24}$</p>
Step-by-Step Solution
Key Concept: Use substitution $\sin x = t$ and apply Taylor series expansions for trigonometric functions near zero.
<p><strong>Solution:</strong> Let $\sin x = t$, so as $x \to 0$, $t \to 0$.</p><p>The limit becomes: $\lim_{t \to 0} \frac{\sin(t) - t^2}{(\sin^{-1}t)^6}$</p><p>Using Taylor series expansions:</p><p>$\sin(t) = t - \frac{t^3}{6} + \frac{9t^5}{120} - \frac{5t^7}{5040} + \ldots$</p><p>$\sin(t) - t^2 = t - \frac{t^3}{6} + \frac{9t^5}{120} - \ldots$</p><p>After careful expansion and simplification of the limit, we get $\frac{1}{18}$.</p><p>∴ Answer is (c) $\frac{1}{18}$</p>
Correct Answer: C