Limits, Continuity & Differentiability
Limits involving piecewise functions
Grade 12
<p><strong>27.</strong> Consider a function \(f: R \to R\) such that \(f(x) = \begin{cases} \sin(\pi x), & \text{if } x \in \mathbb{Q} \\ \tan(\pi\sqrt{|x|}), & \text{if } x \notin \mathbb{Q} \end{cases}\). If \(\displaystyle\lim_{x \to N} f(x)\) exists, then the sum of all positive integers \(N < 100\), is equal to:</p>
Step-by-Step Solution
Key Concept: For a limit to exist at a point, both rational and irrational subsequences must converge to the same value. At integer points, sin(πn) = 0 for all integers n, so we need tan(π√|x|) → 0 as x → N through irrationals, which only occurs when √|N| is an integer.
<p><strong>Step 1: Analyze the rational path</strong></p><p>For x ∈ ℚ: f(x) = sin(πx). As x → N (rational), lim sin(πx) = sin(πN) = 0 for any integer N.</p><p><strong>Step 2: Analyze the irrational path</strong></p><p>For x ∉ ℚ: f(x) = tan(π√|x|). As x → N through irrationals, we need lim tan(π√|x|) to exist and equal 0.</p><p><strong>Step 3: Determine when tan(π√|N|) = 0</strong></p><p>For tan(π√|N|) to be defined and equal 0, we need π√|N| = kπ where k ∈ ℤ. This gives √|N| = k, so |N| = k² for some non-negative integer k.</p><p><strong>Step 4: Find perfect square integers N < 100</strong></p><p>Positive perfect squares less than 100 are: 1, 4, 9, 16, 25, 36, 49, 64, 81.</p><p><strong>Step 5: Verify N = 0</strong></p><p>At N = 0: sin(0) = 0 and tan(0) = 0, so the limit exists. However, we need positive integers only.</p><p><strong>Step 6: Calculate the sum</strong></p><p>Sum = 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 = 285</p><p>∴ Answer: A</p>
Correct Answer: A