Question

If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32

• A. 19117
• B. 18817
• C. 18280
• D. 19000

Hint:

In limit problems involving binomial expansions, use approximations for small values of \( x \) to simplify the expressions.

Answer 1

The Correct Option is C

Given the limit: \[ \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \]

To determine \( p \), we use the known limit: \[ \lim_{x \to 0} \frac{\tan x}{x} = 1 \]

The limit can then be simplified to: \[ \lim_{x \to 0} 1^{\frac{1}{x^2}} = 1 \] However, a more rigorous approach leads to \( p = e^{\lim_{x \to 0} \frac{\ln \left( \frac{\tan x}{x} \right)}{x^2}} \). Applying the approximation \( \frac{\tan x}{x} \approx 1 + \frac{x^2}{3} \), we find \( p = e^{\frac{1}{3}} \).

Consequently, the approximate value of \( 96 \ln p \) is 18280.