Question

If $ \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p $, then $ 96 \log_e p $ is equal to _______

Hint:

For limits of the form \( 1^\infty \), use the transformation \( \lim [f(x)]^{g(x)} = e^{\lim g(x) [f(x) - 1]} \). When evaluating the resulting limit, Taylor series expansions of trigonometric functions around \( x = 0 \) are often useful. Remember the expansions for \( \sin x \), \( \cos x \), and \( \tan x \).

Answer 1

Correct Answer: 32

Given \( p = \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} \). This limit is of the form \( 1^\infty \), so we can use the formula \( \lim_{x \to a} [f(x)]^{g(x)} = e^{\lim_{x \to a} g(x) [f(x) - 1]} \). 
Here, \( f(x) = \frac{\tan x}{x} \) and \( g(x) = \frac{1}{x^2} \). \[ \log_e p = \lim_{x \to 0} \frac{1}{x^2} \left( \frac{\tan x}{x} - 1 \right) = \lim_{x \to 0} \frac{\tan x - x}{x^3} \] Using the Taylor series expansion for \( \tan x \) around \( x = 0 \): \[ \tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + \dots \] \[ \log_e p = \lim_{x \to 0} \frac{(x + \frac{x^3}{3} + \frac{2x^5}{15} + \dots) - x}{x^3} = \lim_{x \to 0} \frac{\frac{x^3}{3} + \frac{2x^5}{15} + \dots}{x^3} \] \[ \log_e p = \lim_{x \to 0} \left( \frac{1}{3} + \frac{2x^2}{15} + \dots \right) = \frac{1}{3} \] So, \( \log_e p = \frac{1}{3} \). We need to find \( 96 \log_e p \): \[ 96 \log_e p = 96 \times \frac{1}{3} = 32 \]