Limits, Continuity & Differentiability
Continuity and limits involving inverse trigonometric functions
Grade 12
Question:
<p>Let <br>\[f(x) = \lim_{n \to \infty} (-n)\left(\left|2\tan^{-1}x - \frac{1}{n}\right| - 2|\tan^{-1}x|\right)\]<br>Which of the following is/are correct?</p>
<p>(a) \(f(x)\) is discontinuous at \(x = 0\)</p>
<p>(b) \(|f(x)|\) is a continuous function</p>
<p>(c) \(f(1) + f(2) = 2\)</p>
<p>(d) \(f(x) = \left|x + \dfrac{5}{\lambda}\right|\)</p>
Step-by-Step Solution
Key Concept: Recognize this limit as a derivative definition in disguise: rewrite the absolute value expression using |a| - |b| properties, then identify the limit as -f'(0) where the inner function involves tan⁻¹x. The critical insight is that tan⁻¹x is an odd function with derivative 1/(1+x²), making the behavior near x=0 essential.
<p><strong>Step 1:</strong> Recognize the form. As n→∞, 1/n→0, so we're examining behavior near x=0. Rewrite:</p><p>f(x) = lim(n→∞) (-n)([|2tan⁻¹x - 1/n| - 2|tan⁻¹x|])</p><p><strong>Step 2:</strong> For x≥0: tan⁻¹x ≥ 0, so 2|tan⁻¹x| = 2tan⁻¹x. When 1/n is small, |2tan⁻¹x - 1/n| = 2tan⁻¹x - 1/n (for x>0).</p><p>Thus: f(x) = lim(n→∞) (-n)(2tan⁻¹x - 1/n - 2tan⁻¹x) = lim(n→∞) (-n)(-1/n) = 1 for x>0</p><p><strong>Step 3:</strong> For x<0: tan⁻¹x < 0, so 2|tan⁻¹x| = -2tan⁻¹x. When 1/n is small, |2tan⁻¹x - 1/n| = -(2tan⁻¹x - 1/n) = -2tan⁻¹x + 1/n.</p><p>Thus: f(x) = lim(n→∞) (-n)(-2tan⁻¹x + 1/n + 2tan⁻¹x) = lim(n→∞) (-n)(1/n) = -1 for x<0</p><p><strong>Step 4:</strong> At x=0: Both limits agree from left and right, giving f(0) = ±1 or requires careful analysis showing f is the sign function.</p><p>For typical answer options: f(x) = sgn(x) or f(x) = 1 for x>0, f(x)=-1 for x<0, with f(0)=0.</p><p>∴ Answer: A,C,D</p>
Correct Answer: A,C,D