<p>The value of \(\lim_{x \to 0^+} x^m (\log x)^n\), \(m, n \in \mathbb{N}\) is</p>
Step-by-Step Solution
Key Concept: For x → 0⁺, logarithmic functions decay slower than any positive power of x grows, so x^m dominates (log x)^n and the product vanishes to 0 for any positive m.
<p><strong>Step 1:</strong> Recognize the form as x → 0⁺. We have x^m → 0 and (log x)^n → (-∞)^n, giving the indeterminate form 0 · ∞.</p><p><strong>Step 2:</strong> Rewrite as limit: lim[x→0⁺] (log x)^n / x^(-m). This is now ∞/∞ form, so apply L'Hôpital's Rule n times.</p><p><strong>Step 3:</strong> After first application: d/dx[(log x)^n] = n(log x)^(n-1) · (1/x) and d/dx[x^(-m)] = -mx^(-m-1).</p><p><strong>Step 4:</strong> This gives lim[x→0⁺] n(log x)^(n-1)/(−mx^(-m-1)) = lim[x→0⁺] −(n/m) · x^m · (log x)^(n-1). Repeating n times reduces the power of (log x) to 0 while keeping x^m factor.</p><p><strong>Step 5:</strong> Eventually we get a constant multiple of x^m as x → 0⁺, which equals 0.</p><p>∴ Answer: <strong>0</strong></p>
Correct Answer: A