Question:

<p><strong>Step 1:</strong> Take logarithm of p to convert the exponential limit into a product form.</p><p>Let $p = \lim_{x \to 0^+} (1 + \tan^2{\sqrt{x}})^{\frac{1}{x}}$</p><p>Then $\ln p = \lim_{x \to 0^+} \frac{1}{x} \ln(1 + \tan^2{\sqrt{x}})$</p><p><strong>Step 2:</strong> Find the Taylor expansion of $\tan^2{\sqrt{x}}$ around $x = 0$.</p><p>We know: $\tan u = u + \frac{u^3}{3} + O(u^5)$</p><p>For $u = \sqrt{x}$: $\tan{\sqrt{x}} = \sqrt{x} + \frac{x^{3/2}}{3} + O(x^{5/2})$</p><p>Therefore: $\tan^2{\sqrt{x}} = \left(\sqrt{x} + \frac{x^{3/2}}{3} + ...\right)^2 = x + \frac{2x^2}{3} + O(x^3)$</p><p><strong>Step 3:</strong> Apply the logarithm expansion $\ln(1 + t) = t - \frac{t^2}{2} + O(t^3)$ where $t = \tan^2{\sqrt{x}}$.</p><p>$\ln(1 + \tan^2{\sqrt{x}}) = \tan^2{\sqrt{x}} - \frac{(\tan^2{\sqrt{x}})^2}{2} + ...$</p><p>$= x + \frac{2x^2}{3} - \frac{x^2}{2} + O(x^3)$</p><p>$= x + \left(\frac{2}{3} - \frac{1}{2}\right)x^2 + O(x^3)$</p><p>$= x + \frac{x^2}{6} + O(x^3)$</p><p><strong>Step 4:</strong> Evaluate the limit.</p><p>$\ln p = \lim_{x \to 0^+} \frac{1}{x}\left(x + \frac{x^2}{6} + O(x^3)\right)$</p><p>$= \lim_{x \to 0^+} \left(1 + \frac{x}{6} + O(x^2)\right)$</p><p>$= 1$</p><p><strong>Step 5:</strong> Therefore $\log p = \ln p = 1$</p><p>∴ The answer is <strong>1</strong>, which corresponds to option <strong>C</strong>.</p><p>Note: The question asks for $\log p$ which in this context means the natural logarithm (ln). If base-10 logarithm were intended, the answer would be different, but given the options and the calculation, natural logarithm is the intended interpretation.</p>