Limits, Continuity & Differentiability
L'Hopital's Rule / Limits
Grade 12
Question:
<p>If \(\lim_{t \to x} \dfrac{\displaystyle\int_x^t \sin^{-1}(nz)\,dz}{t^2 - x^2} = f_n(x)\), then find the value of \(\lim_{x \to 0}\left([f_2(x)] + [f_4(x)]\right)\).</p><p>[Note: [k] denotes greatest integer function less than or equal to k]</p>
Step-by-Step Solution
Key Concept: Apply L'Hôpital's rule to the limit of integral form (0/0 indeterminate), then recognize that the resulting function f_n(x) = sin⁻¹(nx)/(2x) requires careful analysis as x→0 using the Taylor series sin⁻¹(u) ≈ u + u³/6 + ... to find the limiting behavior.
<p><strong>Step 1: Identify the indeterminate form</strong></p><p>As t→x, both numerator ∫ₓᵗ sin⁻¹(nz)dz → 0 and denominator t²-x² → 0, giving 0/0 form.</p><p><strong>Step 2: Apply L'Hôpital's Rule</strong></p><p>Differentiate numerator w.r.t. t and denominator w.r.t. t:</p><p>$$f_n(x) = \lim_{t \to x} \frac{\sin^{-1}(nt)}{2t}$$</p><p><strong>Step 3: Evaluate limit as x→0</strong></p><p>Using Taylor expansion: sin⁻¹(u) = u + u³/6 + O(u⁵)</p><p>$$f_n(x) = \lim_{x \to 0} \frac{\sin^{-1}(nx)}{2x} = \lim_{x \to 0} \frac{nx + (nx)^3/6 + ...}{2x} = \lim_{x \to 0} \left(\frac{n}{2} + \frac{n^3x^2}{12} + ...\right) = \frac{n}{2}$$</p><p><strong>Step 4: Calculate floor values</strong></p><p>$$\lim_{x \to 0} f_2(x) = \frac{2}{2} = 1 \implies [f_2(x)] = [1] = 1$$</p><p>$$\lim_{x \to 0} f_4(x) = \frac{4}{2} = 2 \implies [f_4(x)] = [2] = 2$$</p><p><strong>Step 5: Final answer</strong></p><p>$$[f_2(x)] + [f_4(x)] = 1 + 2 = 3$$</p><p>∴ Answer: <strong>3</strong></p>
Correct Answer: 3