<p>Let function <i>f</i> be defined as <i>f</i>: ℝ<sup>+</sup> → ℝ<sup>+</sup> and function <i>g</i> is defined as <i>g</i>: ℝ<sup>+</sup> → ℝ<sup>+</sup>. Functions <i>f</i> and <i>g</i> are continuous in their domain. Suppose, the function <i>h(x)</i> = <i>lim</i><sub><i>n</i>→∞</sub> \(\frac{f(x) + x^n}{x^n + g(x)}\), <i>x</i> > 0 is continuous in its domain, then <i>f</i>(1) × <i>g</i>(1) is equal to</p>
Step-by-Step Solution
Key Concept: Analyze the limit function behavior at different regions and use continuity condition at x = 1 to establish relationships between f(1) and g(1).
<p><strong>Step 1:</strong> Analyze the limit function <i>h(x)</i> = <i>lim</i><sub><i>n</i>→∞</sub> $\frac{f(x) + x^n}{x^n + g(x)}$ for x > 0.</p><p><strong>Step 2:</strong> For x > 1: as n→∞, x<sup>n</sup>→∞, so $h(x) = \frac{x^n}{x^n} = 1$</p><p><strong>Step 3:</strong> For x = 1: $h(1) = \lim_{n\to\infty} \frac{f(1) + 1}{1 + g(1)}$</p><p><strong>Step 4:</strong> For 0 < x < 1: as n→∞, x<sup>n</sup>→0, so $h(x) = \frac{f(x)}{g(x)}$</p><p><strong>Step 5:</strong> For h(x) to be continuous at x = 1, we need the left limit, right limit, and h(1) to be equal.</p><p><strong>Step 6:</strong> From right continuity: $\lim_{x\to 1^+} h(x) = 1$</p><p><strong>Step 7:</strong> From left continuity: $\lim_{x\to 1^-} h(x) = \frac{f(1)}{g(1)}$</p><p><strong>Step 8:</strong> For continuity: $\frac{f(1)}{g(1)} = 1 = \frac{f(1) + 1}{1 + g(1)}$</p><p><strong>Step 9:</strong> From $\frac{f(1)}{g(1)} = 1$, we get f(1) = g(1).</p><p><strong>Step 10:</strong> Substituting into the second equation: $1 = \frac{f(1) + 1}{1 + f(1)}$, which is always true.</p><p><strong>Step 11:</strong> From $\frac{f(1) + 1}{1 + g(1)} = 1$ and f(1) = g(1), we get f(1) = g(1) = 1.</p><p><strong>Step 12:</strong> Therefore, f(1) × g(1) = 1 × 1 = 1.</p><p>∴ Answer is (b) 1.</p>
Correct Answer: b