1. Key Concepts and Formulas

• Limits of the form $1^\infty$: When evaluating limits of the form $1^\infty$, it is a standard technique to take the natural logarithm of the expression. If $L = \lim_{x \to a} [f(x)]^{g(x)}$ is of the form $1^\infty$, then $\ln L = \lim_{x \to a} g(x) \ln f(x)$.
• Logarithm Properties: $\ln(a^b) = b \ln a$.
• Trigonometric Identities: $\sec^2 \theta = 1 + \tan^2 \theta$, so $\ln(1 + \tan^2 \sqrt{x}) = \ln(\sec^2 \sqrt{x})$.
• L'Hôpital's Rule: If a limit of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ is encountered, L'Hôpital's Rule can be applied: $\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$, provided the latter limit exists.
• Standard Limit: $\lim_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$.

2. Step-by-Step Solution

We are asked to find $\ln p$, where $p = \mathop {\lim }\limits_{x \to {0^ + }} {\left( {1 + {{\tan }^2}\sqrt x } \right)^{{1 \over {2x}}}}$

Step 1: Identify the indeterminate form. As $x \to 0^+$, $\sqrt{x} \to 0^+$. Then $\tan \sqrt{x} \to \tan 0 = 0$. So, $1 + \tan^2 \sqrt{x} \to 1 + 0^2 = 1$. The exponent is $\frac{1}{2x}$. As $x \to 0^+$, $2x \to 0^+$, so $\frac{1}{2x} \to +\infty$. Thus, the limit is of the indeterminate form $1^\infty$.

Step 2: Take the natural logarithm of the expression. Let $p = \mathop {\lim }\limits_{x \to {0^ + }} {\left( {1 + {{\tan }^2}\sqrt x } \right)^{{1 \over {2x}}}}$. We will evaluate $\ln p$: $\ln p = \ln \left( \mathop {\lim }\limits_{x \to {0^ + }} {\left( {1 + {{\tan }^2}\sqrt x } \right)^{{1 \over {2x}}}} \right)$ Since the logarithm function is continuous, we can move the limit outside: $\ln p = \mathop {\lim }\limits_{x \to {0^ + }} \ln \left( {\left( {1 + {{\tan }^2}\sqrt x } \right)^{{1 \over {2x}}}} \right)$ Using the logarithm property $\ln(a^b) = b \ln a$: $\ln p = \mathop {\lim }\limits_{x \to {0^ + }} \frac{1}{2x} \ln \left( {1 + {{\tan }^2}\sqrt x } \right)$

Step 3: Simplify the expression inside the limit using trigonometric identities. We know the identity $1 + \tan^2 \theta = \sec^2 \theta$. Applying this with $\theta = \sqrt{x}$: $1 + \tan^2 \sqrt{x} = \sec^2 \sqrt{x}$. Substituting this into the expression for $\ln p$: $\ln p = \mathop {\lim }\limits_{x \to {0^ + }} \frac{1}{2x} \ln \left( {\sec^2 \sqrt{x} } \right)$ Using the logarithm property $\ln(a^b) = b \ln a$ again: $\ln p = \mathop {\lim }\limits_{x \to {0^ + }} \frac{1}{2x} \cdot 2 \ln \left( {\sec \sqrt{x} } \right)$ $\ln p = \mathop {\lim }\limits_{x \to {0^ + }} \frac{1}{x} \ln \left( {\sec \sqrt{x} } \right)$

Step 4: Check for indeterminate form for L'Hôpital's Rule. As $x \to 0^+$, $\sqrt{x} \to 0^+$, and $\sec \sqrt{x} \to \sec 0 = 1$. So, $\ln(\sec \sqrt{x}) \to \ln(1) = 0$. The denominator is $x$, which also approaches $0$. Thus, the limit is now of the indeterminate form $\frac{0}{0}$, and we can apply L'Hôpital's Rule.

Step 5: Apply L'Hôpital's Rule. Let $f(x) = \ln(\sec \sqrt{x})$ and $g(x) = x$. We need to find the derivatives $f'(x)$ and $g'(x)$. $g'(x) = \frac{d}{dx}(x) = 1$. To find $f'(x)$, we use the chain rule: $f'(x) = \frac{d}{dx} [\ln(\sec \sqrt{x})]$ Let $u = \sec \sqrt{x}$. Then $\frac{df}{dx} = \frac{d}{du}(\ln u) \cdot \frac{du}{dx} = \frac{1}{u} \cdot \frac{du}{dx}$. Now, we find $\frac{du}{dx}$: $\frac{du}{dx} = \frac{d}{dx} (\sec \sqrt{x})$ Let $v = \sqrt{x}$. Then $\frac{du}{dx} = \frac{d}{dv}(\sec v) \cdot \frac{dv}{dx} = (\sec v \tan v) \cdot \frac{dv}{dx}$. $\frac{dv}{dx} = \frac{d}{dx} (x^{1/2}) = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}}$. So, $\frac{du}{dx} = \sec \sqrt{x} \tan \sqrt{x} \cdot \frac{1}{2\sqrt{x}}$. Substituting back into $f'(x)$: $f'(x) = \frac{1}{\sec \sqrt{x}} \cdot \left( \sec \sqrt{x} \tan \sqrt{x} \cdot \frac{1}{2\sqrt{x}} \right)$ $f'(x) = \frac{\tan \sqrt{x}}{2\sqrt{x}}$.

Applying L'Hôpital's Rule: $\ln p = \mathop {\lim }\limits_{x \to {0^ + }} \frac{f'(x)}{g'(x)} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{\frac{\tan \sqrt{x}}{2\sqrt{x}}}{1}$ $\ln p = \mathop {\lim }\limits_{x \to {0^ + }} \frac{\tan \sqrt{x}}{2\sqrt{x}}$

Step 6: Evaluate the resulting limit. We can rewrite the limit as: $\ln p = \frac{1}{2} \mathop {\lim }\limits_{x \to {0^ + }} \frac{\tan \sqrt{x}}{\sqrt{x}}$ Let $y = \sqrt{x}$. As $x \to 0^+$, $y \to 0^+$. The limit becomes: $\ln p = \frac{1}{2} \mathop {\lim }\limits_{y \to 0^ + } \frac{\tan y}{y}$ We know the standard limit $\mathop {\lim }\limits_{y \to 0} \frac{\tan y}{y} = 1$. Therefore, $\ln p = \frac{1}{2} \cdot 1 = \frac{1}{2}$

3. Common Mistakes & Tips

• Incorrectly applying L'Hôpital's Rule: Ensure the limit is indeed in an indeterminate form ($\frac{0}{0}$ or $\frac{\infty}{\infty}$) before applying the rule. Also, be careful with the differentiation of composite functions, especially involving square roots and trigonometric functions.
• Algebraic Errors: Mistakes in simplifying trigonometric expressions or applying logarithm properties can lead to incorrect results. Double-check these steps.
• Ignoring the $1^\infty$ form: If a limit is of the form $1^\infty$, directly substituting values will not work. Always convert it to the form $e^{\lim g(x)\ln f(x)}$ or take the logarithm as shown here.

4. Summary

The problem involves evaluating a limit of the indeterminate form $1^\infty$. We began by taking the natural logarithm of the given expression for $p$, transforming the limit into a form where L'Hôpital's Rule could be applied. After simplifying the expression using trigonometric identities and logarithm properties, we applied L'Hôpital's Rule to differentiate the numerator and denominator. The resulting limit simplified to a standard trigonometric limit, allowing us to find the value of $\ln p$.

The value of $\ln p$ was found to be $\frac{1}{2}$.

The final answer is \boxed{{1 \over 2}}.