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

To solve the limit problem, start by noting the expression for p: $\lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}}$. Use the known limit $\lim_{x \to 0} \frac{\tan x}{x} = 1$.

Substitute this limit value: $\lim_{x \to 0} 1^{\frac{1}{x^2}}$.

Since the base is 1, the expression tends toward indeterminate form $1^\infty$, which requires logarithmic manipulation.

Let $y = \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}}$. Thus, \( \log y = \frac{1}{x^2} \cdot \log \left( \frac{\tan x}{x} \right) \).

Evaluate $\lim_{x \to 0} \frac{\log \left( \frac{\tan x}{x} \right)}{x^2}$. Use Taylor Series: $\tan x = x + \frac{x^3}{3} + \cdots$, so $\log \left( \frac{\tan x}{x} \right) = \log \left( 1 + \frac{x^2}{3} \right) \approx \frac{x^2}{3}$ for small $x$.

Then, $\lim_{x \to 0} \frac{\frac{x^2}{3}}{x^2} = \frac{1}{3}$.

Therefore, $\log y = \frac{1}{3}$, and $y = e^{\frac{1}{3}}$. Thus, $p = e^{\frac{1}{3}}$.

Find $96 \log_e p$: $$96 \log_e \left( e^{\frac{1}{3}} \right) = 96 \cdot \frac{1}{3} = 32.$$

The computed value is 32, which is within the specified range (32, 32).

Answer 2

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 \( 96 \log_e p \): \[ 96 \log_e p = 96 \times \frac{1}{3} = 32 \]