Key Concepts and Formulas

• Indeterminate Forms: Recognizing and handling indeterminate forms such as $1^\infty$.
• Logarithmic Transformation for Limits: The property $a^b = e^{b \log a}$ is crucial for evaluating limits of the form $f(x)^{g(x)}$ when it results in $1^\infty$.
• Taylor Series Expansion: The Taylor series expansion of $\tan x$ around $x=0$ is $\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + \dots$. This is a powerful tool for simplifying expressions in limits.
• Standard Limit: The fundamental limit $\lim_{x \to 0} \frac{\tan x}{x} = 1$.

Step-by-Step Solution

Step 1: Identify the Indeterminate Form We are asked to evaluate the limit $p = \lim\limits_{x \to 0} \left(\frac{\tan x}{x}\right)^{\frac{1}{x^2}}$. As $x \to 0$, we know that $\frac{\tan x}{x} \to 1$ (a standard limit). Also, as $x \to 0$, $\frac{1}{x^2} \to \infty$. Therefore, the limit is of the indeterminate form $1^\infty$.

Step 2: Apply Logarithmic Transformation To evaluate a limit of the form $1^\infty$, we use the property $a^b = e^{b \log a}$. Let $y = \left(\frac{\tan x}{x}\right)^{\frac{1}{x^2}}$. Then, $\log y = \frac{1}{x^2} \log \left(\frac{\tan x}{x}\right)$. So, $p = \lim_{x \to 0} y = e^{\lim_{x \to 0} \frac{\log \left(\frac{\tan x}{x}\right)}{x^2}}$. We now need to evaluate the exponent limit: $L = \lim_{x \to 0} \frac{\log \left(\frac{\tan x}{x}\right)}{x^2}$.

Step 3: Use Taylor Series Expansion for $\tan x$ The Taylor series expansion of $\tan x$ around $x=0$ is given by: $\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + O(x^7)$ Now, let's find the expansion for $\frac{\tan x}{x}$: $\frac{\tan x}{x} = \frac{x + \frac{x^3}{3} + \frac{2x^5}{15} + O(x^7)}{x} = 1 + \frac{x^2}{3} + \frac{2x^4}{15} + O(x^6)$

Step 4: Simplify the Logarithm Term Next, we need to evaluate $\log \left(\frac{\tan x}{x}\right)$. Using the expansion from Step 3: $\frac{\tan x}{x} = 1 + \left(\frac{x^2}{3} + \frac{2x^4}{15} + O(x^6)\right)$ We use the Taylor expansion of $\log(1+u)$ around $u=0$, which is $\log(1+u) = u - \frac{u^2}{2} + O(u^3)$. Let $u = \frac{x^2}{3} + \frac{2x^4}{15} + O(x^6)$. As $x \to 0$, $u \to 0$. $\log \left(\frac{\tan x}{x}\right) = \log \left(1 + \left(\frac{x^2}{3} + \frac{2x^4}{15} + O(x^6)\right)\right)$ $\log \left(\frac{\tan x}{x}\right) = \left(\frac{x^2}{3} + \frac{2x^4}{15} + O(x^6)\right) - \frac{1}{2}\left(\frac{x^2}{3} + \frac{2x^4}{15} + O(x^6)\right)^2 + O(x^6)$ The dominant term as $x \to 0$ is from the first part of the expansion of $\log(1+u)$: $\log \left(\frac{\tan x}{x}\right) = \frac{x^2}{3} + \frac{2x^4}{15} + O(x^6) - \frac{1}{2}\left(\frac{x^4}{9} + O(x^6)\right) + O(x^6)$ $\log \left(\frac{\tan x}{x}\right) = \frac{x^2}{3} + \frac{2x^4}{15} - \frac{x^4}{18} + O(x^6)$ Combining the $x^4$ terms: $\frac{2}{15} - \frac{1}{18} = \frac{12 - 5}{90} = \frac{7}{90}$. So, $\log \left(\frac{\tan x}{x}\right) = \frac{x^2}{3} + \frac{7x^4}{90} + O(x^6)$.

Step 5: Evaluate the Exponent Limit $L$ Now we substitute this back into the limit for the exponent: $L = \lim_{x \to 0} \frac{\log \left(\frac{\tan x}{x}\right)}{x^2} = \lim_{x \to 0} \frac{\frac{x^2}{3} + \frac{7x^4}{90} + O(x^6)}{x^2}$ Divide each term by $x^2$: $L = \lim_{x \to 0} \left(\frac{1}{3} + \frac{7x^2}{90} + O(x^4)\right)$ As $x \to 0$, the terms with $x^2$ and higher powers go to zero. $L = \frac{1}{3}$

Step 6: Determine the Value of $p$ We found that $p = e^L$. So, $p = e^{\frac{1}{3}}$.

Step 7: Calculate $96 \log_e p$ We are asked to find $96 \log_e p$. From $p = e^{\frac{1}{3}}$, we can take the natural logarithm of both sides: $\log_e p = \log_e \left(e^{\frac{1}{3}}\right)$ Using the property $\log_e (e^a) = a$: $\log_e p = \frac{1}{3}$ Now, multiply by 96: $96 \log_e p = 96 \times \frac{1}{3}$ $96 \log_e p = \frac{96}{3}$ $96 \log_e p = 32$

Common Mistakes & Tips

• Incorrect Taylor Expansion: Ensure you use the correct Taylor series for $\tan x$ and $\log(1+u)$. Using only the first term of $\tan x$ expansion (i.e., $\tan x \approx x$) will lead to $\lim_{x \to 0} \frac{\log(1)}{x^2} = 0$, which is incorrect.
• Algebraic Errors: Be meticulous with algebraic manipulations, especially when combining terms or simplifying fractions.
• L'Hôpital's Rule: While L'Hôpital's rule can be used, it often involves more complex derivatives for this type of problem compared to using Taylor series. If using L'Hôpital's rule, be careful with the derivatives of logarithmic and trigonometric functions. For the exponent limit $L = \lim_{x \to 0} \frac{\log(\tan x / x)}{x^2}$, applying L'Hôpital's rule once leads to $\lim_{x \to 0} \frac{\frac{x \sec^2 x - \tan x}{x \tan x}}{2x} = \lim_{x \to 0} \frac{x \sec^2 x - \tan x}{2x^2 \tan x}$, which still requires further simplification or L'Hôpital's rule.

Summary

The problem involves evaluating a limit of the indeterminate form $1^\infty$. The standard approach is to use the logarithmic transformation $p = e^{\lim_{x \to 0} g(x) \log f(x)}$. We then employed the Taylor series expansion of $\tan x$ around $x=0$ to simplify the expression inside the logarithm. By carefully expanding and collecting terms, we evaluated the limit of the exponent to be $\frac{1}{3}$. Consequently, $p = e^{\frac{1}{3}}$, which implies $\log_e p = \frac{1}{3}$. Finally, multiplying by 96 gives the desired result of 32.

The final answer is \boxed{32}.