Question:

<p><strong>Step 1:</strong> Find the slope of the tangent at point $P(a, a^n)$.</p><p>For $y = x^n$, $\frac{dy}{dx} = nx^{n-1}$</p><p>At $P(a, a^n)$: slope of tangent = $na^{n-1}$</p><p><strong>Step 2:</strong> Find the slope of the normal.</p><p>Slope of normal = $-\frac{1}{na^{n-1}}$</p><p><strong>Step 3:</strong> Write the equation of the normal line.</p><p>The normal passes through $P(a, a^n)$ with slope $-\frac{1}{na^{n-1}}$:</p><p>$y - a^n = -\frac{1}{na^{n-1}}(x - a)$</p><p><strong>Step 4:</strong> Find where the normal intersects the Y-axis (set $x = 0$).</p><p>$y - a^n = -\frac{1}{na^{n-1}}(0 - a)$</p><p>$y = a^n + \frac{a}{na^{n-1}} = a^n + \frac{1}{n}a^{2-n}$</p><p>So $b = a^n + \frac{1}{n}a^{2-n}$</p><p><strong>Step 5:</strong> Apply the limit condition.</p><p>$\lim_{a \to 0} b = \lim_{a \to 0} \left(a^n + \frac{1}{n}a^{2-n}\right) = \frac{1}{2}$</p><p>For the limit to be finite and equal to $\frac{1}{2}$, we need the $a^{2-n}$ term to approach a constant.</p><p>This requires $2 - n = 0$, so $n = 2$.</p><p>Then $\lim_{a \to 0} b = 0 + \frac{1}{2}(1) = \frac{1}{2}$ ✓</p><p>∴ $n = 2$</p>